PHP randomly replace words

Question

I have an array called $featuresSEO that has a number of words in it like:

Array (
    [0] => Japan 
    [1] => Japanese 
    [2] => Tokyo 
    [3] => Yokohama 
    [4] => Osaka 
    [5] => Asian 
    [6] => Nagoya 
)

I then have a string like follows:

 Searching the |*-*| Dating membership database is the key to locating |*-*| people
 you would be interested in. You can search for |*-*| Singles including |*-*| Women
 and |*-*| Men in any location worldwide. Join now to search for |*-*| Singles.

I've been trying to replace the instances of |*-*| with random words from the Array. I've tried str_replace() but couldn't get the random aspect working.

Can someone push me in the right direction?

thx

Is every |*-*| random, or is one random word chosen and then used for all |*-*|? — Logan Serman
– Logan Serman, Commented Feb 27, 2012 at 8:30
each one a random word... so different words through the string — Adam
– Adam, Commented Feb 27, 2012 at 8:31
Use the random() function to get a random index, and then replace |*-*| by your Array[your_random_index] — BMN
– BMN, Commented Feb 27, 2012 at 8:31

Rene Pot · Accepted Answer · 2012-02-27 08:38:55Z

2

Replace them one by one. This one will replace each occurence with a random word. You might see the same word from the $wordarray multiple times as it picks 1 at random each time.

for ($i = 0; $i < substr_count($string, '|*-*|'); $i++){
     $string = preg_replace('/\|\*-\*\|/',$wordarray[rand(0,count($wordarray)-1)],$string, 1);
}

Want to use each word only once? Shuffle the array, and loop through it:

shuffle($wordarray);
foreach ($wordarray as $word){
    $string = preg_replace('/\|\*-\*\|/',$word,$string,1);
}

edited Feb 27, 2012 at 8:38

answered Feb 27, 2012 at 8:32

Rene Pot

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4 Comments

Salman Arshad Over a year ago

Will replace all items with the first randomly chosen item, wont it?

Rene Pot Over a year ago

nope. I've put a 1 as 4th parameter. So it will only replace 1 occurence

Salman Arshad Over a year ago

4th parameter gives me a warning!

vdegenne Over a year ago

@Topener the last argument needs to be a reference

Community · Accepted Answer · 2018-02-22 07:32:03Z

2

try this

$string = ' Searching the |*-*| Dating membership database is the key to locating |*-*| people
 you would be interested in. You can search for |*-*| Singles including |*-*| Women
 and |*-*| Men in any location worldwide. Join now to search for |*-*| Singles.';

$words = array('Japanese', 'Tokyo', 'Asian');

$placeholder = '|*-*|';
$pos = null;

while(null === $pos || false !== $pos) {
    $pos = strpos($string, $placeholder);
    $string = substr_replace($string, $words[rand(0, count($words)-1)], $pos, strlen($placeholder));

}

echo $string;

first word become unexpected

edited Feb 22, 2018 at 7:32

CommunityBot

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answered Feb 27, 2012 at 8:38

silly

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Comments

Community · Accepted Answer · 2017-05-23 11:48:01Z

The code for replace-only-first-match is borrowed from here. The following code uses each word from the list only once:

$wordlist = array(
    'Japan',
    'Japanese',
    'Tokyo',
    'Yokohama',
    'Osaka',
    'Asian',
    'Nagoya'
);
$string = "
Searching the |*-*| Dating membership database is the key to locating |*-*| people
you would be interested in. You can search for |*-*| Singles including |*-*| Women
and |*-*| Men in any location worldwide. Join now to search for |*-*| Singles.
";
$replace = '|*-*|';
while(true){
    $index = strpos($string, $replace);
    if($index === false){
        // ran out of place holder strings
        break;
    }
    if(count($wordlist) == 0){
        // ran out of words
        break;
    }
    $word = array_splice($wordlist, rand(0, count($wordlist) - 1), 1);
    $string = substr_replace($string, $word[0], $index, strlen($replace));
}
echo $string;

Mattias · Accepted Answer · 2012-02-27 08:42:53Z

0

Try this

<?php
    $array = array("Japan","Japanese","Tokyo","Yokohama","Osaka","Asian","Nagoya");
    $a = array_rand($array);
    $string= "abc|*-*|";
    echo str_replace("|*-*|", $array[$a], $string);
?>

edited Feb 27, 2012 at 8:42

answered Feb 27, 2012 at 8:37

Mattias

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Comments

mickmackusa · Accepted Answer · 2021-11-22 20:59:24Z

PHP already has a native function which allows you to replace placeholders in a template string with an array of strings -- vprintf() variable print from format if I am not mistaken.

If you had originally set up your template with recognised placeholders, you wouldn't need to call str_replace().

Code: (Demo)

$template = 'Searching the |*-*| Dating membership database is the key to locating |*-*| people
 you would be interested in. You can search for |*-*| Singles including |*-*| Women
 and |*-*| Men in any location worldwide. Join now to search for |*-*| Singles.';

$array = ['Japan', 'Japanese', 'Tokyo', 'Yokohama', 'Osaka', 'Asian', 'Nagoya'];

shuffle($array);
vprintf(str_replace('|*-*|', '%s', $template), $array);

Potential output:

Searching the Asian Dating membership database is the key to locating Japan people
 you would be interested in. You can search for Osaka Singles including Nagoya Women
 and Yokohama Men in any location worldwide. Join now to search for Tokyo Singles.

Anonizmus User · Accepted Answer · 2021-11-22 20:33:38Z

-1

I hope this is still relevant)

$wordarray = [
   'some',
   'random',
   'words',
   'list'
];
shuffle($wordarray);
$string = implode(' ', $wordarray);

You can change implode to sprintf, that was my realization.

answered Nov 22, 2021 at 20:33

Anonizmus User

11 bronze badge

1 Comment

sukalogika Over a year ago

The OP wants random words to be filled into a pattern. In this case, it will replace the character array |*-*|

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