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?
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