PHP - Insert Data From A Form Into Table
HI first timer here so please forgive me if this is in the wrong section.
Anyway, here goes. I have 2 tables Table 1 Levels LevelID LevelName Table 2 Members MemberID FirstName LastName Birthdate LevelID I have a form that has the LevelID populated from table 1 with a query. When I enter all my data and hit submit the data gets written to the members table. However, the LevelID is written as 0 for each and everyone record that I have entered. So the question is what am I missing? As mentioned I am very new at this so any help would be greatly appreciated. Please see below for my code. Thanks in advance <?php require_once 'dblogin1.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); $sql = "SELECT\n" . "levels.LevelID,\n" . "levels.LevelName\n" . "FROM\n" . "levels\n"; $result = mysql_query($sql) or die(mysql_error()); $options=""; while($row = mysql_fetch_array($result)){ $LevelId=$row["LevelID"]; $LevelName=$row["LevelName"]; $options.="<OPTION VALUE=\"$LevelID\">".$LevelName.'</option>'; } ?> <style type="text/css"> <!-- body p { color: #F00; } --> </style> <table width="500" border="1" align="center" cellpadding="0" cellspacing="1" > <tr> <td> <form name="form1" method="post" action="insert_ac.php"> <table width="100%" border="1" cellspacing="1" cellpadding="3"> <tr> <td colspan="3"><div align="center"> <p><strong>Myers Player Registration Form</strong></p> </div></td> </tr> <tr> <td>First Name</td> <td><div align="center">:</div></td> <td><input name="firstname" type="text" id="firstname"></td> </tr> <tr> <td>Last Name</td> <td><div align="center">:</div></td> <td><input name="lastname" type="text" id="lastname"></td> </tr> <tr> <td>Birthdate (yyyy-mm-dd)</td> <td><div align="center">:</div></td> <td><input name="birthdate" type="text" id="birthdate"> </td> </tr> <tr> <td>Level</td> <td><div align="center">:</div></td> <td><select name="LevelID"> <OPTION VALUE="0">Choose <?php echo $options?> </select></td> </tr> </form> </td> </tr> </table> Similar TutorialsI'm creating a report page and can't figure out how to retrieve multiple rows from a table and display them in html. There are 5 data elements in each row and I'm thinking that using a form in the html might be the best way as I'm totally ignorant about tables in html. I'm a newbie to php and can't figure out how to accomplish this. Here's the code that I have so far: <?php $filename = NULL; session_start(); // start of script every time. // setup a path for all of your canned php scripts $php_scripts = '../php/'; // a folder above the web accessible tree // load the pdo connection module require $php_scripts . 'PDO_Connection_Select.php'; //******************************* // Begin the script here // Connect to the database if (!$con = PDOConnect("foxclone")): { echo "Failed to connect to database" ; exit; } else: { $sql = 'SELECT COUNT(IP_ADDRESS) FROM download WHERE FILENAME IS NOT NULL'; $sql1 = 'Update download t2, ip_lookup t1 set t2.country = t1.country, t2.area = t1.area, t2.city = t1.city where ((t2.IP_ADDRESS) = (t1.start_ip) OR (t2.IP_ADDRESS) > (t1.start_ip)) AND ((t2.IP_ADDRESS) = (t1.end_ip) OR (t2.IP_ADDRESS) < (t1.end_ip)) AND (t2.FILENAME is not null and t2.country is null)'; $sql2 = 'SELECT (IP_ADDRESS, FILENAME, country, area, city) from download where FILENAME is not null'; // Update the table $stmt = $con->prepare($sql1); $stmt->execute(); // Get count of rows to be displayed in table $stmt = $con->prepare($sql); $stmt->execute() ; $cnt = $stmt->fetch(PDO::FETCH_NUM); // retrieve one row at a time $i = 1; while($i <= $cnt){ $stmt = $con->prepare($sql2); $row->execute(array('')); // Do I need an array here? // from here on, I'm lost $i++; I'd appreciate any guidance you can provide or understandable tutorials you can point me to. Larry Code: [Select] $myusername = $_POST['myusername']; mysql_query("INSERT INTO logedin ( username ) VALUES ('$myusername')") OR die("Could not send the message: <br>".mysql_error()); wont insert anything no error's given even just trying to insert username into the the loged in table yes i know i spelt loged in wrong but i did that on the db lol Hi guys, Im making an event booking/reservation site, I have 5 tables (Malls, Events, Booking, Reservation and Users) which are all connected and doing fine. I created a function which is this... function select_all_events_by_mall_id($mall_id) { echo "<table id='event_table'>"; echo "<th>Event Name</th>"; echo "<th>Event Location</th>"; echo "<th>Number of Cars</th>"; echo "<th>Description</th>"; echo "<th>Booked</th>"; echo "<th>Reserved</th>"; echo "<th>Reserved Until</th>"; echo "<th>Vacant</th>"; $events_set = select_all_events($mall_id); while($events = mysql_fetch_array($events_set)) { echo "<tr>"; echo "<td>" . $events['event_name'] . "</td>"; echo "<td>" . $events['event_location'] . "</td>"; echo "<td>" . $events['number_of_cars'] . "</td>"; echo "<td>" . $events['description'] . "</td>"; // Booked echo "<td>"; $booked_set = select_all_booking($events['id']); while($booked = mysql_fetch_array($booked_set)) { $user_set = select_all_user_booked($booked['users_id']); while($user = mysql_fetch_array($user_set)) { echo $user['company'] . ", " . $user['branch'] . "<br />"; } } echo "</td>"; // Reserved echo "<td>"; echo "<ol>"; $reserved_set = select_all_reservation($events['id']); while($reserved = mysql_fetch_array($reserved_set)) { if(empty($reserved['events_id'])) { echo "No Reservation's "; } else { $reserved_user = select_all_user_reserved($reserved['users_id']); while($user_set = mysql_fetch_array($reserved_user)) { echo "<li>" . $user_set['brand'] . ", " . $user_set['branch'] . "</li>"; } } } echo "</ol>"; echo "</td>"; // Expiration Date echo "<td>"; echo "<ol>"; $reserved_set = select_all_reservation($events['id']); while($reserved = mysql_fetch_array($reserved_set)) { $reserved_user = select_all_user_reserved($reserved['users_id']); while($user_set = mysql_fetch_array($reserved_user)) { $exp_date = select_all_expiration_date($user_set['id']); while($date = mysql_fetch_array($exp_date)) { echo "<li>" . $date['expiration_date'] . "</li>"; } } } echo "</ol>"; echo "</td>"; //Vacant echo "<td>"; $vacant_set = select_all_booking($events['id']); $booking_count = mysql_num_rows($vacant_set); $reserved_count = mysql_num_rows($reserved_set); $total_count = $booking_count + $reserved_count; $vacant = $events['number_of_cars'] - $total_count; if($vacant < 0) { echo "This Event is fully booked"; } else { echo $vacant; } echo"</td>"; echo "</tr>"; } echo "</table>"; } The function starts by receiving an id from picking a mall name from a select tag. It works fine, it outputs the data on the page like i want it to, but my problem is that i need a way to add a submit or any button to each generated row that will, when clicked will insert the event id and user id (who clicked it) to my booking/reservation table as a new row. I tried adding a form inside my function and saving the event id every loop but when i click the submit button i think it doesn't work (or i made a mistake).. Sorry if my coding sucks, I'm kinda new in php. Sorry if its kinda confusing my head is spinning right now from thinking this stuff the whole day.. Thanks in advance for any comments and ideas. This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=319161.0 Hi? im just a beginner in php i just want to ask how to insert a data into a table from a dropdown list. I have concatenate the itemid and description to form the dropdown list. But when i viewed my item_table the itemid and description columns are null. can you help me with this.. this is my php code for the dropdown list... <?php $query = "SELECT CONCAT(itemid,' ', '-',' ', description) AS Item FROM item_table"; $result = mysql_query($query) or die(mysql_error()); $dropdown = "<SELECT CONCAT(itemid,' ' '-',' ', description) AS Item FROM item_table>"; while($row = mysql_fetch_assoc($result)) { $dropdown .= "\r\n<option value='{$row['Item']}'>{$row['Item']}</option>"; } $dropdown .= "\r\n</select>"; echo $dropdown; ?> this my code for inserting data into the item_table... <?php if(isset($_POST ['submit'])) { $itemid = $_POST['itemid']; $description = $_POST['description']; $datein = $_POST['datein']; $qtyin = $_POST['qtyin']; $unitprice = $_POST['unitprice']; $unit = $_POST['unit']; $category = $_POST['category']; $empid = $_POST['empid']; $message =''; if(($itemid && $description == "")||($itemid && $description == null)) { header("location:IncomingEntry.php?msg=Incorrect"); exit(); } else { $link = mysql_connect('localhost', 'root', '') or die(mysql_error()); $db_selected = mysql_select_db('inventory', $link); $message=''; $query = "INSERT INTO incoming_table (itemid , description, datein, qtyin, unitprice, unit, category, empid) VALUES ('".$itemid."', '".$description."', '".$datein."', '".$qtyin."', '".$unitprice."', '".$unit."', '".$category."', '".$empid."')"; if (!mysql_query($query,$link)) { die('Error: ' . mysql_error()); } header("location: IncomingEntry.php?msg=1 record added"); } } ?> I have tried several attempts and can get data to select and using echo display it, however, I need to take this data and insert it into the database in a separate table. I have the following which does hafl the job, can I get some pointers on the rest. I have looked everywhere and not found a solution, at all. // Selects the data I need <?php mysql_connect("PRIVATE INFO","PRIVATE INFO","PRIVATE INFO") or die("Could not connect: " . mysql_error()); mysql_select_db("wpdb"); $result = mysql_query("SELECT ID FROM wp_posts WHERE post_title LIKE '%future%' AND post_status = 'publish' OR post_title LIKE '%option%' AND post_content LIKE '%fundamental%' AND post_status = 'publish' ORDER BY post_date DESC"); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo $row['ID']; echo "<br />";}; ?