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
– Thamilhan, Commented May 5, 2017 at 11:37
Functionally related: preg match to get text after @ symbol and before next space using php — mickmackusa
– mickmackusa ♦, Commented May 18, 2023 at 13:27
Related: preg_match_all get text between 2 strings and Get string between every two certain characters with PHP — mickmackusa
– mickmackusa ♦, Commented May 18, 2023 at 13:59

Wiktor Stribiżew · Accepted Answer · 2017-05-05 13:48:10Z

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 5, 2017 at 13:48

answered May 5, 2017 at 11:36

Wiktor Stribiżew

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6 Comments

Komal Over a year ago

Thanks wiktor.. but when my string is '#Jhon: Manager #[email protected]: Project Manager #Az: Owner' not working.. sorry for not specify this possibility

Wiktor Stribiżew Over a year ago

@Komal: But I already wrote - use [^:\s]+ in that case, use \s*#[^:\s]+:\s*. I removed the \w-based regex variation.

Komal Over a year ago

One more possibility is space between name '#Jhon M: Manager #[email protected]: Project Manager #Az: Owner'

Wiktor Stribiżew Over a year ago

@Komal: Then how can you differentiate # as the starting point of the key and a # that is part of the value?

Wiktor Stribiżew Over a year ago

Ok, remove \s from the [^:\s]+. See this demo.

|

Sahil Gulati · Accepted Answer · 2017-05-05 12:30:10Z

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 5, 2017 at 12:30

answered May 5, 2017 at 11:37

Sahil Gulati

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4 Comments

Komal Over a year ago

@SahilGulati 2nd Gives me error Undefined offset: 1, and first is working but not worked when my string is '#Jhon: Manager #[email protected]: Project Manager #Az: Owner' sorry for not specifing this possibility..

Sahil Gulati Over a year ago

@Komal Hope this will help you out.. I have updated my post for that for demo you can check this eval.in/787827

Casimir et Hippolyte Over a year ago

This way is indeed better since the pattern starts with a literal character. You can remove the eventual trailing spaces using: regex101.com/r/tgjssx/1

Sahil Gulati Over a year ago

@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...:)

radoAngelov · Accepted Answer · 2017-05-05 12:13:04Z

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 5, 2017 at 12:13

radoAngelov

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Comments

Andrianina OELIMAHEFASON · Accepted Answer · 2017-05-05 12:39:37Z

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 5, 2017 at 12:39

Andrianina OELIMAHEFASON

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