PHP - Problems With Post.php Submitting Mysql Query
I have a post.php that is suposted to run a mysql query, and it used to, but it won't anymore and I don't have the old file.
Can someone just look see if there is something i'm missing? <?php //header("Location: ./?p=UCP"); setcookie("Errors", 0, time()-3600); // Connects to your Database mysql_connect("SERVER", "USER", "PASS") or die(mysql_error()); mysql_select_db("DB") or die(mysql_error()); if (isset($_POST['remove'])) { $id=$_POST['ID']; if($_POST['initals'] == "NLW") { if (is_numeric ($id)) { mysql_query("DELETE FROM `users` WHERE `users`.`ID` = $id LIMIT 1"); $error="<span style="; $error .="color:green"; $error .=">"; $error .= "User Removed."; $error .="</span>"; setcookie(Errors, $error, time()+20); } else { $error="<span style="; $error .="color:red"; $error .=">"; $error .= "Please enter a valid ID"; $error .="</span>"; setcookie(Errors, $error, time()+20); header('Location ./?p=UPC'); } } else { $error="<span style="; $error .="color:red"; $error .=">"; $error .="Initials are not correct"; $error .="<span/>"; setcookie(Errors, $error, time()+20); header('Location ./?p=UPC'); } } elseif (isset($_POST['submit'])) { //This makes sure they did not leave any fields blank if (!$_POST['username'] | !$_POST['pass'] | !$_POST['pass2'] ) { $error="<span style="; $error .="color:red"; $error .=">"; $error .= "You did not complete all of the required fields"; $error .="</span>"; setcookie(Errors, $error, time()+20); header('Location ./?p=UPC'); } // checks if the username is in use if (!get_magic_quotes_gpc()) { $_POST['username'] = addslashes($_POST['username']); } $usercheck = $_POST['username']; $check = mysql_query("SELECT username FROM users WHERE username = '$usercheck'") or die(mysql_error()); $check2 = mysql_num_rows($check); //if the name exists it gives an error if ($check2 != 0) { $error="<span style="; $error .="color:red"; $error .=">"; $error .= "Sorry, the username is already in use."; $error .="</span>"; setcookie(Errors, $error, time()+20); header('Location ./?p=UPC'); } // this makes sure both passwords entered match if ($_POST['pass'] != $_POST['pass2']) { $error="<span style="; $error .="color:red"; $error .=">"; $error .= 'Your passwords did not match.'; $error .="</span>"; setcookie(Errors, $error, time()+20); echo $error; } // here we encrypt the password and add slashes if needed $_POST['pass'] = md5($_POST['pass']); if (!get_magic_quotes_gpc()) { $_POST['pass'] = addslashes($_POST['pass']); $_POST['username'] = addslashes($_POST['username']); $_POST['pass2'] = $_POST['pass2']; } // now we insert it into the database $insert = "INSERT INTO users (username, password, Human-Readable) VALUES ('".$_POST['username']."', '".$_POST['pass']."', '".$_POST['pass2']."')"; mysql_query("INSERT INTO users (username, password, Human-Readable) VALUES ('".$_POST['username']."', '".$_POST['pass']."', '".$_POST['pass2']."')"); $error="<span style="; $error .="color:green"; $error .=">"; $error .= "<h1>User Registered</h1> <p><h2>Thank you, the user has been registered - he/she may now login</a>.</h2></p>"; $error .="</span>"; setcookie(Errors, $error, time()+20); header('Location: ./?p=UCP'); } else { header('Location: ./?p=UCP'); echo $error; } ?> The remove function works but its the submit. Thanks in advanced Similar TutorialsI have a photo album style gallery to build and i'm finding it dificult to list all the table names (these are names of photo albums) and then enter the data into a seperate query for each album name (these will change often so i cant keep updating the file as normal. this will then post all the data to the xml file and show the set of photos in the individual albums in a flash file. can anyone help me where im going wrong at all? <?php $dbname = 'cablard'; if (!mysql_connect('localhost', 'cablard', '')) { echo 'Could not connect to mysql'; exit; } $sql = "SHOW TABLES FROM $dbname"; $result = mysql_query($sql); if (!$result) { echo "DB Error, could not list tables\n"; echo 'MySQL Error: ' . mysql_error(); exit; } while ($row = mysql_fetch_row($result)) { echo "Table: {$row[0]}\n"; } mysql_free_result($result); $query = "SELECT * FROM photo ORDER BY id DESC"; $result2 = mysql_query ($query) or die ("Error in query: $query. ".mysql_error()); while ($row = mysql_fetch_array($result2)) { echo " <image> <date>".