> Thought maybe something like the following would work, but am at a loss: INSERT INTO wp_term_relationships (object_id, term_taxonomy_id, term_order) SELECT ID FROM wp_posts WHERE post_title LIKE '%future%' AND post_status = 'publish' OR post_title LIKE '%option%' AND post_content LIKE '%fundamental%' AND post_status = 'publish' ORDER BY post_date DESC while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) VALUES ($row['ID'], '25', '0') Hello guys, im using this code atm and it's working, but the new rows created are put in the end of the table. I want them to stack up from the beginning pushing older rows down. What should i do to make that work? <?php $con = mysql_connect("localhost","*****","*******"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ananaz_se", $con); $sql="INSERT INTO stuff (adult, namn, url) VALUES ('$_POST[adult]','$_POST[namn]','$_POST[url]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 row added"; mysql_close($con) ?> Hi guys, im inserting data into the table using drop-down list & multi select list,well it works very well. but i need to make sure i should not insert same StudentID & CourseID twice. here my code for you could anyone tell me pls where should i write code to check existing data? <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); $result = mysql_query("SELECT * FROM student") or trigger_error('MySQL error: ' . mysql_error()); echo '<select name="sid">'; while($row = mysql_fetch_array($result)) { echo '<option value="' . $row['StudentID'] . '">' . $row['StudentName'] . '</option>'; } echo '</select>'; // ---------------- ?> </div> <div class="style41" id="Layer7"> <?php $result = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error()); echo '<select name ="cid[]" multiple="multiple" size="10">'; while($row = mysql_fetch_array($result)) { echo '<option value="' . $row['CourseID'] . '">' . $row['CourseName'] . '</option>'; } echo '</select>'; mysql_close($con); ?> ------------------------------------ <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); if (!empty($_POST['sid']) && !empty($_POST['cid'])) { $ct = 0; $student = $_POST['sid']; foreach ($_POST['cid'] as $key => $course) { $sql = "INSERT INTO take (StudentID, CourseID) VALUES('".mysql_real_escape_string($student)."', '".mysql_real_escape_string($course)."')"; $query = mysql_query($sql) or trigger_error('MySQL error: ' . mysql_error()); if (mysql_affected_rows() > 0){$ct++;} } echo $ct . ' rows added.'; } mysql_close($con); ?> Hi, Hope somebody can help me because I am relatively new to PHP and I am currently unsure how to implement the following: Let's assume I have a table like this id_number, name, data1, data2. When displayed on HTML page: Column id_number should be hidden. Column name is a link. When you click on this column then value stored in id_number should be used in a query to database. Result should be displayed below the link on the same page. When clicked again data should hide. I have some idea how to toggle visibility with a simple javascript function. How do i populate the table with hidden data and then extract it later on? Any help or suggestions are appreciated. Thanks. I am not sure if the title is correct; I tried my best.
I'm a PHP/MySQL beginner and I really need some help.
I have a small script that I am using for sending SMS. I recently added a phonebook. The problem with the phonebook right now is that it's available to all users, i.e. they can all update and delete all rows. What I would like to do is make it so that each user can update and delete only their own contacts.
I have a table call contacts. Inside that table there is first name, last name, company and phonenumber.
How can I accomplish this with PHP & MySQL?
CREATE TABLE IF NOT EXISTS `contacts` ( `contact_id` int(10) NOT NULL AUTO_INCREMENT, `firstname` varchar(255) NOT NULL, `lastname` varchar(255) NOT NULL, `company` varchar(255) NOT NULL, `cell_no` text NOT NULL, PRIMARY KEY (`contact_id`) ) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ; CREATE TABLE IF NOT EXISTS `users` ( `user_id` int(11) NOT NULL AUTO_INCREMENT, `users_name` varchar(30) NOT NULL, `uname` varchar(30) NOT NULL, `u_pass` varchar(60) NOT NULL, `utype` varchar(30) NOT NULL, `timezone` varchar(30) NOT NULL, `uapi_user` varchar(30) NOT NULL, `uapi_pass` varchar(60) NOT NULL, PRIMARY KEY (`user_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=20 ; <?php if (isset($_POST['submit'])){ //form has been submitted1 $firstname = trim($_POST['firstname']); $lastname = trim($_POST['lastname']); $company = trim($_POST['company']); $cellno = trim($_POST['cell_no']); if($firstname == ''){ echo '<div class="alert alert-danger">First Name is not Valid!</div>'; exit; }elseif($lastname == ''){ echo '<div class="alert alert-danger">Last Name is not Valid!</div>'; exit; }elseif($company == ''){ echo '<div class="alert alert-danger">Company is not Valid!</div>'; exit; }elseif($cellno == ''){ echo '<div class="alert alert-danger">Cellphone Number is not Valid!</div>'; exit; }else{ $query = "Select cell_no from contacts where cell_no = '".$cellno."' "; $result = mysql_query($query); if (!mysql_num_rows($result)) { $sql = "INSERT INTO contacts(firstname, lastname, company, cell_no) values('{$firstname}','{$lastname}', '{$company}', '{$cellno}')"; $result = mysql_query($sql); confirm_query($result); //echo '<div class="alert alert-success">Successfully added.