jdock1 Posted April 22, 2011 Share Posted April 22, 2011 I am using an if statement. If the user has a credit value of 1 or more it echos a variable thats a form. When the user submits the form, it takes a credit away, so the credit value is -1. I need the form to dissapear, because the form has a button that will execute a query that I dont want to execute. So, usually unset() fixes this, but its not in my case. Heres the code $a='<strong></strong>'; if ($hintcredit<1) $a.='<font style="color: #cc0000;">You have already requested a hint! Once you guess the correct number, you will be able to get another hint. </font>'; else $form= "<form action='blablabla' method='post'>blablablabla</form>"; $a.=$form; $a.='<strong></strong>'; echo $a; if ($_POST['request'] == "Retrieve Hint") { unset($form); $form= '<b>One of the digits in your number is '.substr($num, -1).'</b>'; $newquery= "UPDATE userinfo SET hint_credits = '-1' WHERE username = '".$_SESSION['login_name']."'"; echo $form; mysql_query($newquery,$link); } In this case, the form is still being shown. I need the form completely removed after the form has been submitted. This is simple and idk why its not working for me!? Link to comment https://forums.phpfreaks.com/topic/234413-unset-isnt-working-like-i-want-it-to/ Share on other sites More sharing options...
Pikachu2000 Posted April 22, 2011 Share Posted April 22, 2011 You define $form, then unset it after concatenating it to $a, then you define $form again after you unset it? The logic in that code isn't the easiest to follow. What exactly are you trying to make it do? Link to comment https://forums.phpfreaks.com/topic/234413-unset-isnt-working-like-i-want-it-to/#findComment-1204762 Share on other sites More sharing options...
Drummin Posted April 22, 2011 Share Posted April 22, 2011 Seems like you could just use a "$showform" variable that's is changed if processing doesn't validate. For example. IF ($_POST['Submit']=='Submit){ $showform=false; IF (!empty($_POST['name'])){ $ckname=('good'); } Else{ $ckname=('bad'); $showform=true; } //ect through the rest of the processing IF anything doesn't pass, $showform will be true and you can use this to show the form. Hey, just an idea. Link to comment https://forums.phpfreaks.com/topic/234413-unset-isnt-working-like-i-want-it-to/#findComment-1204780 Share on other sites More sharing options...
jdock1 Posted April 24, 2011 Author Share Posted April 24, 2011 Seems like you could just use a "$showform" variable that's is changed if processing doesn't validate. For example. IF ($_POST['Submit']=='Submit){ $showform=false; IF (!empty($_POST['name'])){ $ckname=('good'); } Else{ $ckname=('bad'); $showform=true; } //ect through the rest of the processing IF anything doesn't pass, $showform will be true and you can use this to show the form. Hey, just an idea. Thanks for the idea. Yeah, that seems like itd be easier. And yes, I know, I have the shittiest code. I can code, but its very messy! Link to comment https://forums.phpfreaks.com/topic/234413-unset-isnt-working-like-i-want-it-to/#findComment-1205381 Share on other sites More sharing options...
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