$row['date']."</date> <title>".$row['title']."</title> <desc>".$row['description']."</desc> <thumb>".$row['thumb']."</thumb> <img>".$row['image']."</img> </image> "; } ?> Thanks James Hi all, I have a query that runs just fine from the command line in MySQL, but when I try to run the exact same query using PHP to query the same database I get an error. The query is: SELECT Street, City, ZipCode, providers.bWebpage, OpHours.Open, OpHours.Close, lat, lng, ( 3959 * acos( cos( radians('41.757787') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('-88.321416') ) + sin( radians('41.757787') ) * sin( radians( lat ) ) ) ) AS distance FROM Providers, OpHours WHERE providers.bName = OpHours.bName AND OpHours.Close > '17:00:00' HAVING distance < '10' ORDER BY distance LIMIT 0 , 30 Like I stated above, when I run this query from the MySQL command line it returns 12 rows, which is what it should. But when I try executing the same query via PHP I get this error message: Invalid query: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''17:00:00' HAVING distance < '10' ORDER BY distance LIMIT 0 , 30' at line 1 The PHP code is thus: $connection=mysql_connect ($server, $username, $password); if (!$connection) { echo 'Connection error'; die("Not connected : " . mysql_error()); } $db_selected = mysql_select_db($database, $connection); if (!$db_selected) { echo 'Database selection error' . '\n'; die ("Can\'t use db : " . mysql_error()); } $query = sprintf("SELECT Street, City, ZipCode, providers.bWebpage, OpHours.Open, OpHours.Close, lat, lng, ( 3959 * acos( cos( radians('%s') ) * cos( radians( lat ) ) * cos( radians( lng ) - radians('%s') ) + sin( radians('%s') ) * sin( radians( lat ) ) ) ) AS distance FROM Providers, OpHours WHERE providers.bName = OpHours.bName %s HAVING distance < '%s' ORDER BY distance LIMIT 0 , 30", mysql_real_escape_string($center_lat), mysql_real_escape_string($center_lng), mysql_real_escape_string($center_lat), mysql_real_escape_string($hours), mysql_real_escape_string($radius)); $result = mysql_query($query); if (!$result) { die("Invalid query: " . mysql_error()); } I've also checked the MySQL logs and they don't report any errors. I'm running this on a Mac using Snow Leopard. MySQL version is 5.1.37 and the PHP version is 5.2.11. If it makes any difference both of these are from MAMP version 1.8.4. Hello, I have a problem with my login form, when i press the submit button, it returns a valitation error that I made. The error message is triggered when the submit button isn't pressed, but the form still tries to submit. session_start(); include "sources/php/class.php"; $e = $_POST['loginname']; $p = $_POST['loginpass']; $s = $_POST['submit']; if(!isset($s)) { header('location: '.$_SESSION['psite'].'.php?p=error&ploca=login&pid=0'); exit(); } Heres a piece of my code, first i define the variables, then check if the button was pressed, but something is wrong there? - But what? Hope you can help me. Thanks in advance. Hello, I'm new to this forum and I need some help. I'm creating a simple database that it submits data from a user input. Unfortunatly, it's not sending any data to mysql also the form is not validating each field. Code: [Select] <?php if (isset($_POST['submitted'])){ $fields = array( 'email', 'state', 'district', 'gender', 'age', 'profession', 'survey', ); foreach($fields as $fieldName) { if(isset($_POST[$fieldName]) and trim($_POST[$fieldName]) !==''){ $fieldName = trim($_POST[$fieldName]); }else { $errors[] = "Please enter your". $fieldName .""; //code to validate fields } } if(isset($errors)){ require_once('Connections/encourage.php'); $query = "INSERT INTO participants (email, state, district, gender, age, profession, survey) VALUES ('$email', '$state', '$district', '$gender', '$age', '$profession','$survey')"; //databasse connection $result = mysql_query ($query); if ($result){ echo '<h1 id="mainhead">Thanks for submitting</hl> <p>You are now registered</p>'; exit(); }else{ echo '<h1 id="mainhead">System Error</hl> <p>Your registration could not be completed due to a system error We apologize for any incovience</p>';//gives system error echo 'p' . mysql_error(). '<br /><br />Query: ' . $query . '</p>'; exit(); } mysql_close(); } else { echo '<h1 id="mainhead">Error!