How do I check if a string contains a specific word?

Question

Consider:

$a = 'How are you?';

if ($a contains 'are')
    echo 'true';

Suppose I have the code above, what is the correct way to write the statement if ($a contains 'are')?

Answer 1

You can use the strpos() function which is used to find the occurrence of one string inside another one:

$a = 'How are you?';

if (strpos($a, 'are') !== false) {
    echo 'true';
}

Note that the use of !== false is deliberate (neither != false nor === true will return the desired result); strpos() returns either the offset at which the needle string begins in the haystack string, or the boolean false if the needle isn't found. Since 0 is a valid offset and 0 is "falsey", we can't use simpler constructs like !strpos($a, 'are').

Now with PHP 8 you can do this using str_contains:

if (str_contains('How are you', 'are')) { 
    echo 'true';
}

RFC

Answer 2

You could use regular expressions as it's better for word matching compared to strpos, as mentioned by other users. A strpos check for are will also return true for strings such as: fare, care, stare, etc. These unintended matches can simply be avoided in regular expression by using word boundaries.

A simple match for are could look something like this:

$a = 'How are you?';

if (preg_match('/\bare\b/', $a)) {
    echo 'true';
}

On the performance side, strpos is about three times faster. When I did one million compares at once, it took preg_match 1.5 seconds to finish and for strpos it took 0.5 seconds.

Edit: In order to search any part of the string, not just word by word, I would recommend using a regular expression like

$a = 'How are you?';
$search = 'are y';
if(preg_match("/{$search}/i", $a)) {
    echo 'true';
}

The i at the end of regular expression changes regular expression to be case-insensitive, if you do not want that, you can leave it out.

Now, this can be quite problematic in some cases as the $search string isn't sanitized in any way, I mean, it might not pass the check in some cases as if $search is a user input they can add some string that might behave like some different regular expression...

Also, here's a great tool for testing and seeing explanations of various regular expressions Regex101

To combine both sets of functionality into a single multi-purpose function (including with selectable case sensitivity), you could use something like this:

function FindString($needle,$haystack,$i,$word)
{   // $i should be "" or "i" for case insensitive
    if (strtoupper($word)=="W")
    {   // if $word is "W" then word search instead of string in string search.
        if (preg_match("/\b{$needle}\b/{$i}", $haystack)) 
        {
            return true;
        }
    }
    else
    {
        if(preg_match("/{$needle}/{$i}", $haystack)) 
        {
            return true;
        }
    }
    return false;
    // Put quotes around true and false above to return them as strings instead of as bools/ints.
}

One more thing to take in mind, is that \b will not work in different languages other than english.

The explanation for this and the solution is taken from here:

\b represents the beginning or end of a word (Word Boundary). This regex would match apple in an apple pie, but wouldn’t match apple in pineapple, applecarts or bakeapples.

How about “café”? How can we extract the word “café” in regex? Actually, \bcafé\b wouldn’t work. Why? Because “café” contains non-ASCII character: é. \b can’t be simply used with Unicode such as समुद्र, 감사, месяц and .

When you want to extract Unicode characters, you should directly define characters which represent word boundaries.

The answer: (?<=[\s,.:;"']|^)UNICODE_WORD(?=[\s,.:;"']|$)

So in order to use the answer in PHP, you can use this function:

function contains($str, array $arr) {
    // Works in Hebrew and any other unicode characters
    // Thanks https://medium.com/@shiba1014/regex-word-boundaries-with-unicode-207794f6e7ed
    // Thanks https://www.phpliveregex.com/
    if (preg_match('/(?<=[\s,.:;"\']|^)' . $word . '(?=[\s,.:;"\']|$)/', $str)) return true;
}

And if you want to search for array of words, you can use this:

function arrayContainsWord($str, array $arr)
{
    foreach ($arr as $word) {
        // Works in Hebrew and any other unicode characters
        // Thanks https://medium.com/@shiba1014/regex-word-boundaries-with-unicode-207794f6e7ed
        // Thanks https://www.phpliveregex.com/
        if (preg_match('/(?<=[\s,.:;"\']|^)' . $word . '(?=[\s,.:;"\']|$)/', $str)) return true;
    }
    return false;
}

As of PHP 8.0.0 you can now use str_contains

<?php
    if (str_contains('abc', '')) {
        echo "Checking the existence of the empty string will always 
        return true";
    }

Answer 3

Here is a little utility function that is useful in situations like this

// returns true if $needle is a substring of $haystack
function contains($needle, $haystack)
{
    return strpos($haystack, $needle) !== false;
}

Answer 4

While most of these answers will tell you if a substring appears in your string, that's usually not what you want if you're looking for a particular word, and not a substring.

