How can I check that an array key exist if the array does not exist in PHP

Question

I am in a bind, multiple times on a page for different form items, I insert a css div class into a form item ONLY if an error_array key exists, this is how I highlight a form item that has an error in it.

Works, great if I have an error because then my error array is set, problem is before an error is set, it tries to look for an array key that does not exist. I was thinking using php's isset function first would be the ticket but I guess you cannot combine isset with another function?

<?php
//this works
$search_array = array('first' => 1, 'second' => 4);

if (array_key_exists('first', $search_array)){
    echo "good";
}


// this does not work, will give write errors
$search_array = array('first' => 1, 'second' => 4);

if (isset(array_key_exists('first', $search_array))){
    echo "good";
}

// Here is  1 example how I need to make the end result work
$country_class = (array_key_exists('country', $signup_errors)) ? ' signup_error' : ' signup_good';
echo '<select name="country" id="country" class="textarealong ' .$country_class. '"/>';
?>

In other parts I use it like this

<?PHP echo(array_key_exists('password', $signup_errors)) ? ' signup_error' : ' signup_good';?>

And I need to have it be a 1 line code if possible

Tammo · Accepted Answer · 2009-09-03 19:48:09Z

4

If isset is false, the second statement wont get executed, because the parsers knows that both statements have to be true to get the whole statement true.

   $country_class = ( isset($signup_errors) && array_key_exists('country', $signup_errors)) ? ' signup_error' : ' signup_good';

BUT i would suggest you to initialize every variable you are using...

answered Sep 3, 2009 at 19:48

Tammo

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2 Comments

JasonDavis Over a year ago

Also about your comment, doesn't this initialize this variable?

Guss Over a year ago

No it wont because even if array_key_exists auto-vivifies $signup_errors (and I'm pretty sure it doesn't), then the failure of isset() will cause it not to run at all. isset() obviously is not allowed to vivify the variables it tests.

Exp HP · Accepted Answer · 2009-09-03 19:49:50Z

0

I'm not familiar enough with PHP syntax, but this sounds like a job for short-circuit evaluation, i.e. || or &&, where the second term is not evaluated if the first term alone can determine the result (if it's True in || or False in &&).

answered Sep 3, 2009 at 19:49

Exp HP

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Comments

AbraCadaver · Accepted Answer · 2015-01-30 22:25:48Z

0

This is an old question, but it's an even simpler answer. isset checks the array and the key. This works great and doesn't generate any notices if the array is not set:

if (isset($search_array['first'])){
    echo "good";
}

Or:

$country_class = isset($signup_errors['country']) ? ' signup_error' : ' signup_good';

edited Jan 30, 2015 at 22:25

answered Jan 30, 2015 at 21:31

AbraCadaver

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