array_fill with increasing $i values

Question

Is there a function like array_fill, that lets me fill an array with the same string ("field_"), combined with an increasing $i variable like this:

private function fields() {
    $result = [];

    foreach(range(1,44) as $i) {
        $result[] = "field_$i";
    }

    return $result;
}

I know there is a bunch of functions in php and I would love to have this in one line (if possible).

Anytime you see `field_1` `field_2` etc, you should consider an array. — AbraCadaver, Sep 26 '17 at 17:26

score 4 · Answer 1 · answered Sep 26 '17 at 16:57

4

You could use array_map to remove the need for an explicit loop, but I don't think putting it on one line gains you much in terms of readability:

$results = array_map(function ($i) { return "field_{$i}"; }, range(1, 44));

=

Array
(
  [0] => field_1
  [1] => field_2
  ...

answered Sep 26 '17 at 16:57

iainn

you are right, thanks for your response! I was hoping for something shorter.. like a single method call: array_fill($i = 1, 44, "field_$i"); or so :) – Basti Sep 26 '17 at 17:04
I mean, you've basically got the function you need in the question - just move the prefix and the range parameters into function arguments, and declare it as `array_prefix($from, $to, $prefix)` or something similar. Then you can use it wherever you like as `array_prefix(1, 44, 'field_');` – iainn Sep 26 '17 at 17:10

score 1 · Accepted Answer · answered Sep 26 '17 at 18:23

1

I found this, seems to be 3 times faster than foreach and it's the shortest solution I could find so far.

$prefixed_array = preg_filter('/^/', 'prefix_', $array);

answered Sep 26 '17 at 18:23

Basti

score 0 · Answer 3 · answered Sep 26 '17 at 17:18

0

Php's foreach loops actually allow you to access the current index

private function fields() {
    $result = [];

    foreach(range(1,44) as $k => $i) {
        $result[] = "field_$k";
    }

    return $result;
}

answered Sep 26 '17 at 17:18

William Perron

3 Answers3