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PHP: Undefined Variable errors. |
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Mar 8th 2004 | #144694 Report |
Member since: Jul 5th 2003 Posts: 80 |
I'm having a bit of a problem. Here is the scenario: I have two pages. The first one is an HTML page containing a form with several text input fields and a checkbox (our focus here). The form submits the info to the second page, which is a .php page. Now, the php script is written so that if the checkbox is checked, thus returning a value of "on" it will display one line of text, and if the checkbox was unchecked (which I assumed returned no value) then it would display a different line. Pretty basic, but I'm learning. The problem is that when the checkbox is unchecked, I get a variable undefined error. The tags for the form and checkbox are form method=post action="car.php"> and input name="License" type="checkbox"> (*Edit - removed the opening tags) The php script is: if ($License=="on" AND $Age>20) echo "Your car hire has been accepted."; if ($License=="" OR $Age<21) echo "Your car hire has not been accepted."; This is following the guidelines of a PHP book, so I just went ahead and used their text. Any help would be appreciated, as according to the book it should work. |
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Mar 8th 2004 | #144699 Report |
Member since: May 22nd 2003 Posts: 315 |
you need to use $_POST['variableName'] to get the data from a Form... unless you have globals on... for example: input type="text" name="text1"> to get that ont he next page you'd use $text1 = $_POST['text1']; assuming that the form method was 'opst' |
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Mar 8th 2004 | #144703 Report |
Member since: Nov 26th 2001 Posts: 2586 |
also anytime you allow a form variable to not be set, check that it is indeed not set by using [php]isset()[/php] So an example in your case: [php] if ( isset($_POST['some_checkbox_var']) ) { mmmm doughnuts.... (execute some code.) } [/php] of course you can also use the not set version in some cases: [php] if ( !isset($someVar) ) { ..... [/php] If it's not set it will just ignore it. |
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