bob2006 Posted January 13, 2011 Share Posted January 13, 2011 now i am trying to get this to work it should change the variable if the statement is ture and it not in that and any one help include'db.php'; $sql2=mysql_query("SELECT * FROM site_config WHERE id=1")or die(mysql_error()); while($row2=mysql_fetch_array($sql2)) { $paypal_email=$row2['paypal_email']; $price=$row2['price']; $sandbox=$row2['sandbox']; $account_number=$row2['account_number']; $quantity_1=$row2['quantity_1']; $product=$row2['product']; $return_url=$row2['return_url']; $demo=$row2['demo']; if($sandbox==1) { $link='http://www.sambob.info'; if(empty($sandbox)) { $link='http://www.bob.tdsnet.org'; } echo"$link<br>$sandbox"; } } Link to comment https://forums.phpfreaks.com/topic/224373-if-statement-not-working/ Share on other sites More sharing options...
Pikachu2000 Posted January 13, 2011 Share Posted January 13, 2011 Misplaced curly braces? Currently, your check to see if $sandbox is empty is inside the if($sandbox == 1) { conditional. Obviously, if $sandbox == 1, it can never be empty. Link to comment https://forums.phpfreaks.com/topic/224373-if-statement-not-working/#findComment-1159124 Share on other sites More sharing options...
bob2006 Posted January 14, 2011 Author Share Posted January 14, 2011 the 1 come form a checkbox when i uncheck it the databases has noting in it Link to comment https://forums.phpfreaks.com/topic/224373-if-statement-not-working/#findComment-1159130 Share on other sites More sharing options...
Pikachu2000 Posted January 14, 2011 Share Posted January 14, 2011 Then if $sandbox doesn't have a value of 1, the conditional that checks if $sandbox is empty is never reached, leaving $link most likely undefined. Indent your code properly and it should be fairly obvious what's happening with it. Link to comment https://forums.phpfreaks.com/topic/224373-if-statement-not-working/#findComment-1159135 Share on other sites More sharing options...
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