</div>'; //exit; ?> <script type="text/javascript"> window.location = "contact_list.php"; </script> <?php } else{ echo '<div class="alert alert-danger">Username. already exist!.</div>'; echo '<p><a href="new_contact.php" class="btn btn-success"> Back </a></p>'; exit; }} }else{ $firstname = ""; $lastname = ""; $company = ""; $cellno = ""; } ?> Hi. I'm making a private webpage for my self where i can post news, games, music titles and movies. I'm made it so that i can post and delete them from my page. Also i wanted to have the option to edit them. So i made a action that shows me a list og the posts in each category. And then i want the option to hit UPDATE, and then it should open the form with the info's from MySQL. I've made those two options as the following: Code: [Select] elseif ( $action == "EditNews" ) { // EDIT NEWS START if ( !$_GET['id'] ) { ?> <div id="content"> <?php $sql = mysql_query("SELECT * FROM news") or die(mysql_error()); while ( $row = mysql_fetch_assoc($sql) ) { echo "<a href=\"?page=Admin&action=UpdateNews&id=" . $row['news_id'] . "\">Update</a> | " . $row['subject'] . "<br>"; } echo "</div>"; } else { $type = $_GET['type']; $id = $_GET['id']; } // EDIT NEWS END } elseif ( $action == "UpdateNews" ) { // UPDATE NEWS START if ( !$_POST['submit'] ) { $id = $_GET['id']; $sql = mysql_query("SELECT * FROM news WHERE news_id='$id'") or die(mysql_error()); $user_id = $_SESSION['uid']; $subject = $_row['subject']; $body = nl2br($_row['body']); ?> <div id="content"> <form action="?page=Admin&action=EditNews" method="post"> <div id="">Subject</div> <div id=""><input id="input" type="text" name="subject"></div> <div id="">Message</div> <div id=""><textarea id="input" name='body' cols='15' rows='4'></textarea></div> <div id=""><input id="submit" class="input" type="submit" name="submit" value="Post News"></div> </form> </div> <?php } else { ?> <div id="content">Not Working!</div> <?php } // UPDATE NEWS END } - I know i miss something, but i've tried to get to the point, where i only have the form as blank, and need some help from there. I hope someone can help me with my problem. MOD EDIT: [code] . . . [/code] tags added. I almost have it but I am missing something. The form is being sent and the row is being created but there is not data. It is blank. I know it is something simple I am missing but I cannot figure it out
If anyone can look at my code below and tell me what I am missing to make the inputed info be seen, I would sure appreciate it.
<?php include_once('class/class_email.php'); // contact to database $connect = mysql_connect("localhost", "admin", "password") or die ("Error , check your server connection."); mysql_select_db("database"); $fname = $_POST['fname']; $lname = $_POST['lname']; $email = $_POST['email']; $company = $_POST['company']; $telephone = $_POST['telephone']; $comments = $_POST['comments']; $EID = $_POST['eid']; extract ($_POST); // Pick up the form data and assign it to variables // $id = intval($id); $fname = strip_tags($fname); $lname = strip_tags($lname); $email = strip_tags($email); $company = strip_tags($company); $telephone = intval($telephone); $query="INSERT INTO users(`id`, `fname`, `lname`, `email`,`company`,`telephone`) VALUES('$id','$fname','$lname','$email','$company','$telephone')"; echo $query; mysql_query($query) or die(mysql_error()); echo mysql_error(); $SQL_GetEquipment = "SELECT * FROM `new_equip` WHERE `id`='$EID' LIMIT 1;"; $result = mysqli_query($connect,$SQL_GetEquipment); $row = mysqli_fetch_assoc($result); $EmailBody = "$fname $lname has requested a quote from NAPE on Item $EID\n Information on quote request: \n Name: $fname $lname \n Email: $email \n Company: $company \n Number: $telephone \n Comments: $comments \n \n Information Requested for: {$row['itemname']}\n The URL to {$row['itemname']} is: http://www.mydomain.com.com/new-product.php?Item=$EID \n Click to send a quote now:\n http://www.mydomain.com.com/Admin/send-quote.php?id=$EID "; $e = new email(); //First value is the URL of your server, the second the port number $e->set_server( 'mail.mydomain.com.com', 26); //First value is your username, then your password $e->set_auth('noreply@mydomain.com', '112233'); //Set the "From" setting for your e-mail. The Name will be base64 encoded $e->set_sender( 'Quote Requested', 'noreply@mydomain.com' ); //for one recipient //$send_to = array('myemail@mydomain.com','myemail2@mydomain.com'); $send_to = ('myemail@gmail.com'); //you may also specify multiple recipients by creating an array like this: //$send_to = array('foo1@localhost.local', 'foo2@localhost.local', 'foo3@localhost.local'); $subject = 'Quote Request from NAPE'; $body = "$EmailBody"; if( $e->mail($send_to, $subject, $body, $headers) == true ) { //message was received by the smtp server //['last'] tends to contain the queue id so I like to save that string in the database echo 'last: '.htmlspecialchars($e->srv_ret['last']).''; }else{ //something went wrong echo 'all: '.nl2br(htmlspecialchars($e->srv_ret['all'])).''; echo 'full:'.nl2br(htmlspecialchars($e->srv_ret['full'])).''; } ?