Need help with Syntax

Richardknox · #1 23-Nov-2007, 19:34

I am uncertain what I am doing wrong with this.

The 1st set of code does not work, but the second set of code works fine, the only difference is the 'where `ip` = xxxxxxx' in both pieces of code.

PHP Code:


// 1st set of code
$ip = '71.206.122.144';

$query = 'SELECT * FROM `wcvisitor_ip` WHERE `ip` = \'$ip\'';
$result = mysql_query($query);
$row = mysql_fetch_array($result);


// 2nd set of code
$query = 'SELECT * FROM `wcvisitor_ip` WHERE `ip` = \'71.206.122.144\''; 
$result = mysql_query($query);
$row = mysql_fetch_array($result);

Please show me what I am doing wrong, and what to do to fix this.

Thanks,
Richard

filth · #2 26-Nov-2007, 08:04

in the first piece of code $ip is within single quotes therefore it is a part of the string. To fix this you can do either of the following:-

PHP Code:


$query = 'SELECT * FROM `wcvisitor_ip` WHERE `ip` = \''.$ip.'\'';

OR

PHP Code:


$query = "SELECT * FROM `wcvisitor_ip` WHERE `ip` = \'$ip\'";

In the first instance we are using something called concatenation. Ie. we have come out of the single quotes and used a . to show that we are still assigning data to the variable.

In the 2nd example we use double quotes ("). Variables within double quotes are evaluated and output their content to the string.

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Need help with Syntax

Re: Need help with Syntax