Could you help here is the query cannot figure out where the problem is
Code:
if(isset($_GET['n'])){
$s=$_GET['n'];
$query="SELECT FROM student WHERE name LIKE '$s%' ";
$result=mysql_query($query,$db) or die (mysql_error()
Mary Meer
Last Activity: Aug 12 '11, 06:43 AM
Joined: Feb 27 '11
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Error in SQL Syntax
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 'FROM student WHERE name LIKE 'a%'' at line 1
Could you help here is the query cannot figure out where the problem is
...Last edited by Dormilich; Jun 9 '11, 02:48 PM. Reason: please use [CODE] [/CODE] tags when posting code -
that is a headache, i did copy all code and see nothing wrong.
i will go through once again -
that is gr8,but if i click on first img several times then it goes on and on till last img,it should show only one img by one click and then stop.
i have all tags closed still the same problem
thanks for helpLeave a comment:
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i can get only the default img the other two does not show upLeave a comment:
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How to change images on click event?
Hello Every one,
i have three small images while clicking on each one the image of background should be changed. I have used javascript Dom along div, for the first two images it works but i don't know how to get right the third one.
Could anyone help?
...Code:<style type="text/css"> #div1 { width:100px; height:100px; background-image:url(../photo/cake.jpg); }Last edited by Niheel; Feb 27 '11, 02:37 PM.
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