> Not sure exactly what my problem is here, but i have looked at this at least a hundred times and I just can't figure out what is going on. Everything works great except for the sending the email part. It displays the error "There was a problem sending the mail". The form field checking works fine, the insert into mysql works fine - - but it won't send the email. I have tried using double quotes and single quotes for the email information ($to, $subject, etc...) I have even eliminated the form data from the email information and it still doesn't send. I am hopelessly stuck at this point Any ideas?? Code below: <html> <body bgcolor = 'blue'> <div align = 'center'> <h1>test FORM</h1> </div> <p> <?php If ($_POST['submit']) //if the Submit button pressed { //collect form data $fname = $_POST['FName']; $lname = $_POST['LName']; $email = $_POST['Email']; $tel = $_POST['Tel']; $mess = $_POST['Mess']; $errorstring = ""; if (!$fname) $errorstring = $errorstring."First Name<br>"; if (!$lname) $errorstring = $errorstring."Last Name<br>"; if (!$email) $errorstring = $errorstring."Email<br>"; if ($errorstring !="") echo "<div align = 'center'><b>Please fill out the following fields:</b><br><font color = 'red'><b>$errorstring</b></font></div>"; else { $fname = mysql_real_escape_string($fname); $lname = mysql_real_escape_string($lname); $email = mysql_real_escape_string($email); $tel = mysql_real_escape_string($tel); $mess = mysql_real_escape_string($mess); $con = mysql_connect("localhost","user","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("mec", $con); $sql="INSERT INTO formtest (FName, LName, Title, Tel, Email, Mess) VALUES ('$fname','$lname','$title','$tel','$email','$mess')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } $to = 'myemailaddress'; $subject = 'test form'; $message = 'Hello'; $headers = 'From Me'; $sent = mail($to, $subject, $message, $headers); if($sent) {print "Email successfully sent";} else {print "There was an error sending the mail";} mysql_close($con); } } ?> <p> <form action= 'testmecform.php' method= 'POST'> <table width = '640' border = '0' align = 'center'> <tr> <td align = 'right'><b>First Name</b></td> <td><input type = 'text' name = 'FName' value = '<?php echo $fname; ?>' size = '25'></td> <td><div align = 'right'><b>Telephone</b></div></td> <td><input type = 'text' name = 'Tel' value = '<?php echo $tel; ?>' size = '25'></td> </tr> <tr> <td align = 'right'><b>Last Name</b></td> <td><input type = 'text' name = 'LName' value = '<?php echo $lname; ?>' size = '25'></td> <td> </td> <td> </td> </tr> <tr> <td align = 'right'><b>Email</b></td> <td><input type = 'text' name = 'Email' value = '<?php echo $email; ?>' size = '25'></td> <td> </td> <td> </td> </tr> <tr> <th colspan = '4'><b>Please enter any additional information he </b></th> </tr> <tr> <th colspan = '4'><textarea name = 'Mess' cols = '50' rows = '10'><?php echo $mess; ?></textarea></th> </tr> <tr> <th colspan = '4'><b>Please make sure all information is correct before submitting</b></th> </tr> <tr> <th colspan = '4'><input type = 'submit' name = 'submit' value = 'Submit Form'></th> </tr> </table> </form> </body> </html> Need a little help with this code. I'm running Apache 2.21 / PHP 5.3.5 on Windows. This is a V6 application. I am running Windows 7. I am new, and trying to learn. What am I doing wrong in line 8 with the 'if' statement? The idea is to create a table in mySQL using a form interface. Thank you in advance. Parse error: syntax error, unexpected '{' in C:\website\do_createtable.php on line 8 <? $db_name="booster"; $connection=mysql("localhost", "user", "password") or die (mysql_error()); $db=mysql_select_db($db_name, $connection) or die (mysql_error()); $sql="CREATE TABLE $_POST[table_name] ("; for ($i=0; $i < count($_POST[field_name]); $i++) { $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i]; if ($_POST[field_length][$i] != "" { $sql .= " (".$_POST[field_length][$i]."),"; } else { $sql .= ","; } } $sql=substr($sql,0,-1); $sql .= ")"; hello i have a db with name which populates a form dropdown list there are various variables like this in the form. the user selects their choices and then submits the form to a new db table at the moment it does not capture the dropdown data to the insert sql help please here is the code: =============================== <?php include ('connect.php'); // Insert any new image into database if(isset($_POST['xsubmit']) && $_FILES['imagefile']['name'] != "") { $fileName = $_FILES['imagefile']['name']; $fileSize = $_FILES['imagefile']['size']; $fileType = $_FILES['imagefile']['type']; $content = addslashes (file_get_contents($_FILES['imagefile']['tmp_name'])); $jeweltype = $_POST['jeweltype']; $jewelsize = $_POST['jewelsize_in']; $jewelcolour = $_POST['jewelcolour_in']; $jewelmaterial = $_POST['jewelmaterial_in']; $jewelgender = $_POST['jewelgender_in']; if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } // Checking file size if ($fileSize < 150000) { mysql_query ("INSERT into jewel_images (name,size,type,content,jeweltype,jewelsize,jewelcolour,jewelmaterial,jewelgender) " . "values ('$fileName','$fileSize','$fileType','$content','$jeweltype','$jewelsize','$jewelcolour','$jewelmaterial','$jewelgender')"); } else { $err = "The Image file to too large!"; } } // Find out about latest image $gotten = mysql_query("select * from jewel_images order by row_id desc"); $row = mysql_fetch_assoc($gotten); $bytes = $row['content']; // If this is the image request, send out the image if ($_REQUEST['pic'] == 1) { header("Content-type: $row[type];"); print $bytes; } ?