Wickedself

PHP Syntax (Toggle Plain Text) $result = mysql_query("SELECT COUNT(*) FROM TEST Where Name LIKE 'Smith%'") or die(mysql_error()); // Print out result while($row = mysql_fetch_array($result)){ echo "There are ". $row['COUNT(Surname)'] ."; <-- This is the problem I guess } There should be another double quote right.

Hi. Huma. We need the source of only one of you pages.

Apple is trying to avoid 3rd party products. They want to be totally independent but in todays world if you don't collaborate with other companies you won't get anywhere.

Declare a character. Then assign integer values(ASCII values) to it. I am not sure about the exact ascii of the boxes but thats how it's done

You are getting an error in the else line because you are trying to use VB type if else in PHP. But PHP is originally based on C language so that if else will surely return an error to learn more about the if else statement in PHP Follow this …

Most probably they both have the same id or class names. and they might even contain the same variable names which is another issue

I am new at Java. I have used Microsoft's Visual J++ but I have heard that it doesnot follow the java standard. Can you please give me the name of a good standard compiler that will work fine in windows XP.