</h1> <p class="error">The following error(s) occurred:<br />'; foreach($errors as $msg) { echo " - $msg<br/>\n"; } echo '</p><p>Please try again.</p><p><br/></p>'; } } ?> <form id="form1" name="form1" method="post" action"registration.php"> <fieldset class="first"> <label class="lableone" for="email">Email:* </label> <input name="email" value="<?php if(isset($_POST['email'])) echo $_POST['name'];?>"/> <label for="state"/>State:* </label> <input name="state" value="<?php if(isset($_POST['state'])) echo $_POST['state'];?>"/> <label for="schooldistrict"/>School District:* </label> <input name="schooldistrict" value="<?php if(isset($_POST['district'])) echo $_POST['district'];?>" /> <label for="gender">Gender:* </label> <select name="gender"> <option>Choose Your Gender</option> <option value="male" <?php echo ($form['gender'] == 'male' ? ' selected' : ''); ?>>Male</option> <option value="female"<?php echo ($form['gender'] == 'female' ? ' selected' : ''); ?>>Female</option> </select> <label for="age"/>Your Age:* </label> <input name="age" type="text" class="age" maxlength="2" value="<?php if(isset($_POST['age'])) echo $_POST['age'];?>" /> <label for="profession"/>Profession:* </label> <input name="profession" value="<?php if(isset($_POST['age'])) echo $_POST['age'];?>" /> <label for="surveys"/>Willingness to participate in future surveys: </label> <input name="surveys" type="checkbox" id="surveys" value="yes" <?php echo ($form['survey'] == 'yes' ? 'checked' : '');?>/> </fieldset> <fieldset> <input class="btn" name="submit" type="submit" value="Submit" /> <input class="btn" name="reset" type="reset" value="Clear Form" /> <input type="hidden" name="submitted" value="TRUE" /> </fieldset> </form> Can someone help me out? Thanks in advanced! hello. i have an issue where the data stored with an image is not saving to a mysql table. the image data is ok, just not the selections from the dropdown lists. 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> ============================================== here is the query =============================================== <html> <head><title>Your Page Title</title></head> <body> <?php $database="josh_jewel"; mysql_connect ("localhost", "xxxxxxxxx", "yyyyyyyyyyyy"); @mysql_select_db($database) or die( "Unable to select database"); $result = mysql_query( "SELECT jewelcolour FROM jewel_images" ) or die("SELECT Error: ".mysql_error()); $num_rows = mysql_num_rows($result); print "There are $num_rows records.<P>"; print "<table width=400 border=1>\n"; while ($get_info = mysql_fetch_row($result)){ print "<tr>\n"; foreach ($get_info as $field) print "\t<td><font face=arial size=1/>$field</font></td>\n"; print "</tr>\n"; } print "</table>\n"; ?> </body> </html> Here is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 If you also have any feedback on my code, please do tell me. I wish to improve my coding base. Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference. Code: Code: [Select] mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); Full code: Code: [Select] <?php include_once("includes/config.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $name = mysql_real_escape_string($_POST['name']); $server_type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1"); if(mysql_num_rows($check1) < 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> Here there,
This is a post for all of you ajax/php pimps who don't mind helping a beginner. I have been working on a little ajax script to filter a table called "recipes2" in a database called "recipes" using a dropdown menu. Once I get this script to work, I can run queries without having to refresh the page.
The problem, something is not right, the php script does not seem to be getting the posted data. even though the ajax script returns success when i check for it. What am I doing wrong?
Also please note that, if possible, I'd like all the code to stay on one page.
This is the SQL script if you want to create the database and table:
[attachment=3349:recipesDB.txt]
Here is the code, the file should be named 'filterRecipes.php'.