> <html> <head> <title>Upload an image to a database</title> </head> <body> <font color="#FF3333"><?php echo $err ?></font> <table> <form name="Upload" enctype="multipart/form-data" method="post"> <tr> <td>Upload <input type="file" name="imagefile"><br /> Jewelery Type: <select> <?php $sql="SELECT jeweltype FROM jeweltypes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jeweltype" ><?php echo $data['jeweltype'] ?></option> <?php } ?> </select> <br /> Jewelery Size: <select> <?php $sql="SELECT jewelsize FROM jewelsizes"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelsize" ><?php echo $data['jewelsize'] ?></option> <?php } ?> </select> <br /> Jewelery Colour: <select> <?php $sql="SELECT jewelcolour FROM jewelcolours"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelcolour_in" ><?php echo $data['jewelcolour'] ?></option> <?php } ?> </select> <br /> Jewelery Material: <select> <?php $sql="SELECT jewelmaterial FROM jewelmaterials"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelmaterial_in" ><?php echo $data['jewelmaterial'] ?></option> <?php } ?> </select> <br /> Jewelery Gender: <select> <?php $sql="SELECT jewelgender FROM jewelgenders"; $result =mysql_query($sql); while ($data=mysql_fetch_assoc($result)){ ?> <option value ="jewelgender_in" ><?php echo $data['jewelgender'] ?></option> <?php } ?> </select> <br /> <input type="submit" name="xsubmit" value="Upload"> </td> </tr> <tr> <td>Latest Image</td> </tr> <tr> <td><img src="?pic=1"></td> </tr> </form> </table> </body> </html> First page adds a new job number, then the order page loaded with the job number id as a get id. Basically there is a while loop in the order page which shows products/services client can order and he chooses what he requires 'one or eight services' (8 in total) and some other variables like date of order and client name etc. Here is the order item code The first sql statement executes fine, but in the second sql query nothing happens $sql="insert into job_order(order_num,order_date,order_customer_id, order_remarks) values(".$_GET['id'].",NOW(),".$_POST['companyBox'].",'".$_POST['remarkBox']."');"; $res=mysql_query($sql); $id=mysql_insert_id(); foreach($_POST as $key => $value) { if(!empty($value)) { $key.' => '.trim(strip_tags($value)); $order="INSERT INTO orderprod (order_num,prod_id,order_amount,teeth_amount) VALUES ('$_GET[id]','$value','$value','$value');"; $orderres=mysql_query($order) or die(mysql_error()); } }?>$sql="insert into job_order(order_num,order_date,order_customer_id, order_remarks) values(".$_GET['id'].",NOW(),".$_POST['companyBox'].",'".$_POST['remarkBox']."');"; $res=mysql_query($sql); $id=mysql_insert_id(); foreach($_POST as $key => $value) { if(!empty($value)) { $key.' => '.trim(strip_tags($value)); $order="INSERT INTO orderprod (order_num,prod_id,order_amount,teeth_amount) VALUES ('$_GET[id]','$value','$value','$value');"; $orderres=mysql_query($order) or die(mysql_error()); } } ?> ============= This is the formI have removed parts which are irrelevant. Please note that dateBox and companyBox are not required to be looping as they are only for first table, echo "<form action=".$config_basedir."./vieworder.php?id=".$_GET['id']." name=form1 method=post>";?><table><tr><td><h4>JOB ORDER</H4></TD><TD></TD></TR><TR><TD>ORDER NUMBER</TD><TD><?PHP ECHO $_GET['id'] ?></td></tr><tr><td>ORDER DATE</td><td><input type=text name=dateBox></td></tr> <tr><td>COMPANY NAME</td><td><?PHP$sql="select * from customers";$res=mysql_query($sql);echo "<select name=companyBox><option value=''>Please select</option>";WHILE($fetch=mysql_fetch_assoc($res)){ echo "<option value='".$fetch['id']."'>".$fetch['cust_name']."</option>";} echo "</select>"; echo "</td></tr>"; ?> </table> <table><tr><th>ITEM</th><th>QUANTITY</th><th>N0. of Teeths</th></tr><tr><?PHP$sql="select * from products";$res=mysql_query($sql); WHILE($fetch=mysql_fetch_assoc($res)){ echo "<td><input type=text name=desBox value='".$fetch['prod_id']."'>".$fetch['prod_name']."</td><td><input type=text name=quantBox></td><td><input type=text name=teethBox></td>";echo "</tr>";}echo "</table>";?>echo "<form action=".$config_basedir."./vieworder.php?id=".$_GET['id']." name=form1 method=post>"; ?> <table> <tr> <td><h4>JOB ORDER</H4></TD><TD></TD> </TR> <TR> <TD>ORDER NUMBER</TD><TD><?PHP ECHO $_GET['id'] ?></td> </tr> <tr> <td>ORDER DATE</td><td><input type=text name=dateBox></td> </tr> <tr> <td>COMPANY NAME</td><td> <?PHP $sql="select * from customers"; $res=mysql_query($sql); echo "<select name=companyBox><option value=''>Please select</option>"; WHILE($fetch=mysql_fetch_assoc($res)){ echo "<option value='".