Use set post value as php variable

Question

In a form I have a radio, of which one value is a specific API Key and the other radio opens a text input to enter a custom api key.

In order to distinguish from the two inputs in the action, they are set under different names.

My question is how do I set a PHP variable to equal the post value that is not empty.

For example, if input "a" has a set value then $apikey = $_POST['a'] but if "a" is empty and "p" has a value then $apikey = $_POST['p']

Here is the code used in the form for the two values that will be used:

<input type="radio" name="a" id="apibeta" value="###APIKEY###" onclick="javascript:hide();" required />

<input id='yes' name="p" class="form-control" placeholder="Personal API Key" type="text">

Thank you for any help!

Nico · Accepted Answer · 2016-11-19 03:01:21Z

1

Using the ternary operator:

$apiKey = !empty($_POST['a']) ? $_POST['a'] : $_POST['p'];

Which is equivalent to:

if (!empty($_POST['a'])) {
    $apiKey = $_POST['a'];
} else {
    $apiKey = $_POST['p'];
}

answered Nov 19, 2016 at 3:01

Nico

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2 Comments

TimBrownlaw Over a year ago

I would have thought @Nytrix answer was better as it deals with the case where $_POST['p'] and $_POST['a] are both empty. A condition that could happen.

Nico Over a year ago

In that case $apiKey will end up being empty as well, which in many cases would be ok. If we're making assumptions I'd also say that sending an API key using a form probably isn't a good idea in the first place. But this is getting further and further away from the original question ;)

Nytrix · Accepted Answer · 2016-11-19 03:01:04Z

0

That is what if statements are for or switch statements.

Here is an example with your problem

<?php
if(isset($_POST['a'])){
    //do something with that
}elseif(isset($_POST['p'])){
    //p has the value
}else{
    //neither have a value. 
}
?>

answered Nov 19, 2016 at 3:01

Nytrix

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