I have a key stored in a variable like so:
$key = 4;
I tried to get the relevant value like so:
$value = $array[$key];
but it failed. Help.
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2What do you mean "failed," any errors?Sampson– Sampson2010-02-17 14:43:21 +00:00Commented Feb 17, 2010 at 14:43
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2Are you sure there's a value in $array[4]? Because your syntax is correct - look elsewhere for the problem.Erik– Erik2010-02-17 14:44:15 +00:00Commented Feb 17, 2010 at 14:44
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2Could you add some more details, for example which is the contents of $array ?Dominik– Dominik2010-02-17 14:44:51 +00:00Commented Feb 17, 2010 at 14:44
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4 Answers
Your code seems to be fine, make sure that key you specify really exists in the array or such key has a value in your array eg:
$array = array(4 => 'Hello There');
print_r(array_keys($array));
// or better
print_r($array);
Output:
Array
(
[0] => 4
)
Now:
$key = 4;
$value = $array[$key];
print $value;
Output:
Hello There
Comments
$value = ( array_key_exists($key, $array) && !empty($array[$key]) )
? $array[$key]
: 'non-existant or empty value key';
Comments
As others stated, it's likely failing because the requested key doesn't exist in the array. I have a helper function here that takes the array, the suspected key, as well as a default return in the event the key does not exist.
protected function _getArrayValue($array, $key, $default = null)
{
if (isset($array[$key])) return $array[$key];
return $default;
}
hope it helps.
Comments
It should work the way you intended.
$array = array('value-0', 'value-1', 'value-2', 'value-3', 'value-4', 'value-5' /* … */);
$key = 4;
$value = $array[$key];
echo $value; // value-4
But maybe there is no element with the key 4. If you want to get the fiveth item no matter what key it has, you can use array_slice:
$value = array_slice($array, 4, 1);