$fetch['id']."'>".$fetch['cust_name']."</option>";} echo "</select>"; echo "</td></tr>"; ?> </table> <table> <tr> <th>ITEM</th><th>QUANTITY</th><th>N0. of Teeths</th> </tr> <tr> <?PHP $sql="select * from products"; $res=mysql_query($sql); WHILE($fetch=mysql_fetch_assoc($res)){ echo "<td><input type=text name=desBox value='".$fetch['prod_id']."'>".$fetch['prod_name']."</td> <td><input type=text name=quantBox></td> <td><input type=text name=teethBox></td>"; echo "</tr>";} echo "</table>"; ?> HERE IS THE IMAGE showing the populated services. http://dubads.com/images/order.jpg Gang, I'm trying to create a form that allows me to edit table data within a MySQL database. I've been able to display the data, no problem. I want to be able to edit the fields in the database tables but have had no luck. I'm using session variables to connect and gather the information I need. Here's the code for collecting the table data: <?php mysql_select_db($database); if(empty($database)) { echo "<p>You must be connected to a database in order to view any table data.</p>"; }else{ // Show table data start $sql = "SHOW TABLES FROM $database"; $result = mysql_query($sql); $table = array(); while ($row = mysql_fetch_row($result)) { $table[] = $row[0]; } if (count($table) == 0) { echo "<p>The database '" . $database . "' contains no tables.</p>\n"; } else { foreach($table AS $aTable) { echo "<p style='text-align:left;float:left;width:100%;margin-top:15px;'>Table: <font color='green'>$aTable</font>"; if (!mysql_connect($hostname, $user, $passwd)) die("Can't connect to database"); if (!mysql_select_db($database)) die("Can't select database"); // sending query $result = mysql_query("SELECT * FROM {$aTable}"); if (!$result) { die("Query to show fields from table failed"); } $fields_num = mysql_num_fields($result); echo "<div style='text-align:left;width:100%;'><table border='0'><tr><td></td>"; // printing table headers for($i=0; $i<$fields_num; $i++) { $field = mysql_fetch_field($result); echo "<td style='text-align:left;padding:3px;font-size:11px;color:green;'>{$field->name}</td>"; } echo "</tr>\n"; // printing table rows while($row = mysql_fetch_row($result)) { echo "<tr><td style='text-align:center;padding:3px;font-size:10px;border:1px solid #888;background:#fff;'><a href='edit_data.php?=$row[0]' style='color:red;text-decoration:none;'>edit</a></td>"; // $row is array... foreach( .. ) puts every element // of $row to $cell variable foreach($row as $cell) echo "<td style='text-align:left;padding:3px;font-size:10px;border:1px solid #888;background:#fff;clear:both;float:left;'>$cell</td>"; echo "</tr>\n"; } mysql_free_result($result); echo "</table></div>"; } } } //end show table data include("includes/footer.php"); ?> I'm not sure how to create a form using the session variables in order to be able to edit the correct information since I want to edit numerous databases. Any help would be great! Thanks hi, I am displaying mysql data in a table form using php. The first attached pic shows how is it looking with the current code. As you can see there are heading on the left and top of the table. So user can make booking on a specific day (heading is on the top for days) and for specific duration of time(headings are on the left). As in first pic, there has been a booking on Wed at 10 o clock, but the problem is, it is displaying duration just for half an hour. As this duration is from 10 o clock to 11 o clock. So what I want to do is in the second pic. Please view the second pic. In the second pic I want to display the highlighted area in blie because all this time is booked or reserved. Here is my current code. Code: [Select] <?php require("dbconnect/dbconnect.php"); $building = $_POST['building']; $block = $_POST['block']; $d_m = $_POST['d_m']; $choose_one = $_POST['choose_one']; function get_data($table, $choose_one) { $bool = 1; $get_days = mysql_query("SELECT * FROM days"); $get_times = mysql_query("SELECT * FROM time"); $days = mysql_num_rows($get_days); //$row_day_id = mysql_fetch_assoc($get_days); //$day_id_db = $row['dayid']; echo "<table cellspacing='0' border='0' cellpadding='5'> <tr> <th scope='col' width='100'></th>"; while($row_time = mysql_fetch_assoc($get_times)) { while($bool <= $days) { $row_days = mysql_fetch_assoc($get_days); $dayname = $row_days['days']; echo " <th scope='col' width='100'>$dayname</th> "; $bool++; } $day_id = 1; $timeid = $row_time['timeid']; $time = $row_time['time']; echo " <tr> <th scope='row' width='100' height='50'>$time</th>"; while($day_id <= $days) { $get = mysql_query(" SELECT * FROM $table WHERE dayid='$day_id' AND timeid_from='$timeid' AND d_m='$choose_one' ") or die(mysql_error()); if(mysql_numrows($get) > 0) { $row = mysql_fetch_assoc($get); $block = $row['block']; $apt = $row['apartment_no']; $room = $row['room_no']; $d_m = $row['d_m']; $timeid_to = $row['timeid_to']; echo " <td width='110' height='50' bgcolor='lightblue'> <b>Block:</b> $block<br /> <b>Apartment:</b> $apt<br /> <b>Room:</b> $room </td> "; //echo "</tr>"; } else { echo " <td width='110' height='50' bgcolor='green'></td> "; } $day_id++; } //end of second inner while loop echo "</tr>"; } //end of outer while loop echo "</table>"; } if($building) { if($block) { if($d_m) { if($choose_one) { if($building == 'mik') { if($block == 'Block A' || $block == 'Block B') { get_data('reserve_mik_ab', $choose_one); } else if ($block == 'Block C' || $block == 'Block D') { get_data('reserve_mik_cd', $choose_one); } } if($building == 'p') { get_data('reserve_p',$choose_one); } } else { echo "Choose one!"; } } } } ?