[SOLVED] help with an if statement

[marksie1988]

i am trying to create an if statement which basically says

 

if ('order' in table1 == 'order' in table2){

 

echo data

 

}

else{

dont echo data

}

 

i dont have a clue how to compare 2 tables like this help appreciated or a point in the right direction

[teng84]

are you asking for SQL statement or what ? explain it better

[Asperon]

<?php

$query = "SELECT order1 FROM table1 WHERE something='condition' LIMIT 1"; //this is your mysql statement that will get your order1 info
$result = mysql_query($query) or die('Query Failed: '.mysql_error());

$order1 = mysql_result($result,0);

$query = "SELECT order2 FROM table2 WHERE something='condition' LIMIT 1"; //this is your mysql statement that will get your order2 info
$result = mysql_query($query) or die('Query Failed: '.mysql_error());

$order2 = mysql_result($result,0);

if($order1 == $order2){

echo 'We are the same';

}
else{

echo 'We are not the same';

}
?>

[marksie1988]

ok i intergrated the code but i get this error

 

Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 21 in /home/******/public_html/blacklime/track/index.blc on line 54

[marksie1988]

here is the code where the problem is if this helps

 

$id = $_POST['on'];
$query = "SELECT `on` FROM `track_website` WHERE `on`='".$id."' LIMIT 1"; //this is your mysql statement that will get your order1 info
$result = mysql_query($query) or die('Query Failed: '.mysql_error());

$order1 = mysql_result($result, 0);

$query = "SELECT `order` FROM `track_website_post` WHERE `order`='".$id."' LIMIT 1"; //this is your mysql statement that will get your order2 info
$result = mysql_query($query) or die('Query Failed: '.mysql_error());

$order2 = mysql_result($result, 0);

[marksie1988]

*BUMP*