[attachment=3350:filterRecipes.txt]
hi ..
hit little snag Ive got html <form name="formx" method="POST" action="connect.php" class='ajaxform'> and about 4 varibles in that form .. now how to properly
query table users so i can find out which prefix correspond with that user
Insert into table "upis" in this case that prefix with 3 variables..
everything in single,.. in this case connect.php file
I know how to do QUERY and INSERT INTO but i don't know how to connect with 2 different tbl or db and do the query and input of data.. every time I put 2 of them together .. never works out..
this is example.. i dont know is this right way to do it .. I am retrieving a rowfrom a table and when I post the row variable it doesnt read it. ___ $query = "SELECT * FROM $tbl_name"; $result = mysql_query($query) or die(mysql_error()); while($row = mysql_fetch_array($result)){ echo $row['name']; echo "<br />"; $postinfo = 'p_doctor_name=' . $row .'&p_name_type=A&p_search_type=BEGINS'; __ This outputs p_entity_name=&p_name_type=A&p_search_type=BEGINS Note that it is missing $row Do I need to put it in an array? This topic has been moved to PHP Freelancing. http://www.phpfreaks.com/forums/index.php?topic=334867.0 Hi, I'm new to PHP/MySQL and need some help getting my query to work for my selection list: The selection list is built with: <form action='processformmissing.php' method='POST'> <fieldset> <legend>Choose Department</legend> <select name='depart'> <option value=''></option> <?php while ($row = mysqli_fetch_array($result)) { extract($row); echo "<option value='$department'>$department</option>\n"; } ?> </select> <p><input type='submit' value='Select Department' /></p> </fieldset> </form> The data is then sent to: $depart = $_POST['depart']; $deptlike = "%".$depart."%"; echo "<p>$depart</p>"; echo "<p>$deptlike</p>"; $query = "SELECT * FROM lifecerts INNER JOIN employees ON lifecerts.cid = employees.cid WHERE department LIKE '$deptlike' ORDER BY employees.name"; Hitting the submit button from my selection list form seems to be working fine because when I echo my data ($depart and $deptlike) it is giving me the correct value, but the query doesn't give me any results. However, if my post data comes from a text box instead of a selection list, my query works fine. Any thoughts on what I'm doing wrong??? Many thanks! I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas? I have a query which pulls all the data i need and it works fine. But when there is more than one topic in a forum it displays the forum each time. Say there are 4 topics in a forum. The forum name is displayed 4 times. Heres the code: $query = $db->query("SELECT f.forum_id, f.forum_name, f.forum_description, f.forum_topics, f.forum_posts, f.forum_last_poster, f.forum_last_post_time, f.forum_last_post, p.parent_id, p.parent_name, m.user_id, m.user_username, m.user_group, t.topic_id, t.topic_name, po.post_id, po.post_subject FROM ".DB_PREFIX."forums as f JOIN ".DB_PREFIX."parents as p ON f.parent_id = p.parent_id LEFT JOIN ".DB_PREFIX."members as m ON f.forum_last_poster = m.user_id LEFT JOIN ".DB_PREFIX."topics as t ON f.forum_id = t.forum_id LEFT JOIN ".DB_PREFIX."posts as po ON po.post_id = f.forum_last_post ORDER BY p.parent_id") or trigger_error("SQL", E_USER_ERROR); while ($forum_info = mysql_fetch_object($query)) { I am new to table joins but someone helped me set this up so it should work but it doesnt. Any ideas why? This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=321853.0 Hi all, Firstly here is my code Code: [Select] <?php if (isset($_POST['update'])) { include('../config.php'); $name = $_POST['name']; $email = $_POST['email']; $id = $_POST['id']; $uquery=mysql_query("UPDATE customers SET name='$name' AND email='$email' WHERE id='$id'"); if($uquery) { echo mysql_error(); } else { echo mysql_error(); } } ?> <link href="../styles/clientbox.css" rel="stylesheet" type="text/css"> <link href="cancelform.css" rel="stylesheet" type="text/css" /> <link href="../styles/form_dark.css" rel="stylesheet" type="text/css" /> <body><br> <h3>My Details</h3> <div class="text"> Please keep your details updated below, for security reasons you cannot change your password.<br> <?php $id = $_GET['id']; include('../config.php'); $query=mysql_query("SELECT * FROM customers WHERE id='$id'"); while($row = mysql_fetch_assoc($query)) { $name = $row['name']; $email = $row['email']; } ?