PHP - Insert Into Not Working With Two Variables
I have no idea and there is no reason why this should not be working.
im simply trying to add three variables into a database, and only one works. the other two do not work for any reason i can find. can someone point out my error, if any? code: <?php $date = date("Y-m-d"); $dbc = mysqli_connect('localhost', 'root', '', 'timer') or die('Error connecting to DB'); $query = @"INSERT INTO sessions (date, user, sessiontime) VALUES ('$date', '$user', '$sessiontime')"; $user = @$_GET['user']; $sessiontime = @$_GET['clock']; if (@$_GET['addDB'] == "Session Complete") { mysqli_query($dbc, $query) or die( '<br>Query string: ' . $query . '<br>Produced error: ' . mysqli_error($dbc) ); } ?> Form: Code: [Select] <label for="user"><b><em>Your name: </b></em></label><br /><input type="text" name="user" value="Admin/User" /> <input id="clock" name="clock" type="text" value="00:00:0" readonly><br> <input id="startstopbutton" type="button" value="S t a r t" onClick="startstop();" style="font-weight:bold"><br> <input type="submit" name="addDB" value="Session Complete" /> See, all the variables match up!? I dont get what im doing wrong? Similar TutorialsI am new to PHP and am currently learning the ropes. I have writen some code to insert a line into a mySQL database. I have created 3 fields in the mySQL database and am passing values two them. I can use a declared variable to pass information to the filed ID (Index filed in mySQL) but if I use a variable to pass purely text values the database is not updated. mysql_query( "INSERT INTO $tbl_name (category_Name, ID, test) VALUES ($category, $internalId, $category2)" ); If I replace the variables with specific text, the rows are added successfully. mysql_query( "INSERT INTO $tbl_name (category_Name, ID, test) VALUES ('werwerwer', $internalId, 'werrr')" ); It must be something stupid, I am sure. Full code below ob_start(); $host="localhost:8888"; // Host name $username=""; // Mysql username $password=""; // Mysql password $db_name="expenses"; // Database name $tbl_name="expense_category"; // Table name $category="test3"; $internalId="7"; $category2="test3"; // Connect to server and select databse. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); mysql_query( "INSERT INTO $tbl_name (category_Name, ID, test) VALUES ('werwerwer', $internalId, 'werrr')" ); echo "Done now 6!"; ?> I am trying to modify a header redirect similar to this: header("Location: index.html"); exit; I would like to do something like this: header("Location: index.html?user=' . $user . '"); exit; Ofcourse, this does not work. Any hints? Hello. I have a simple enough code that takes information from one table and drops it into another. This is great, but I have 2a new complexities that I have been unable to code correctly. A. 'lastname', 'firstname' on table1 need to be combined into 'name' How & Where do I combine these strings and then pass them? Code: [Select] <?php include('dbconfig.php'); // Make a MySQL Connection mysql_connect("localhost", "$user", "$password") or die(mysql_error()); mysql_select_db("$database") or die(mysql_error()); $result = mysql_query( "INSERT INTO table2 (lastname, firstname, email) SELECT lastname, firstname, email FROM table1 WHERE email='someemail@gmail.com' ") or die(mysql_error()); ?