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posting the value of an auto submit drop down list to another page

ajoo

By ajoo
in PHP Coding Help

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ajoo Prolific Member

ajoo

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Posted

Hi all !

 

I wish to post the value of an auto submit dropdown value to another page, be redirected to that page, and  use it on that page. I am unable to achieve this in php and so I tried as :-follows:-

 

dropdown.php

<form method="post" action = dropdownaction.php>
    <select name="myselect" onchange="this.form.submit();">
        <option>blue</option>
        <option>red</option>
    </select>
</form>

and now I need the equivalent of 

 

dropdownaction.php

<?php

if(isset($_POST(['myselect']))) echo " I am selected".$_POST['myselect'];

?>

Please can someone tell me how I may retrieve the value of 'myselect' in the dropdownaction.php after being redirected to it.

Thanks loads.

 

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Ch0cu3r Newbie

Ch0cu3r

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Posted (edited)

Your if statement should be

if(isset($_POST['myselect'])) echo " I am selected".$_POST['myselect'];

Also always enclose HTML attribute values in quotes too

<form method="post" action="dropdownaction.php">
...
Edited by Ch0cu3r
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ajoo Prolific Member

ajoo

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Posted

Hi Ch0cu3r, 

 

Thanks for the reply. I'll remember to enclose the HTML attributes in quotes. Thanks.

 

However this does not work. ( How I wished it would ). But so far as I have read this does not work and we have to use javascript and we have to use the onChange event, like I have done in dropdown.php. The rest ( envoking the value of myselect) in and after being redirected to dropdownaction.php , however is still a mystery to me. 

 

Thanks !!

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ajoo Prolific Member

ajoo

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Hi ! Ok so I have tried the code again as well. Just to be doubly sure that I have not made a mistake.  When I select the color Red from the drop down menu, I get this error message. 

 

( ! ) Fatal error: Cannot use isset() on the result of a function call (you can use "null !== func()" instead) in D:\xampp\htdocs\xampp\MagicOnn\testers\dropdownaction.php on line 3.

 

Thus I do not get message "I am selected" or the value of $_POST['myselect'].  Fastsol are you getting the message as well as the value of myselect?

Thanks all ! 

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fastsol Regular Member

fastsol

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Posted (edited)

Honestly I don't know what that error means.  Are you sure you changed the isset code like Ch0cu3r showed?  You had to many () in your original code.  This is what I have on my end that tested good.
 
select.php

<form method="post" action="dropdownaction.php">
    <select name="myselect" onchange="this.form.submit();">
        <option>blue</option>
        <option>red</option>
    </select>
</form>

dropdownaction.php

<!DOCTYPE html>
<html>

<head>
<meta content="text/html; charset=utf-8" http-equiv="Content-Type">
<title>Untitled 1</title>
</head>

<body>
<?php
if(isset($_POST['myselect'])) echo " I am selected".$_POST['myselect'];
?>
</body>

</html>
Edited by fastsol
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Ch0cu3r Newbie

Ch0cu3r

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Posted (edited)

@ajoo you get that error message because you have not change your if statement to the one I have posted!

 

This is because there is an error in your code. I have highlighted what is wrong below

if(isset($_POST(['myselect']))) echo " I am selected".$_POST['myselect'];
//             ^            ^
//             |            |
//            remove two braces, these are causing the error
Edited by Ch0cu3r
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ajoo Prolific Member

ajoo

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Yes Ch0cu3r, It has worked. I don't know why I got that message. I had removed the extra parenthesis. However I have noticed that it does not work for the first value of the list. Please suggest how that can be accomplished. 

 

fastsol thanks to you too. 

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