PHP: array_pop returns the empty string

Question

I'm using explode to break a FQDN and store it in an array. Calling pop on that array however returns a blank or empty string, and I can't figure out why.

 $domains = explode(".",str_replace("example.com","","test.example.com"));
 print "Test: " . $domains[0] . "<br>";
 $target = array_pop($domains);
 print "Target: " . $target . "<br>";

Running the above code results in:

 Test: test
 Target:

Why doesn't $target contain "test"?

Didn't checked that as you.. But, now its clear, try `var_dump($domains)` to see.. It contains **`2`** elements.. `test` and an empty string. However, this is not a reason for a downvote (imo) — hek2mgl, Nov 29 '13 at 22:46

score 2 · Accepted Answer · answered Nov 29 '13 at 22:42

2

In effect, here's what you're actually doing:

var_dump(explode('.', 'test.'));

array(2) {
  [0]=>
  string(4) "test"
  [1]=>
  string(0) ""
}

You get two elements in the array: "test" and what's after the period i.e. an empty string.

answered Nov 29 '13 at 22:42

Nev Stokes

9,051
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score 1 · Answer 2 · answered Nov 29 '13 at 22:43

1

array_pop() pops and returns the last value of the array

Use array_shift

answered Nov 29 '13 at 22:43

Scony

4,068
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I actually want the last element of the array, it was the whitespace as mentioned by Nev Stokes that threw me off. Thanks for the reply! – rthur Nov 29 '13 at 22:48

PHP: array_pop returns the empty string

2 Answers2