Need to pass value of php in jquery script

Question

I need to pass the PHP retrieved value from database to Ajax function for further querying in database my code is given below.

The PHP Script:

$query = mysql_query("SELECT * FROM employee") or die(mysql_error());

while($res=mysql_fetch_assoc($query)){  
echo $res['emp_id'];   
$emp_id=$res['emp_id']

and submit button with id name-submit

The Javascript

$('input#name-submit').on('click',function(){ 
var empid = $('php echo $emp_id; ');
alert(empid);

the script is running just below the php script.

you never opened a PHP tag, so how would your server know it's supposed to do some php'ing? If you look at your source code, you'll notice it's just printing the letters you wrote. — d-_-b
– d-_-b, Commented Sep 17, 2012 at 20:28
Where do all of you new guys come here from using mysql_query? I wanna send a message to whoever is teaching you this to update his instructions and teach you to use PDO instead. — xception
– xception, Commented Sep 17, 2012 at 20:29
@xception Then send a message to internet. Google mysql php tutorial and you'll be surprised. — Tchoupi
– Tchoupi, Commented Sep 17, 2012 at 20:31
done bt the in alret box im having [object object] which means it is in array now how cld i xtract the value from it :X help plz — fahad
– fahad, Commented Sep 17, 2012 at 20:49

Josh Mein · Accepted Answer · 2012-09-17 20:28:54Z

4

Exactly as you are doing, but by putting the full opening tag in there. Change the second JS line to:

var empid = $('<?php echo $emp_id; ?>');

edited Sep 17, 2012 at 20:28

Josh Mein

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answered Sep 17, 2012 at 20:27

slugonamission

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1 Comment

fahad Over a year ago

josh acctually iwasnt ablt to print the php tag here thats y and thanx thing is done but thats was not the problem the main issue i was facing was using old version of jquery library nyway thanx but got new issue in alret box im having undefined

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