wizzle

i already fix the problem,... thank you guys..

i already put a space., but still same error and problem occur.. :'( :'(

switch($select) { case 1: $result=mysql_query("SELECT * FROM tb_addressbook WHERE lastname LIKE'%$text%'"); break; case 2: $result=mysql_query("SELECT * FROM tb_addressbook WHERE firstname='$text'"); break; case 3: $result=mysql_query("SELECT * FROM tb_addressbook WHERE address='$text'"); break; } if (!$result=) { echo "Could not successfully run query ($result) from database" . mysql_error($con); } if (mysql_num_rows($result) == 0) //(im having a problem in this line. { echo "<br>No rows found, Please try again! <br> <a href='searchform.php'>search</a>"; echo "<br><br>"; exit; }

here' the function for mysql_error(): mysql_error($con) $con=mysql_connect("localhost",$username,$password); if(!$con) die("couldn't connect to MySQL: ".mysql_error()); mysql_select_db($db_name,$con) or die("couldn't connect to $db_name: ".mysql_error());

hi! i badly need your help.. i'm having an error in my addressbook system.. Here's the error: mysql_num_rows() expects parameter 1 to be resource, null given in /Applications/XAMPP/xamppfiles/htdocs/sampol3/design/edit.php on line 74. please give the right code.. thank you.. 17348_.php

pls help me.. i cant fix the problem Could not successfully run query () from database Warning: mysql_num_rows() expects parameter 1 to be resource, null given in on line 78 17293_.php 17294_.php