> Hello! How do I insert the content of two different variables into the SAME field in a table? Let's say, that variable A contains the word "blue" and variable B contains the word "sky". How do I "merge" these to variables into only one variable containing the words "Blue Sky"? Best regards Morris 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 hello, i am returning the error in the following code. im i doing something wrong? Code: [Select] { $sql = "UPDATE jobs SET contact = ' . mysql_real_escape_string{$_POST['contact']} .', contactphone = ' . mysql_real_escape_string{$_POST['contactphone']} .', customer = ' . mysql_real_escape_string{$_POST['customer']} .', initial = ' . mysql_real_escape_string{$_POST['initial']} .', address = ' . mysql_real_escape_string{$_POST['address']} .', city = ' . mysql_real_escape_string{$_POST['city']} .', postal = ' . mysql_real_escape_string{$_POST['postal']} .', province = ' . mysql_real_escape_string{$_POST['province']} .', description = ' . mysql_real_escape_string{$_POST['description']} .', mechanic = ' . mysql_real_escape_string{$_POST['mechanic']} .', ponumber = ' . mysql_real_escape_string{$_POST['ponumber']} .', status = ' . mysql_real_escape_string{$_POST['status']} .' WHERE id = '$id"; $result = mysql_query($sql) or die('Error, updating job failed. Check you fields and try again.'); echo "<center> You have successfully updated the new job. You can see your changes below."; } EvilSpider 123456 FakeEmail@yahoo.com Connected successfullyYou have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@yahoo.com)' at line 2 mysql_query("INSERT INTO user (username, password,email) VALUES($user, $pass, $email) ") and when i dont do the @ sign i still get the error EvilSpider 123456 FakeEmail Connected successfully Unknown column 'EvilSpider' in 'field list' I am trying to do an INSERT on my "bio_answer" table which is a junction table in between the "member" and "bio_question" tables. My PHP just tries to do the INSERT into "bio_answer" and doesn't touch the parent tables, which I assume is okay?! Here is a snippet of my code... if ($_SERVER['REQUEST_METHOD']=='POST'){ // Form was Submitted (Post). // Initialize Errors Array. $errors = array(); // Trim all form data. $trimmed = array_map('trim', $_POST); // ************************ // Validate Form Data. * // ************************ // Validate Answer1. if (strlen($trimmed['answer01']) >= 2 && strlen($trimmed['answer01']) <= 1024){ // Valid Answer1. $answerArray[0] = $trimmed['answer01']; $questionID = 1; echo '<p>$memberID = ' . $memberID . '</p>'; // Resolves to 19 which exists in the "member" table echo '<p>$questionID = ' . $questionID . '</p>'; // Resolves to 1 which exists in the "bio_question" table echo '<p>$answerArray[0] = ' . $answerArray[0] . '</p>'; // Resolves to whatever I type in my form, e.g. "This is a test..." }else{ // Invalid Answer1. $errors['question01'] = 'Answer must be 2-1024 characters.'; }//End of VALIDATE ANSWER1 // ****************************** // Attempt to Create Thoughts. * // ****************************** if (empty($errors)){ // Valid form data. // Build query. $q1 = "INSERT INTO bio_answer(member_id, question_id, response, created_on) VALUES(?, ?, ?, NOW())"; // Prepare statement. $stmt1 = mysqli_prepare($dbc, $q1); // Bind variables to query. mysqli_stmt_bind_param($stmt1, 'iis', $memberID, $questionID, $answerArray[0]); // Execute query. mysqli_stmt_execute($stmt1); // Verify Insert. if (mysqli_stmt_affected_rows($stmt1)==1){ // Insert Succeeded. $_SESSION['resultsCode'] = 'THOUGHTS_NEW_THOUGHTS_CREATED_2138'; }else{ // Insert Failed. $_SESSION['resultsCode'] = 'THOUGHTS_NEW_THOUGHTS_FAILED_2139'; }//End of UPDATE MEMBER RECORD // Close prepared statement. mysqli_stmt_close($stmt1); // Set Error Source. $_SESSION['errorPage'] = $_SERVER['SCRIPT_NAME']; // Redirect to Display Outcome. header("Location: " . BASE_URL . "/members/results.php"); // End script. exit(); }//End of ATTEMPT TO CREATE THOUGHTS My script keeps failing and errors to 'THOUGHTS_NEW_THOUGHTS_FAILED_2139' What is wrong with my Script/SQL?? Thanks, Debbie Hello PHP freaks. Having a few issues with my PHP scripts, specifically the $query = sprintf and mysql_real_escape_string functions. Kind of new to this, so if you do reply, explain it to me like I am a complete moron... Oh, before I forget. My specific problem is that I can click the "Post" button and follow the header to "Location: view.php", but the actual text in the Subject and Message fields is not being sent to the database. Finally managed to get rid of all the error messages I was getting, and now I get this... Thanks in advance! Here is my script for the entire page: _________________________________________________ ____________ <?php require_once('auth.php'); ?> <?php mysql_connect('xxxxxx', 'xxxxxx', 'xxxxxx'); $subject = $_POST['subject']; $message_text = $_POST['message_text']; // add entry to the databse if the form was submitted and // the necessary information was supplied in the form if (isset($_POST['submitted']) && $subject && $message_text) { $query = sprintf('INSERT INTO FORUM_MESSAGE (SUBJECT, MSG_TEXT) VALUES ($subject, $message_text)', Ryan_iframe, mysql_real_escape_string($subject), mysql_real_escape_string($message_text)); mysql_query($query); // redirect user to list of forums after new record has been stored header('Location: view.php'); } // form was submitted but not all the information was correctly filled in else if (isset($_POST['submitted'])) { $message = '<p>Not all information was provided. Please correct ' . 'and resubmit.</p>'; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Member Index</title> <link href="loginmodule.css" rel="stylesheet" type="text/css" /> </head> <body> <h1>Welcome <?php echo $_SESSION['SESS_FIRST_NAME'];?></h1> Home | <a href="member-profile.php">My Profile</a> | <a href="logout.php">Logout</a> <br /><br /> <form method="post"> <div> <label for="subject">Subject: </label> <input type="text" id="subject" name="subject" value="<?php echo htmlspecialchars($subject); ?>" /><br /> <label for="message_text">Message: </label> <input type="text" id="message_text" name="message_text" value="<?php echo htmlspecialchars($message_text); ?>" /><br /> <input type="hidden" name="submitted" value="true" /> <input type="submit" value="Post" /> </div> </form> </body> </html> I have a mysql insert statement generated with php that is not populating the table. I've echoed the statement and if I copy and paste into phpmyadmin it works fine. The result of the mysql_query function is true. I've emptied the table so there are no primary key conflicts. I've put the statement in a try catch and it does not display a exception. What else can I try? Here's the statement INSERT INTO `wp_term_relationships` (object_id, term_taxonomy_id, term_order) VALUES (1597,83,0) Works absolute fine if I copy and paste into phpmyadmin. Does not populated the table if run through mysql_query Basically I need to input data into two tables. I am running 3 different query's but only 2 of them work. The other one doesn't. None working query: mysql_query("INSERT INTO users(username, password, email, pin, key) VALUES('$username', '$password', '$email', '$key', '$pin')"); Working querys: mysql_query("DELETE FROM beta_keys WHERE keys_new='$key'"); **and** mysql_query("INSERT INTO beta_keys(keys_used) VALUES('$key')"); So any ideas why the top one doesn't work but the bottom two do? I have a class built for an INSERT query but it is passing two sets of records into the database rather than one. Code: [Select] class DatabaseInsert { function DatabaseConnectionRequire() { include("../scrips/php/database.connection.class.php"); include("../scrips/php/database.settings.php"); include("../scrips/php/database.connection.class.invoke.php"); } function ArticleInsert($values,$fields,$table) { $values_imploded = implode(" ",$values); $fields_imploded = implode(" ",$fields); $i = "INSERT INTO $table ($fields_imploded) VALUES ($values_imploded)"; mysql_query($i) or die(mysql_error()); if (!mysql_query($i)) { echo "Sorry, something whent wrong there..."; } else { echo "<strong><p style='color:green;'>Content added sucessfully!!!</p></strong>"; } } } Hello all, so I created an insert function and it seems no matter what I try that it won't add values using the query function inside a table from the respective variables, I would like to know why is this happening? Here is the code can you tell me why it doesn't insert anything in the database? It shows no errors when it runs but then again when I check the tables they're empty!
function insert(){ $user = $_POST['user']; $pass = md5($_POST['pass']); $priv = "User"; $mail = $_POST['mail']; $avatar = $_FILES['avatar']['name']; $date="now()"; $submit = $_POST['submit']; $query = "INSERT INTO user(user,pass,priv,mail,avatar,date) VALUES(`$user`,`$pass`,`$priv`,`$mail`,`$avatar`,`$date`);"; if($submit){ $res = mysqli_query($con,$query) or die(mysqli_error($con)); } } <?php error_reporting(E_ALL^E_NOTICE); $connect = mysqli_connect("");//removed $doc = $_GET["doctor"]; $username = $_GET["username"]; $sql = "SELECT fname, lname from newpatient where username = '$username'"; $result = mysqli_query($connect, $sql); $value = mysqli_fetch_row($result); $fname = $value[0]; $lname = $value[1]; $totalcost = $_GET["totalcost"]; $reason1 = $_GET["reason1"]; $reason2 = $_GET["reason2"]; $reason3 = $_GET["reason3"]; $reason4 = $_GET["reason4"]; $reason5 = $_GET["reason5"]; $reason6 = $_GET["reason6"]; $reason7 = $_GET["reason7"]; $reason8 = $_GET["reason8"]; $date = $_GET["date"]; $reasons = array($reason1,$reason2,$reason3,$reason4,$reason5,$reason6,$reason7,$reason8); rsort($reasons); $reason1 = $reasons[0]; $reason2 = $reasons[1]; $reason3 = $reasons[2]; $reason4 = $reasons[3]; if(isset($_REQUEST["yes"])) { $sql1 = "SELECT * FROM appointments where doctor_name = '$doc' and time = '$time'"; $result1 = mysqli_query($connect, $sql1); $num_rows = mysqli_num_rows($result1); if($num_rows > 0) { echo "Appointment Time already chosen. Select another time."; echo "<script language = 'javascript'>document.location.href='make_appointment.php?doc=$doc&username=$username'</script>"; } else { $sql2 = "INSERT INTO appointments (username, time, doctor_name, cost, reason1_for_visit, reason2_for_visit,reason3_for_visit,reason4_for_visit, fname, lname) values ('$username','$date','$doc',$totalcost,'$reason1','$reason2','$reason3','$reason4','$fname','$lname')"; $result2 = mysqli_query($connect, $sql2); if($result2) echo "This worked."; else echo "Insert did not work."; //echo "<script language = 'javascript'>document.location.href='registered_login_page.php?username=$username'</script>"; } } mysqli_close($connect); ?> Can anyone help me out with this. I have been struggling on and off it for weeks. Heres the description of the problem. I insert into a MySQL database from a query that is performed using an array that is posted from a page. The incoming array contains selected users (unique id's) that are used to lookup the users telephone numbers in a query and then insert into a DB. This code works intermittently. It will insert the records fine and then on the next attempt it may not? There is not pattern to it working and failing. Does anyone have any idea why this is happening? I have tried a loop using the posted array data count and also a loop based on the count of records brought back from the query but to no avail, it is still intermittently working. The '$stripped_message' data is just a text string. Code: [Select] <?php $date = date('Y-m-d H:i:s'); for ($i=0;$i<count($_POST['recipient']);$i++) // loop based on count of selected users $uk_mob_number = "44".substr($row_selected_recipients['User_mobile'], 1); // take off the first digit and replace with 44 $unique = makeRandomstring(); // Create a unique string for each entry $values.="('$unique','$date','$stripped_message','$uk_mob_number','sent','$date','$userid'),"; // usleep(50000); // Tried a delay to try to fix, did not work } // End of loop $values=substr($values,0,-1); // to remove last comma $query="INSERT INTO `sms` (`sms_unique`, `sms_sent`, `sms_body`, `sms_to`, `sms_status`, `sms_db_entry_time`, `sms_user`) VALUES $values".";"; ?> my SQL Query wont execute on on following lines: Code: [Select] $result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')") or die (mysql_error()); i get the following error: 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 ''gallery' ('image', 'memberid', 'caption') VALUES ('gallery/9074849_1.jpg', '1',' at line 1 here is my full code: Code: [Select] <?php require_once('connect.php'); $rand = mt_rand(1,9999999); $rand2 = mt_rand(1,9999999); $member_id = $_SESSION['SESS_MEMBER_ID']; $caption = $_POST["caption"]; if(isset($_FILES['uploaded']['name'])) { $allowed_filetypes = array('.jpg','.gif','.bmp','.png','.jpeg'); $max_filesize = 524288; // Maximum filesize in BYTES (currently 0.5MB) $fileName = basename($_FILES['uploaded']['name']); $errors = array(); $target = "gallery/"; $fileBaseName = substr($fileName, 0, strripos($fileName, '.')); // Get the extension from the filename. $ext = substr($fileName, strpos($fileName,'.'), strlen($fileName)-1); //$newFileName = md5($fileBaseName) . $ext; $newFileName = $target . $rand . "_" . $member_id.$ext; // Check if filename already exists if(file_exists("gallery/" . $newFileName)) { $errors[] = "The file you attempted to upload already exists, please try again."; } // Check if the filetype is allowed. if(!in_array($ext,$allowed_filetypes)) { $errors[] = "The file you attempted to upload is not allowed."; } // Now check the filesize. if(!filesize($_FILES['uploaded']['tmp_name']) > $max_filesize) { $errors[] = "The file you attempted to upload is too large."; } // Check if we can upload to the specified path. if(!is_writable($target)) { $errors[] = "You cannot upload to the specified directory, please CHMOD it to 777."; } //Here we check that no validation errors have occured. if(count($errors)==0) { //Try to upload it. if(!move_uploaded_file($_FILES['uploaded']['tmp_name'], $newFileName)) { $errors[] = "Sorry, there was a problem uploading your file."; } } //Lets INSERT database information here //Here we check that no validation errors have occured. if(count($errors)==0) { $result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')") or die (mysql_error()); { $errors[] = "SQL Error."; } } //If no errors show confirmation message if(count($errors)==0) { echo "<div class='notification success png_bg'> <a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a> <div> The file {$newFileName} has been uploaded<br>\n </div> </div>"; //echo "The file {$fileName} has been uploaded"; echo "<br>\n"; echo "<a href='gallery.php'>Go Back</a>\n"; } else { //show error message echo "<div class='notification attention png_bg'> <a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a> <div> Sorry your file was not uploaded due to the following errors:<br>\n </div> </div>"; //echo "Sorry your file was not uploaded due to the following errors:<br>\n"; echo "<ul>\n"; foreach($errors as $error) { echo "<li>{$error}</li>\n"; } echo "</ul>\n"; echo "<br>\n"; echo "<a href='gallery.php'>Go Back</a>\n"; } } else { //Show the form echo "Use the following form below to add a new image to your gallery;<br />\n"; echo "<form enctype='multipart/form-data' action='' method='POST'>\n"; echo "Please choose a file: <input name='uploaded' type='file' /><br />\n"; echo "Caption: <input name='caption' type='text' /><br />\n"; echo "<input type='submit' value='Upload' />\n"; echo "</form>\n"; //Echo Tests! echo "<br /><br />Random FileName: "; echo $rand; echo "<br />"; echo "member ID: #"; echo $member_id; } ?> any help appreciated. its prob something simple. my table has the following fields: "gallery" id (primary Key, AUTO_INC) memberid (fetched from session) image (will store image name including extension) caption (from "caption" text field in form) Hello All, In below script, I'm using another script called "SimpleImage.php" to upload and resize photos for a news site. The problem comes with renaming the photos. I need to name the photo based off of the story id. I have everything right to get the story id, but saving it as the variable is what's confusing me. Here's what I have: <?php include ("newsconnection.php");?> <?php $picturename = $_POST['newstoryid']; include('SimpleImage.php'); $image = new SimpleImage(); $image->load($_FILES['uploaded_image']['tmp_name']); $image->resizeToWidth(300); $image->save('$picturename'); ?> It's the $image->save('$picturename'); line that's messing me up. Any help would be great!! Hello everybody, I can't seem to figure out why this insert code isn't working. I'm trying to create a database of US zip codes. I created this user interface (form) with nothing but a submit button to execute the insert query <div id="right_content" class=""> <h3> Insert Zips </h3> <form action = "insertzip1.php" method = "post"> <input type = "submit" name = "submit" value = "submit"/> </form> </div> <!--closes right content--> Well here is the insert query which is supposed to accomplish the task. I have just included a tiny subsets of all the zipcodes (the insertzip1.php page which is the value of the action attribute of the form). <php? if (isset($_POST['submit'])) { require ('config.php'); $query = "INSERT INTO zips (zip, lat, lon, city, state, county, z_type, xaxis, yaxis, zaxis, z_primary, worldregion, country, locationtext, location, population, housingunits, income, landarea, waterarea, decommisioned, militaryrestrictioncodes, decommisionedplace) VALUES ('00501', 40.81, -73.04, 'HOLTSVILLE', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'Yes', 'NA', 'US', 'Holtsville, NY', 'NA-US-NY-HOLTSVILLE', '', 0, 0, '', '', 'No', '', ''), ('00501', 40.81, -73.04, 'I R S SERVICE CENTER', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'No', 'NA', 'US', 'I R S Service Center, NY', 'NA-US-NY-I R S SERVICE CENTER', '', 0, 0, '', '', 'No', '', ''), ('00544', 40.81, -73.04, 'HOLTSVILLE', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'Yes', 'NA', 'US', 'Holtsville, NY', 'NA-US-NY-HOLTSVILLE', '', 0, 0, '', '', 'No', '', ''), ('00544', 40.81, -73.04, 'IRS SERVICE CENTER', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'No', 'NA', 'US', 'Irs Service Center, NY', 'NA-US-NY-IRS SERVICE CENTER', '', 0, 0, '', '', 'No', '', '') "; $result = mysql_query($query); header("Location: insertzipsuccess.php"); }else{ die ("Could not insert data because" . mysql_error());} ?> The insertzipsuccess.php page is simply a page that prints out a success message if the query is successfully executed. Well when I hit the submit button, I just get redirected to a blank insertzip1.php page Can anyone show me what I'm not doing right here? PS I already created the table with fields that correspond to all the fields I'm trying insert. Sorry got it! my apologizes Hello, guys. I am experiencing some problems with an INSERT statement in this page. It simply won't write to the database! I added echo at the bottom to check my variables and they print the values just fine. I checked the database, table and datafield names and everything is correct, plus I don't have any issues with the other 25 tables of my database. I'm using XAMPP btw... Any help would be appreciated! Code: [Select] <!DOCTYPE html> <html> <head> <meta content="text/html; charset=utf-8" http-equiv="content-type"> <title>Doctors</title> <link rel="stylesheet" href="style.css" media="screen" /> </head> <body > <?php session_start(); $inc_code=$_SESSION['incident']; $doc_code=$_REQUEST['doctor_code']; $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } $mdb = "registry_db"; mysql_select_db($mdb, $con); mysql_query("SET NAMES 'utf8'", $con); ?> <div id="myform"> <p> <h2>Doctor in charge</h2> </p> <?php $sql="INSERT INTO doctors_per_incident(Incident_code, doctor_code) VALUES ('$inc_code', '$doc_code')"; echo "1 record added"." ".$inc_code." ".$doc_code; mysql_close($con); ?> </div> </body> </html> |