PHP Get first array value that ends with ".jpg"

Question

I have some arrays and they might all be formatted like so:

Array (
    [0] => .
    [1] => ..
    [2] => 151108-some_image-006.jpg
    [3] => high
    [4] => low
)

I know they have 5 values, but I can not be certain where each value is being placed.

I am trying to get only the image out of this array

$pos = array_search('*.jpg', $main_photo_directory);
echo $main_photo_directory[$pos];

But as we all know, it's looking for a literal *.jpg which it can't find. I'm not so super at regex and wouldn't know how to format an appropriate string.

What is the best way to get the first image (assuming there may be more than one) out of this array?

ADDITION

One of the reasons I am asking is to find a simple way to search through an array. I do not know regex, though I'd like to use it. I wrote the question looking for a regex to find '.jpg' at the end of a string which no search results had yielded.

Right - there's my problem. How do I search for a string ending in .jpg? — ntgCleaner, Nov 17 '15 at 20:06
Just read the last four characters of each array element in a loop to see if they equal `.jpg` — j08691, Nov 17 '15 at 20:06
Or http://stackoverflow.com/questions/4764859/how-to-know-if-a-string-ends-with or http://stackoverflow.com/questions/619610/whats-the-most-efficient-test-of-whether-a-php-string-ends-with-another-string — PeeHaa, Nov 17 '15 at 20:07
Also possible duplicate of: http://stackoverflow.com/questions/5592120/php-search-array-using-wildcard — War10ck, Nov 17 '15 at 20:08
2k rep doesn't mean I know everything. Most of the rep is from javascript (jQuery really) — ntgCleaner, Nov 17 '15 at 20:08
Thanks. So for all of the criticism on a brain fart, I guess I'll close the question. Love this community! — ntgCleaner, Nov 17 '15 at 20:09
it should mean you can use google and or the sites search engine. — , Nov 17 '15 at 20:11
@dagon, Yes, I googled for 15 minutes not getting a specific answer. The voted up answer below is what I was looking for, but I couldn't figure out the regex to place in there. — ntgCleaner, Nov 17 '15 at 20:11
@PeeHaa: Bad dupe. I recommend http://stackoverflow.com/questions/5592120/php-search-array-using-wildcard — AbraCadaver, Nov 17 '15 at 20:14
ok last comment, a whole 15 minutes before asking? i would spend days - but i realise some people have a very short attention span (its that MTV right?) — , Nov 17 '15 at 20:15
@Dagon, Yes, you're perfectly in-line to start the insults and the "I am holier than thou". A question that took less than 3 minutes to answer and wasted 10 minutes of your time making stupid comments deserved 15 minutes of looking. - You must be bitter with 11/12 of your questions having down votes. — ntgCleaner, Nov 17 '15 at 20:16
@PeeHaa: The question was only marked as a dupe of _startsWith() and endsWith() functions in PHP_. Not the best one IMHO. — AbraCadaver, Nov 17 '15 at 20:23

score 7 · Accepted Answer · answered Nov 17 '15 at 20:06

7

This should work:

echo current(preg_grep('/\.jpg$/', $array));

answered Nov 17 '15 at 20:06

AbraCadaver

78,200
7
66
87

Cleanest, simplest, most straight forward answer. Thank you. - Will accept in 7 minutes – ntgCleaner Nov 17 '15 at 20:11
Indeed, impressive. You only have to take care how to use the result, since you cannot simply use `current()` in function calls, for example. – arkascha Nov 17 '15 at 20:14
@arkascha That's fine, I am using the literal string to push back through an ajax call. What has been supplied works perfectly for my needs. – ntgCleaner Nov 17 '15 at 20:17
@arkascha: What? Why not? `echo strtoupper(current(preg_grep('/\.jpg$/', $array)));` – AbraCadaver Nov 17 '15 at 20:19

score 0 · Answer 2 · answered Nov 17 '15 at 20:07

You can loop over the array until you find a string ending in jpg:

function endsWith($haystack, $needle) {
    return $needle === "" || (($temp = strlen($haystack) - strlen($needle)) >= 0 && strpos($haystack, $needle, $temp) !== FALSE);
}

$array = [];
$result = false;
foreach($array as $img) {
  if(endsWith($img, '.jpg')) {
    $result = $img;
    break;
  }
}

echo $result;

PHP Get first array value that ends with ".jpg"

2 Answers2