What's the difference? Substrings can appear within other words:

• The "are" at the beginning of "area"
• The "are" at the end of "hare"
• The "are" in the middle of "fares"

One way to mitigate this would be to use a regular expression coupled with word boundaries (\b):

function containsWord($str, $word)
{
    return !!preg_match('#\\b' . preg_quote($word, '#') . '\\b#i', $str);
}

This method doesn't have the same false positives noted above, but it does have some edge cases of its own. Word boundaries match on non-word characters (\W), which are going to be anything that isn't a-z, A-Z, 0-9, or _. That means digits and underscores are going to be counted as word characters and scenarios like this will fail:

• The "are" in "What _are_ you thinking?"
• The "are" in "lol u dunno wut those are4?"

If you want anything more accurate than this, you'll have to start doing English language syntax parsing, and that's a pretty big can of worms (and assumes proper use of syntax, anyway, which isn't always a given).

Answer 5

To determine whether a string contains another string you can use the PHP function strpos().

int strpos ( string $haystack , mixed $needle [, int $offset = 0 ] )`

<?php

$haystack = 'how are you';
$needle = 'are';

if (strpos($haystack,$needle) !== false) {
    echo "$haystack contains $needle";
}

?>

CAUTION:

If the needle you are searching for is at the beginning of the haystack it will return position 0, if you do a == compare that will not work, you will need to do a ===

A == sign is a comparison and tests whether the variable / expression / constant to the left has the same value as the variable / expression / constant to the right.

A === sign is a comparison to see whether two variables / expresions / constants are equal AND have the same type - i.e. both are strings or both are integers.

Answer 6

Look at strpos():

<?php
    $mystring = 'abc';
    $findme   = 'a';
    $pos = strpos($mystring, $findme);

    // Note our use of ===. Simply, == would not work as expected
    // because the position of 'a' was the 0th (first) character.
    if ($pos === false) {
        echo "The string '$findme' was not found in the string '$mystring'.";
    }
    else {
        echo "The string '$findme' was found in the string '$mystring',";
        echo " and exists at position $pos.";
    }
?>

Answer 7

Using strstr() or stristr() if your search should be case insensitive would be another option.

Answer 8

Peer to SamGoody and Lego Stormtroopr comments.

If you are looking for a PHP algorithm to rank search results based on proximity/relevance of multiple words here comes a quick and easy way of generating search results with PHP only:

Issues with the other boolean search methods such as strpos(), preg_match(), strstr() or stristr()

1. can't search for multiple words
2. results are unranked

PHP method based on Vector Space Model and tf-idf (term frequency–inverse document frequency):

It sounds difficult but is surprisingly easy.

If we want to search for multiple words in a string the core problem is how we assign a weight to each one of them?

If we could weight the terms in a string based on how representative they are of the string as a whole, we could order our results by the ones that best match the query.

This is the idea of the vector space model, not far from how SQL full-text search works:

function get_corpus_index($corpus = array(), $separator=' ') {

    $dictionary = array();

    $doc_count = array();

    foreach($corpus as $doc_id => $doc) {

        $terms = explode($separator, $doc);

        $doc_count[$doc_id] = count($terms);

        // tf–idf, short for term frequency–inverse document frequency, 
        // according to wikipedia is a numerical statistic that is intended to reflect 
        // how important a word is to a document in a corpus

        foreach($terms as $term) {

            if(!isset($dictionary[$term])) {

                $dictionary[$term] = array('document_frequency' => 0, 'postings' => array());
            }
            if(!isset($dictionary[$term]['postings'][$doc_id])) {

                $dictionary[$term]['document_frequency']++;

                $dictionary[$term]['postings'][$doc_id] = array('term_frequency' => 0);
            }

            $dictionary[$term]['postings'][$doc_id]['term_frequency']++;
        }

        //from http://phpir.com/simple-search-the-vector-space-model/

    }

    return array('doc_count' => $doc_count, 'dictionary' => $dictionary);
}

function get_similar_documents($query='', $corpus=array(), $separator=' '){

    $similar_documents=array();

    if($query!=''&&!empty($corpus)){

        $words=explode($separator,$query);

        $corpus=get_corpus_index($corpus, $separator);

        $doc_count=count($corpus['doc_count']);

        foreach($words as $word) {

            if(isset($corpus['dictionary'][$word])){

                $entry = $corpus['dictionary'][$word];


                foreach($entry['postings'] as $doc_id => $posting) {

                    //get term frequency–inverse document frequency
                    $score=$posting['term_frequency'] * log($doc_count + 1 / $entry['document_frequency'] + 1, 2);

                    if(isset($similar_documents[$doc_id])){

                        $similar_documents[$doc_id]+=$score;

                    }
                    else{

                        $similar_documents[$doc_id]=$score;

                    }
                }
            }
        }

        // length normalise
        foreach($similar_documents as $doc_id => $score) {

            $similar_documents[$doc_id] = $score/$corpus['doc_count'][$doc_id];

        }

        // sort from  high to low

        arsort($similar_documents);

