How to generate set of array from string in php

Question

This is my string

#Jhon: Manager #Mac: Project Manager #Az: Owner

And I want array something like this

$array = ['0' => 'Manager', '1' => 'Project Manager', '2' => 'Owner']

I tried this but each time return only 'Manager'

$string = '#Jhon: Manager #Mac: Project Manager #Az: Owner';
getText($string, ':', ' #')
public function getText($string, $start, $end)
{
  $pattern = sprintf(
      '/%s(.+?)%s/ims',
      preg_quote($start, '/'), preg_quote($end, '/')
  );

  if (preg_match($pattern, $string, $matches)) {
      list(, $match) = $matches;
      echo $match;
  }
}

Try using `preg_match_all`. But remeber, owner doesn't have the end character `#`. So only `Manager` and `Project Manager` will be matched — Thamilhan, May 05 '17 at 11:37
Functionally related: [preg match to get text after @ symbol and before next space using php](https://stackoverflow.com/q/25914018/2943403) — mickmackusa, May 18 '23 at 13:27
Related: [preg_match_all get text between 2 strings](https://stackoverflow.com/q/6976707/2943403) and [Get string between every two certain characters with PHP](https://stackoverflow.com/q/53721443/2943403) — mickmackusa, May 18 '23 at 13:59

Wiktor Stribiżew · Accepted Answer · 2017-05-05T13:48:10.313

8

You may preg_split the contents and use the following solution:

$re = '/\s*#[^:]+:\s*/';
$str = '#Jhon: Manager #Mac: Project Manager #Az: Owner';
$res = preg_split($re, $str, -1, PREG_SPLIT_NO_EMPTY);
print_r($res);

See the PHP demo and a regex demo.

Pattern details:

\s* - 0+ whitespaces
# - a literal # symbol
[^:]+ to match 1+ chars other than :
: - a colon
\s* - 0+ whitespaces.

Note that -1 in the preg_split function is the $limit argument telling PHP to split any amount of times (as necessary) and PREG_SPLIT_NO_EMPTY will discard all empty matches (this one may be removed if you need to keep empty matches, that depends on what you need to do further).

edited May 05 '17 at 13:48

answered May 05 '17 at 11:36

Wiktor Stribiżew

607,720
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Thanks wiktor.. but when my string is '#Jhon: Manager #Mac@gmail.com: Project Manager #Az: Owner' not working.. sorry for not specify this possibility – Komal May 05 '17 at 12:12
@Komal: But I already wrote - use `[^:\s]+` in that case, use [`\s*#[^:\s]+:\s*`](https://regex101.com/r/P0s10e/2). I removed the `\w`-based regex variation. – Wiktor Stribiżew May 05 '17 at 12:14
One more possibility is space between name '#Jhon M: Manager #Mac@gmail.com: Project Manager #Az: Owner' – Komal May 05 '17 at 13:37
@Komal: Then how can you differentiate `#` as the starting point of the key and a `#` that is part of the value? – Wiktor Stribiżew May 05 '17 at 13:40
Ok, remove `\s` from the `[^:\s]+`. See [**this demo**](https://regex101.com/r/P0s10e/3). – Wiktor Stribiżew May 05 '17 at 13:46
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/143507/discussion-between-komal-and-wiktor-stribizew). – Komal May 05 '17 at 13:48

Sahil Gulati · Answer 2 · 2017-05-05T12:30:10.840

6

Here we are using preg_match to achieve desired output.

Regex: #\w+\s*\:\s*\K[\w\s]+

1. #\w+\s*\:\s*\K this will match # then words then spaces and then : \K will reset current match.

2. [\w\s]+ This will match your desired output which contains words and spaces.

Solution 1:

<?php
ini_set('display_errors', 1);
$string="#Jhon: Manager #Mac: Project Manager #Az: Owner";
preg_match_all("/#\w+\s*\:\s*\K[\w\s]+/", $string,$matches);
print_r($matches);

Output:

Array
(
    [0] => Array
        (
            [0] => Manager 
            [1] => Project Manager 
            [2] => Owner
        )

)

Here we are using array_map and explode to achieve desired out. Here we first we are exploding string on # and then exploding its array values on : and pushing its first index in the resultant array.

Solution 2:

Try this code snippet here

<?php

$string="#Jhon: Manager #Mac: Project Manager #Az: Owner";
$result=  array_map(function($value){
    return trim(explode(":",$value)[1]);
}, array_filter(explode("#", $string)));
print_r(array_filter($result));

Output:

Array
(
    [1] => Manager
    [2] => Project Manager
    [3] => Owner
)

edited May 05 '17 at 12:30

answered May 05 '17 at 11:37

Sahil Gulati

15,028
4
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@SahilGulati 2nd Gives me error Undefined offset: 1, and first is working but not worked when my string is '#Jhon: Manager #Mac@gmail.com: Project Manager #Az: Owner' sorry for not specifing this possibility.. – Komal May 05 '17 at 12:10
@Komal Hope this will help you out.. I have updated my post for that for demo you can check this https://eval.in/787827 – Sahil Gulati May 05 '17 at 12:31
1

This way is indeed better since the pattern starts with a literal character. You can remove the eventual trailing spaces using: https://regex101.com/r/tgjssx/1 – Casimir et Hippolyte May 05 '17 at 19:52
@Casimir just for my safe side i added this so that OP will not face any further issues with other string input which may contain spaces.. Anyways thanks sir...:) – Sahil Gulati May 05 '17 at 19:55

score 0 · Answer 3 · answered May 05 '17 at 12:13

0

You can use explode() function to split the string by char. After that just replace the elements of new array with preg_replace().

answered May 05 '17 at 12:13

radoAngelov

684
5
12

score 0 · Answer 4 · answered May 05 '17 at 12:39

0

Try this

$string = '#Jhon: Manager #Mac: Project Manager #Az: Owner';
$res = explode("#", $string);
$result = array();
for($i = 1; $i < count($res); $i++)  {
    $result[] = preg_replace("/(^[A-Z][a-z]*: )/", "", $res[$i]);
}

answered May 05 '17 at 12:39

Andrianina OELIMAHEFASON

3
1

How to generate set of array from string in php

4 Answers4