Convert array element to variable

Question

Given,

$required_fields = array('name', 'location', 'email');                               
foreach ($required_fields as $fieldname) {
    if (isset($_POST[$fieldname]) || !empty($_POST[$fieldname])) {      


        }

Now in the if statement, I need something like

$fieldname = $_POST[$fieldname];

So that i get

$name = "name", $location = "location" and $email = "email"

I tried $fieldname = $_POST[$fieldname] But it don't work. How can I do this dynamically?

When you think you need variable variables, it is time to step away from the keyboard and have a rethink. Sanitize user submission data using whitelisted keys: Yes. Variable variables: No. — mickmackusa
– mickmackusa ♦, Commented Oct 5, 2024 at 6:40

Shiplu Mokaddim · Accepted Answer · 2012-12-25 15:47:00Z

9

You do it like this,

$$fieldname = $_POST[$fieldname];

This $$ notation is called variable variable.

answered Dec 25, 2012 at 15:47

Shiplu Mokaddim

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Comments

goat · Accepted Answer · 2012-12-25 16:12:39Z

3

you could use extract function for this. To filter array you can use array_diff function

edited Dec 25, 2012 at 16:12

goat

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answered Dec 25, 2012 at 15:47

Artem L

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3 Comments

Nobwyn Over a year ago

could be also useful to take a look to - Is using extract($_POST) insecure?

ianace Over a year ago

there's a possibility that extract() will overwrite your existing variables with new ones and that's debugging hell

Artem L Over a year ago

@ianace Assigning variables like in question is a debugging hell.

goat · Accepted Answer · 2012-12-25 17:02:38Z

1

$required_fields = array('name', 'location', 'email');
extract(array_intersect_keys($_POST, array_flip($required_fields)));

This will create local variables for each of the variables named in $required_fields only if it exists in the $_POST array. Otherwise, the variable will be undefined.

array_flip
array_intersect_key
extract

edited Dec 25, 2012 at 17:02

answered Dec 25, 2012 at 16:17

goat

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Comments

Tomi · Accepted Answer · 2012-12-25 16:01:57Z

-2

You can also use

eval("$".$fieldname."='".$_POST["fieldname"]."';");

edited Dec 25, 2012 at 16:01

answered Dec 25, 2012 at 15:55

Tomi

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2 Comments

Shiplu Mokaddim Over a year ago

why invoke a new PHP interpreter?

Tomi Over a year ago

This is another example of how the issue can be solved. Although is also dangerous if the values aren't validated properly. The final result is same as your given solution.

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