    }   

    return $similar_documents;
}

CASE 1

$query = 'are';

$corpus = array(
    1 => 'How are you?',
);

$match_results=get_similar_documents($query,$corpus);
echo '<pre>';
    print_r($match_results);
echo '</pre>';

RESULT

Array
(
    [1] => 0.52832083357372
)

CASE 2

$query = 'are';

$corpus = array(
    1 => 'how are you today?',
    2 => 'how do you do',
    3 => 'here you are! how are you? Are we done yet?'
);

$match_results=get_similar_documents($query,$corpus);
echo '<pre>';
    print_r($match_results);
echo '</pre>';

RESULTS

Array
(
    [1] => 0.54248125036058
    [3] => 0.21699250014423
)

CASE 3

$query = 'we are done';

$corpus = array(
    1 => 'how are you today?',
    2 => 'how do you do',
    3 => 'here you are! how are you? Are we done yet?'
);

$match_results=get_similar_documents($query,$corpus);
echo '<pre>';
    print_r($match_results);
echo '</pre>';

RESULTS

Array
(
    [3] => 0.6813781191217
    [1] => 0.54248125036058
)

There are plenty of improvements to be made but the model provides a way of getting good results from natural queries, which don't have boolean operators such as strpos(), preg_match(), strstr() or stristr().

NOTA BENE

Optionally eliminating redundancy prior to search the words

• thereby reducing index size and resulting in less storage requirement

• less disk I/O

• faster indexing and a consequently faster search.

1. Normalisation

• Convert all text to lower case

2. Stopword elimination

• Eliminate words from the text which carry no real meaning (like 'and', 'or', 'the', 'for', etc.)

3. Dictionary substitution

• Replace words with others which have an identical or similar meaning. (ex:replace instances of 'hungrily' and 'hungry' with 'hunger')

• Further algorithmic measures (snowball) may be performed to further reduce words to their essential meaning.

• The replacement of colour names with their hexadecimal equivalents

• The reduction of numeric values by reducing precision are other ways of normalising the text.

RESOURCES

• http://linuxgazette.net/164/sephton.html
• http://snowball.tartarus.org/
• MySQL Fulltext Search Score Explained
• http://dev.mysql.com/doc/internals/en/full-text-search.html
• http://en.wikipedia.org/wiki/Vector_space_model
• http://en.wikipedia.org/wiki/Tf%E2%80%93idf
• http://phpir.com/simple-search-the-vector-space-model/

Answer 9

Make use of case-insensitve matching using stripos():

if (stripos($string,$stringToSearch) !== false) {
    echo 'true';
}

Answer 10

If you want to avoid the "falsey" and "truthy" problem, you can use substr_count:

if (substr_count($a, 'are') > 0) {
    echo "at least one 'are' is present!";
}

It's a bit slower than strpos but it avoids the comparison problems.

Answer 11

if (preg_match('/(are)/', $a)) {
   echo 'true';
}

Answer 12

Another option is to use the strstr() function. Something like:

if (strlen(strstr($haystack,$needle))>0) {
// Needle Found
}

Point to note: The strstr() function is case-sensitive. For a case-insensitive search, use the stristr() function.

Answer 13

I'm a bit impressed that none of the answers here that used strpos, strstr and similar functions mentioned Multibyte String Functions yet (2015-05-08).

Basically, if you're having trouble finding words with characters specific to some languages, such as German, French, Portuguese, Spanish, etc. (e.g.: ä, é, ô, ç, º, ñ), you may want to precede the functions with mb_. Therefore, the accepted answer would use mb_strpos or mb_stripos (for case-insensitive matching) instead:

if (mb_strpos($a,'are') !== false) {
    echo 'true';
}

If you cannot guarantee that all your data is 100% in UTF-8, you may want to use the mb_ functions.

A good article to understand why is The Absolute Minimum Every Software Developer Absolutely, Positively Must Know About Unicode and Character Sets (No Excuses!) by Joel Spolsky.

Answer 14

In PHP, the best way to verify if a string contains a certain substring, is to use a simple helper function like this:

function contains($haystack, $needle, $caseSensitive = false) {
    return $caseSensitive ?
            (strpos($haystack, $needle) === FALSE ? FALSE : TRUE):
            (stripos($haystack, $needle) === FALSE ? FALSE : TRUE);
}

Explanation:

• strpos finds the position of the first occurrence of a case-sensitive substring in a string.
• stripos finds the position of the first occurrence of a case-insensitive substring in a string.
• myFunction($haystack, $needle) === FALSE ? FALSE : TRUE ensures that myFunction always returns a boolean and fixes unexpected behavior when the index of the substring is 0.
• $caseSensitive ? A : B selects either strpos or stripos to do the work, depending on the value of $caseSensitive.

Output:

var_dump(contains('bare','are'));            // Outputs: bool(true)
var_dump(contains('stare', 'are'));          // