if have checked this query several times but it aint passing the data to database..in other way every thing seems fine in this query kindly please sort out ths problem.
$qry="Insert into `postjob` (jobtitle,positions,category) VALUES('".$_POST['title']."','".$_POST['position']."','".$_POST['cate']."')"; printf mysql_query($qry) or die(mysql_error()); What error do you get ?
Parse error: syntax error, unexpected 'mysql_query' (T_STRING) in D:\xampp\htdocs\jobtest\post.php on line 10
Try replacing your code with the following. If it works, tell me and I'll explain why.
$qry = mysql_query('INSERT INTO `postjob` (jobtitle, positions, category) VALUES ("'.$_POST['title'].'", "'.$_POST['position'].'", "'.$_POST['cate'].'")'); if(!$qry) { die(mysql_error()); } 
The printf function expects a string passed to it as a parameter to output. i.e. printf('Hello Word') you don't need it, you can just run the query.
$qry="Insert into `postjob` (jobtitle,positions,category) VALUES('".$_POST['title']."','".$_POST['position']."','".$_POST['cate']."')"; mysql_query($qry) or die (mysql_error());
yehuda2001 its not working my bro..:(
Post all of your code, including the code for the method I suggested.
im gonna upload the files which i have actually created
This is post.php file
<?php include "db.php"; $action =isset($_POST['action']) ? trim ($_POST['action']):NULL; if($action=='postjob') { $qry="Insert into `postjob` (jobtitle,positions,category) VALUES('".$_POST['title']."','".$_POST['position']."','".$_POST['cate']."')"; mysql_query($qry) or die (mysql_error()); header("Location:jobpost.php"); } ?> this is db.php file
<?php $con = mysql_connect("localhost","root",""); if($con) { mysql_select_db("jobtest") or die (mysql_error()); } else { exit("Error connecting to server"); } ?> this my index.php file
<html> <title>My Test</title> <head> </head> <body> <form action="post.php" method="post"> Job title:<input type="text" name="title" value=""> <br> Positions: <input type="text" name="position" value=""> <br> <select name ="cate"> <option>Accounting</option> <option>Maths</option> <option>Banking</option> </select> <input type ="submit" name="" value="submit"> <input type ="hidden" name="action" value="postjop"> </form> </body> </html> Are you getting a different error now that you got rid of the printf?
GliderPilot bro i had written the printf to display the actual mysql error.if dnt use it i get the blank page.
some how others told me to echo the query to get a error.although ths query seems fine.
but creating the hidden error which is not being shown
You cannot echo a "mysql_query", it isn't a text string. To get the error, you need to do as I said:
$qry = mysql_query('SELECT id FROM tableName'); if(!$qry) { die(mysql_error()); } Check your server's error logs for the error that's occuring or turn on error reporting to E_ALL
hi,
what others meant by echo your query is something like this.
<?php $title = 'this title'; $position = 'this position'; $cate = 'this cate'; $qry="Insert into postjob (jobtitle,positions,category) VALUES('".$title."','".$position."','".$cate."')"; //printf mysql_query($qry) or die(mysql_error()); echo $qry; Isolate the codes where you are having problems with. On the codes above... disable the database connection first, and make sure to replace the $title, position, and cate with the actual values from the form on post.
try removing the quote on postjob first, before echoing your query. I just don't like how the single quoute is formatted. Sometimes they do create an error like that.
$qry="Insert into postjob (jobtitle,positions,category) VALUES('".$_POST['title']."','".$_POST['position']."','".$_POST['cate']."')"; mysql_query($qry) or die (mysql_error()); Just noticed something. Change this line:
$action =isset($_POST['action']) ? trim ($_POST['action']):NULL;
To this:
$action = (isset($_POST['action']) ? trim($_POST['action']) : NULL);
The malformed if statement was causing your mysql query to never run
Query is working by this method but not by above mentioned method
<?php
include "db.php";
$title =isset($_POST['title']) ? trim ($_POST['title']):NULL;
$position =isset($_POST['position']) ? trim ($_POST['position']):NULL;
$cate =isset($_POST['cate']) ? trim ($_POST['cate']):NULL;
$qry="Insert into postjob (jobtitle,positions,category) VALUES('".$title."','".$position."','".$cate."')"; mysql_query($qry) or die (mysql_error()); echo $qry; ?>
What get's outputted when you echo $qry?
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.