> <form class="dark" action="" method="post"> <ol> <li> <fieldset> <legend>My Details</legend> <ol> <li> <label for="name">Account Holder</label> <input type="text" id="name" name="name" value="<?php echo $name;?>" /> </li> <li> <label for="email">Contact/Login Email Address</label> <input type="text" id="email" name="email" value="<?php echo $email;?>" /> <input type="hidden" id="id" name="id" value="<?php echo $id;?>" /> </li> </ol> </fieldset> </li> </ol> <p style="text-align:right;"> <input type="reset" value="CANCEL" /> <input type="submit" value="UPDATE" name="update" /> </p> </form> Basically what is happening is that whatever you enter into the name box returns a 0 if its text or 1 if numbers are entered as opposed to storing the value inputed, if the email address is changed then nothing happens at all. Can someone see if I have made a noob mistake somewhere? The name and email fields are both varchar(256) in my database Thanks Hi, i have the following problem. I use a script to enter data in a mysql database and for now i only have three variable... Quote <h2>Invoeren</h2> <form action="insert.php" method="post" class="cmxform"> <fieldset> <legend>Gegevens</legend> <table border="2"> <tr> <td>1. </td> <td><label for="merk">merk:</label></td> <td><input id="merk" name="merk" /></td> </tr> <tr> <td>2. </td> <td><label for="cpu">cpu:</label></td> <td><select name="cpu" id="cpu" /> <option value="">Maakt niet uit</option> <option value="amd">amd</option> <option value="intel">intel</option> </select></td> </tr> <tr> <td>3. </td> <td><label for="hdd">hdd:</label></td> <td><select name="hdd" id="hdd" /> <option value="">Maakt niet uit</option> <option value="160GB">160GB</option> <option value="320GB">320GB</option> <option value="500GB">500GB</option> </select></td> </tr> </table> <input type="submit" value="Verstuur" /> </fieldset> </form> Data goes in to the database correctly but now i use the following search script... Quote <h2>zoeken</h2> <form action="search.php" method="post" class="cmxform"> <fieldset> <legend>Gegevens</legend> <table border="2"> <tr> <td>1. </td> <td><label for="merk">merk:</label></td> <td><input id="merk" name="merk" /></td> </tr> <tr> <td>2. </td> <td><label for="cpu">cpu:</label></td> <td><select name="cpu" id="cpu" /> <option value="0">Maakt niet uit</option> <option value="amd">amd</option> <option value="intel">intel</option> </select></td> </tr> <tr> <td>3. </td> <td><label for="hdd">hdd:</label></td> <td><select name="hdd" id="hdd" /> <option value="0">Maakt niet uit</option> <option value="160GB">160GB</option> <option value="320GB">320GB</option> <option value="500GB">500GB</option> </select></td> </tr> </table> <input type="submit" value="Verstuur" /> </fieldset> </form> The values go correct to the search script because i checked this with an echo function. The search.php loks like this Quote <?php $con = mysql_connect("localhost","admin","t1tan1um"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("model", $con); $qry = "SELECT * FROM eigenschappen WHERE 1=1"; if (isset($_POST['merk']) && $_POST['merk'] != 0) { $qry .= " AND merk='" . mysql_real_escape_string($_POST['merk']). "'"; } if (isset($_POST['cpu']) && $_POST['cpu'] != 0) { $qry .= " AND cpu='" . mysql_real_escape_string($_POST['cpu']). "'"; } if (isset($_POST['hdd']) && $_POST['hdd'] != 0) { $qry .= " AND hdd='" . mysql_real_escape_string($_POST['hdd']). "'"; } $result = mysql_query($qry) or die(mysql_error()); // lelijke foutmelding, zelf mooier maken of $result door een if statement halen while($row = mysql_fetch_array($result)) { echo $row['merk'] . " " . $row['cpu'] . " " . $row['hdd']; echo "<br />"; } ?> When i search for hdd wich is last in de search list the query only returns the records wich contain that value. When the condition has to be for example Intel cpu AND 500GB it returns everyting with 500GB but also every kind of CPU. Anyone? Hi.. I think I have bad code in three different similar query's in my code. The first query is: $sql="SELECT * FROM invoice as d INNER JOIN members as c ON d.buyer=c.usernum order by " . $orderBy . " " . $order; $result = $con->query($sql); Actually in this one I didn't even know I had a problem other than it was slow working until I put this error trap in:
