PHP: Replace certain parts of string with array values

Question

I want to replace values in a string if a particular string exists in an array.

$str = 'My name is {{name}}. I live in {{city}}. I love to {{hobby}}. {{ops...}}';

$array = array(
    'name' => '010 Pixel',
    'city' => 'USA',
    'hobby' => 'code',
    'email' => '[email protected]'
);

I want to replace {{name}} with the value of name in $array. If the string inside curly brackets doesn't exist in the $array then let that string stay as it is.

Expected result:

My name is 010 Pixel. I live in USA. I love to code. {{ops...}}

The reason I'm concern about this is, when any value coming from form contains any {{field-name}} then it shouldn't get replaced. I want to replace only what is set in $str.

This is why I like StackOverflow, here we have working 4 answers with all different approaches ! — Birla
– Birla, Commented Feb 25, 2014 at 9:46

xdazz · Accepted Answer · 2014-02-25 09:39:11Z

3

There is strtr function.

$str = 'My name is {{name}}. I live in {{city}}. I love to {{hobby}}. {{ops...}}';

$array = array(
    '{{name}}' => '010 Pixel',
    '{{city}}' => 'USA',
    '{{hobby}}' => 'code',
    '{{email}}' => '[email protected]'
);
echo strtr($str, $array);

answered Feb 25, 2014 at 9:39

xdazz

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2 Comments

010 Pixel Over a year ago

What if my array values are coming from $_POST?

xdazz Over a year ago

@010Pixel Wrap {{}} to your array index should not be hard work.

Birla · Accepted Answer · 2014-02-25 09:40:30Z

1

Try this:

<?
$str = 'My name is {{name}}. I live in {{city}}. I love to {{hobby}}. {{ops...}}';

$array = array(
    'name' => '010 Pixel',
    'city' => 'USA',
    'hobby' => 'code',
    'email' => '[email protected]'
);

if (preg_match_all("/{{(.*?)}}/", $str, $m)) {
  foreach ($m[1] as $i => $varname) {
    $str = str_replace($m[0][$i], $array[$varname], $str);
  }
}

answered Feb 25, 2014 at 9:40

Birla

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1 Comment

010 Pixel Over a year ago

I have already tried this but I want some more simple way haha. Thanks :)

gontrollez · Accepted Answer · 2014-02-25 09:43:34Z

1

$str = 'My name is {{name}}. I live in {{city}}. I love to {{hobby}}. {{ops...}}';

$array = array(
    'name' => '010 Pixel',
    'city' => 'USA',
    'hobby' => 'code',
    'email' => '[email protected]'
);

$callback = function($match) use ($array) {
    if (array_key_exists($match[1], $array)) {
        return $array[$match[1]];
    } else {
        return $match[0];
    }
};

$str = preg_replace_callback('/\{\{(.*?)\}\}/', $callback, $str);

answered Feb 25, 2014 at 9:43

gontrollez

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Comments

Community · Accepted Answer · 2023-11-17 19:55:19Z

1

Using preg_replace_callback works-

$str = 'My name is {{name}}. I live in {{city}}. I love to {{hobby}}. {{ops...}}';
$array = array(
    'name' => '010 Pixel',
    'city' => 'USA',
    'hobby' => 'code',
    'email' => '[email protected]'
);
$res = preg_replace_callback('/\{{2}(.*?)\}{2}/',
        function($matches)use($array){
            $key = $matches[1];
            if(isset($array[$key])){
                return "{{".$array[$key]."}}";
            };
            return $matches[0];
        },
        $str);
var_dump($res);
/*
    OUTPUT-
    string 'My name is {{010 Pixel}}. I live in {{USA}}. I love to {{code}}. {{ops...}}' (length=75)
*/

edited Nov 17, 2023 at 19:55

CommunityBot

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answered Feb 25, 2014 at 9:50

Kamehameha

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Comments

SpencerX · Accepted Answer · 2014-02-25 09:41:29Z

0

$string = preg_replace('/{{([a-zA-Z\_\-]*?)}}/ie','$array',$str);

Hope this could help

answered Feb 25, 2014 at 9:41

SpencerX

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3 Comments

SpencerX Over a year ago

Using the e modifier you can have any PHP expression in the replacement part of preg_replace.

010 Pixel Over a year ago

Can you please show me a demo of this? It gives me some errors.

Toto Over a year ago

/e modifier is deprecated. Use preg_replace_callback instead.

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