when am using mysqlfetcharray() that display the following error. the code look like this..
<?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("samp", $con); $Name=$_POST['unames']; $data=mysql_query("SELECT * FROM tbl_sample where Name='$Name'"); /*$info=mysql_fetch_array($data);*/ while($info=mysql_fetch_array($data)) { echo"<b>NAME</b>:".$info['Name']."<br><b>KEYSKILLS</b>:".$info['Keyskills']."<br><b>EXPERIENCE</b>:".$info['Experience']."<br><b>CURRENT ADDRESS</b>".$info['CurrentAddress']."<br><b>PERMANANT ADDRESS</b>".$info['PermanantAddress']."<br><b>COMPANY ADDRESS</b>".$info['CompanyAddress']."<br><b>MOBILE NO</b>".$info['MobileNo']."<br><b>QUALIFICATION</b>".$info['Qualification']."<br><b>CURRENT LOCATION</b>".$info['CurrentLocation']."<br><b>STATE</b>".$info['State']."<br><b>NATIONALITY</b>".$info['Nationality']."<br><b>DATE OF BIRTH</b>".$info['DOB']; } mysql_close($con); ?>
this shows the error like this....
Warning: mysqlfetcharray(): supplied argument is not a valid MySQL result resource in C:\wamp\www\sample\related.php on line 15
plz give suggestion..
Thanks in advance
Ads