Define Variable Variable and use it in $_POST

Question

Simple questions I hope but I just can't get the solution.

I have a foreach loop producing variable $checkbox.

I then have a field name, for the example approvereject

I want to create a new variable that combines the fieldname and the forecah value.

So:

$joinedvariable=approvereject.$checkbox

This does not work, I have also tried:

$joinedvariable=${approvereject_.$checkbox};

but when I echo $joinedvariable I simply get a blank result.

Lastly, I want to use $joinedvariable in a $_POST statement:

$_POST[$joinedvariable ] - will this work as I cant get it right?

Has nothing to do with variable variables. You just need a concated array key for $_POST ($_POST is an array variable, not a "statement", or "function"). — mario
– mario, Commented Feb 20, 2013 at 12:04

Prasanth Bendra · Accepted Answer · 2013-02-20 11:44:45Z

1

Try this :

$joinedvariable="approvereject".$checkbox;

$_POST[$joinedvariable];

answered Feb 20, 2013 at 11:44

Prasanth Bendra

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Comments

sanj · Accepted Answer · 2013-02-20 11:45:41Z

1

$joinedvariable='approvereject'.$checkbox

answered Feb 20, 2013 at 11:45

sanj

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Sober · Accepted Answer · 2013-02-20 11:52:28Z

1

If you have some code like this:

foreach ($checkboxes as $checkbox) {
    echo '<input type="checkbox" name="approvereject_'. $checkbox .'" value="1">';
}

in a <form> with method="post", than you can try to get values of checkboxes like this:

foreach ($checkboxes as $checkbox) {
    print_r(!empty($_POST['approvereject_'. $checkbox]));
}

answered Feb 20, 2013 at 11:52

Sober

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