if (!$check1_res) { The other two pass $id from the previous script they a $sql=mysqli_query($con,"SELECT * FROM invoice_items as d inner JOIN items as c ON d.itemnum=c.itemnum where invnumber = '$id'"); $row = mysqli_fetch_array($sql); and $sql=mysqli_query($con,"SELECT * FROM invoice as d inner JOIN members as c ON d.buyer=c.usernum where invnumber = '$id'"); $row = mysqli_fetch_array($sql); To be honest I didn't know I had a problem with the first two until The third would not return the correct data. I only got partial or none of the invoice data I was expecting. But I realized all three have problems when I used the error trap which come back with this:
Notice: Undefined variable: check1_res in C:\Apache24\htdocs\choo\tc_invoice.php on line 37 The error I get is identical for the first query. At first I thought it was the way that I was passing the variable, but $id is valid when echoed just below the query. So I am hoping that this is going to be something I did wrong with the querys or possibly the fetch statement after the query is wrong??? Really appreciate any help you can give me. If you need to see more of the code on either script please let me know. original sql Code: [Select] -- -------------------------------------------------------- -- -- Table structure for table `countries` -- CREATE TABLE `countries` ( `id` int(6) NOT NULL auto_increment, `value` varchar(250) NOT NULL default '', PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=243 ; -- -- Dumping data for table `countries` -- INSERT INTO `countries` VALUES (1, 'Vancouver'); New Sql Code: [Select] -- -------------------------------------------------------- -- -- Table structure for table `countries` -- CREATE TABLE `countries` ( `id` int(6) NOT NULL auto_increment, `value` varchar(250) NOT NULL default '', `code` varchar(12) NOT NULL default '', PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=243 ; -- -- Dumping data for table `countries` -- INSERT INTO `countries` VALUES (1, 'Vancouver', 'BC-50'); PHP code for the job Code: [Select] <?php // PHP5 Implementation - uses MySQLi. // mysqli('localhost', 'yourUsername', 'yourPassword', 'yourDatabase'); $db = new mysqli('localhost', 'root' ,'password', 'weather'); if(!$db) { // Show error if we cannot connect. echo 'ERROR: Could not connect to the database.'; } else { // Is there a posted query string? if(isset($_POST['queryString'])) { $queryString = $db->real_escape_string($_POST['queryString']); // Is the string length greater than 0? if(strlen($queryString) >0) { // Run the query: We use LIKE '$queryString%' // The percentage sign is a wild-card, in my example of countries it works like this... // $queryString = 'Uni'; // Returned data = 'United States, United Kindom'; // YOU NEED TO ALTER THE QUERY TO MATCH YOUR DATABASE. // eg: SELECT yourColumnName FROM yourTable WHERE yourColumnName LIKE '$queryString%' LIMIT 10 $query = $db->query("SELECT your_column FROM your_db_table WHERE your_column LIKE '$queryString%' LIMIT 10"); if($query) { // While there are results loop through them - fetching an Object (i like PHP5 btw!). while ($result = $query ->fetch_object()) { // Format the results, im using <li> for the list, you can change it. // The onClick function fills the textbox with the result. // YOU MUST CHANGE: $result->value to $result->your_colum echo '<li onClick="fill(\''.$result->value.'\');">'.$result->value.'</li>'; } } else { echo 'ERROR: There was a problem with the query.'; } } else { // Dont do anything. } // There is a queryString. } else { echo 'There should be no direct access to this script!'; } } ?> What the original code does 1) you start typing your city (eg. Van) 2) when you type the first letter (eg. V), it looks into mysql and auto fills a dropdown menu with all possible cities What i need it to do 1) you start typing your city (eg. Van) 2) when you type the first letter (eg. V), it looks into mysql and auto fills a dropdown menu with all possible cities 3) when you find your city you click it or press enter, and it POST's the city code as well now how do i munipulate the script to do that... another thing, when i put the extra sql entry in "code", the auto fill stopped working, why? Thanks |