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While fetching multiple records you should use array try this [CODE] $result = mysql_query("CALL SelectContacts(@mobileNumber)"); while([B]$row[][/B] = mysql_fetch_assoc($result)) print_r($row); [/CODE]

Hi I want to remove the first <LI> item and last three <LI> item from <UL> list. How to write code in JQuery. I need the below list [CODE] <ul> <li>1</li> <li>2</li> <li>3</li> <li>4</li> <li>5</li> <li>6</li> <li>7</li> </ul> [/CODE] as below result [CODE] <ul> <li>1</li> <li>5</li> <li>6</li> <li>7</li> </ul> [/CODE] …

I came across the below link looked nice to me. [URL="http://www.smashingmagazine.com/2009/06/21/50-fresh-javascript-tools-that-will-improve-your-workflow/"]http://www.smashingmagazine.com/2009/06/21/50-fresh-javascript-tools-that-will-improve-your-workflow/[/URL]

it may be because of the url that you are passing. make sure your query string should be lowercase it will cause problem in Linux server and also check the condition inside your code is correct also if you redirecting to any where else.

try this [CODE]if(document.catsearch.submit()) { document.searchForm.submit(); }[/CODE]

try this [CODE] $page = isset($GET ['page']) : $GET ['page'] : ''; if($page) { $path="inc/".$page.".php"; if(isset($path)) { if (file_exists($path)) { include($path); } else { echo "Page doesnt exist."; } } [/CODE]

do you get any error? if yes what is that or try to use Firefox Firebug to find what error and where is it?

can you paste what type of error you are getting...

write this [CODE] <?php $server_name="localhost"; $db_user="sql_user"; $db_pass="password"; $db="mydb"; mysql_connect($server_name, $db_user, $db_pass); mysql_select_db($db); $sql = "SELECT city FROM posts"; $result = mysql_query($sql) or die ('Query Error: ' . mysql_error()); while ($results = mysql_fetch_array($result)) { $gender[] = $results; } ?> [/CODE] instead of [CODE]<?php $server_name="localhost"; $db_user="sql_user"; $db_pass="password"; $db="mydb"; $sql = "SELECT city …

better use JQuery. Write select box elements and use unique ID as well as value. and use [CODE] $("#IDValue").change(funtion(){ // here write code for open or close new form. }); [/CODE]

It's very simple. If you use well defined database table for each registering user. When they login you get it from table and display it on the page.

I have data value of a table column. I want to know in which database and in which table and in which column the data value appeared in mysql.

use JQuery to accomplish this. In your configuration menu assign ID on the red surface image or link On click of image link write [CODE]$('#frame_id').hide('slow');[/CODE]

Basically ajax will work in the background. I thing you might call ajax through java script file or JQuery. In this case I thing you are writing java script inside the server side script. So don't write a java script server side. Before you call an ajax file write a …

use ... select count(*) from TABLE_NAME

If you want get the [cart_id + 1], after you inserted your value in your stored procedure better you write [B]SELECT LAST_INSERT_ID() + 1;[/B]