How to avoid undefined index

Question

How can you easily avoid getting this error/notice in PHP?

Notice: Undefined index: test in /var/www/page.php on line 21

The code:

$table = 'test';
$preset = array();
method($preset[$table]);

The array $preset exists but not with the specified index

Check http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index/16971723#16971723 for IMHO a much better answer I gave in this question — gts, Jun 06 '13 at 20:39

score 14 · Accepted Answer · answered Jul 04 '11 at 21:34

14

Check if it exists using array_key_exists:

$table = 'test';
$preset = array();
if(array_key_exists($table, $preset)) {
    method($preset[$table]);
}else{
    // $table doesn't exist in $preset
}

Alternatively, you could use isset:

$table = 'test';
$preset = array();
if(isset($preset[$table])) {
    method($preset[$table]);
}else{
    // $table doesn't exist in $preset
}

answered Jul 04 '11 at 21:34

icktoofay

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8

+ 1 on this, isset() does not return TRUE for array keys that correspond to a NULL value, while array_key_exists() does. – Stephane Gosselin Jul 04 '11 at 21:39

score 3 · Answer 2 · answered Jul 04 '11 at 21:35

3

Use if (isset($preset[$table]))

answered Jul 04 '11 at 21:35

Fabio

18,856
9
82
114

score 0 · Answer 3 · answered Jul 04 '11 at 21:35

0

Or you can check first if key exists by using isset().

if ( isset($preset[$table]) )

Return true if exists, otherwise return false.

answered Jul 04 '11 at 21:35

Puzo

596
2
11
26

How to avoid undefined index

3 Answers3

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