CODEWITHSUNDEEP
phpisset
tuespetre
tuespetre

Reputation: 1887

PHP isset() returning false when should return true?

Alright... so on my forms, I am setting all the fields to something like this:

name="formdata['name']" and name="formdata['active']".

Of course, that means whatever is entered in those fields should be stored in $_POST['formdata']['name'] and $_POST['formdata']['active'].

I do my values that way because I pass the 'formdata' to a couple of functions I have written (which work as they should.) Now let's say I'm using this to edit an item or add a new item -- the name will go through those functions as it should and the item will save with its name, but 'active' will never save as it should. The entry field is a checkbox -- if it's checked, the value is "1".

If I print_r($_POST['formdata']) after entering 'Name' and checking 'active', I get this: Array ( ['name'] => Name ['active'] => 1 ).

Looks fine, right? But when I do the following:

if (!isset($_POST['formdata']['active']) echo "Error 1";
if (empty($_POST['formdata']['active']) echo "Error 2";
if ($_POST['formdata']['active'] != 1) echo "Error 3";

They all return errors! I am baffled by this. Am I overlooking something very simple? I have thought about this for at least 2 hours now.

Upvotes: 2

Views: 3044

Answers (2)

webbiedave
webbiedave

Reputation: 48887

Remove the single quotes from the input names in your HTML so it reads:

name="formdata[active]"

Adding the single quotes would mean you would have to access the array in PHP as:

$_POST['formdata']['\'active\'']

or 

$_POST['formdata']["'active'"]

which is highly inconvenient.

Upvotes: 4

Shackrock
Shackrock

Reputation: 4701

missing parenthesis:

if (!isset($_POST['formdata']['active'])) echo "Error 1";
if (empty($_POST['formdata']['active'])) echo "Error 2";

Upvotes: 0

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