Calling PHP explode and access first element? [duplicate]

Question

Possible Duplicate:
PHP syntax for dereferencing function result

I have a string, which looks like 1234#5678. Now I am calling this:

$last = explode("#", "1234#5678")[1]

Its not working, there is some syntax error...but where? What I expect is 5678 in $last. Is this not working in PHP?

Luik

You'll be able to do this (Array Dereferencing) in PHP 5.4, not in the current 5.3 – Spain

Answer 1

Array dereferencing is not possible in the current PHP versions (unfortunately). But you can use list ^[docs] to directly assign the array elements to variables:

list($first, $last) = explode("#", "1234#5678");

UPDATE

Since PHP 5.4 (released 01-Mar-2012) it supports array dereferencing.

Grapher

and if you have many values in string then you can use $last = end(explode('#', $string)); – Superpower

Alternate for list is list(,$last) = ..... Then you won't have the first variable hanging when you don't need it. – Ferretti

Answer 2

Most likely PHP is getting confused by the syntax. Just assign the result of explode to an array variable and then use index on it:

$arr = explode("#", "1234#5678");
$last = $arr[1];

Bhili

Wow, that sucks :D ... ok, good I know now. – Luik

Answer 3

Here's how to get it down to one line:

$last = current(array_slice(explode("#", "1234#5678"), indx,1));

Where indx is the index you want in the array, in your example it was 1.

Ovate

Answer 4

You can't do this:

explode("#", "1234#5678")[1]

Because explode is a function, not an array. It returns an array, sure, but in PHP you can't treat the function as an array until it is set into an array.

This is how to do it:

 $last = explode('#', '1234#5678');
 $last = $last[1];

Shotgun

Answer 5

PHP can be a little dim. You probably need to do this on two lines:

$a = explode("#", "1234#5678");
$last = $a[1];

Nape