> Please help! Hi, I am extracting data from a mysql db to then edit and update back into the db and use below to list items first - Code: [Select] <?php // Connect to server and select database. mysql_connect('localhost', 'xxxxxxxxx', 'xxxxxxxxxxxxx') or die("Error: ".mysql_error()); mysql_select_db("xxxxxxxxx"); $sql="SELECT * FROM properties"; $result=mysql_query($sql); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="400" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="4"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Name</strong></td> <td align="center"><strong>Lastname</strong></td> <td align="center"><strong>Email</strong></td> <td align="center"><strong>Update</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td><? echo $rows['Property_Rating']; ?></td> <td align="center"><a href="update.php?id=<? echo $rows['ID']; ?>">update</a></td> </tr> <?php } ?> </table> </td> </tr> </table> <?php mysql_close(); ?> I then go next page where I show data and allow editing - <?php // Connect to server and select database. mysql_connect('localhost', 'xxxxxxxxx', 'xxxxxxxxxxxxx') or die("Error: ".mysql_error()); mysql_select_db("xxxxxxxxx"); // get value of id that sent from address bar $id=$_GET['id']; // Retrieve data from database $sql="SELECT * FROM properties WHERE id='$id'"; $result=mysql_query($sql); $rows=mysql_fetch_array($result); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <form name="form1" method="post" action="update_ac.php"> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td> </td> <td colspan="3"><strong>Update data in mysql</strong> </td> </tr> <tr> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> </tr> <tr> <td align="center"> </td> <td align="center"><strong>Property_Name</strong></td> <td align="center"><strong>roperty_Rating</strong></td> <td align="center"><strong>Property_Address</strong></td> </tr> <tr> <td> </td> <td align="center"><input name="Property_Rating" type="text" id="Property_Rating" value="<? echo $rows['Property_Rating']; ?>"></td> </tr> <tr> <td> </td> <td><input name="id" type="hidden" id="id" value="<? echo $rows['id']; ?>"></td> <td align="center"><input type="submit" name="Submit" value="Submit"></td> <td> </td> </tr> </table> </td> </form> </tr> </table> <? // close connection mysql_close(); ?> And then the update page - <?php // Connect to server and select database. mysql_connect('localhost', 'xxxxxxxxx', 'xxxxxxxxxxxxx') or die("Error: ".mysql_error()); mysql_select_db("xxxxxxxxx"); // update data in mysql database $sql="UPDATE properties SET Property_Name='$Property_Name' WHERE id='$id'"; $result=mysql_query($sql); // if successfully updated. if($result){ echo "Successful"; } else { echo "ERROR"; } ?> Everything seems ok but when I edit the data and submit I get the success message but data not changed in db. Any ideas? Thanks. MOD EDIT: code tags added. how i want to display data from database to look like this : <table width="633" height="224" border="1"> <tr bgcolor="#999900"> <td width="45">Bil</td> <td width="121">Course_name</td> <td width="83">session</td> <td width="83">start_date</td> <td width="83">end_date</td> <td width="83">notes</td> <td width="89">pre-req</td> </tr> <tr bgcolor="#6A7AEA"> <td rowspan="2">1.</td> <td rowspan="2" bgcolor="#6A7AEA">Math</td> <td>1st session </td> <td>1 jan 11 </td> <td>6 jan 11 </td> <td rowspan="2"> </td> <td rowspan="2"><image icon that will link to the oter site> </td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session </td> <td bgcolor="#6A7AEA">8 jan 11 </td> <td bgcolor="#6A7AEA">15 jan 11 </td> </tr> <tr> <td bgcolor="#0066CC">2.</td> <td bgcolor="#0066CC">English</td> <td bgcolor="#0066CC">1st session </td> <td bgcolor="#0066CC">1 feb 11 </td> <td bgcolor="#0066CC">6 feb 11 </td> <td bgcolor="#0066CC"> </td> <td bgcolor="#0066CC"><image icon that will link to the oter site></td> </tr> <tr> <td rowspan="2" bgcolor="#6A7AEA">3.</td> <td rowspan="2" bgcolor="#6A7AEA">Science</td> <td height="29" bgcolor="#6A7AEA">1st session </td> <td bgcolor="#6A7AEA">8 march 11 </td> <td bgcolor="#6A7AEA">15 march 11 </td> <td rowspan="2" bgcolor="#6A7AEA"> </td> <td rowspan="2" bgcolor="#6A7AEA"><image icon that will link to the oter site></td> </tr> <tr> <td bgcolor="#6A7AEA">2nd session</td> <td bgcolor="#6A7AEA">16 march 11 </td> <td bgcolor="#6A7AEA">21 march 11 </td> </tr> </table> ** all the view data is called from database including the icon image thanks... |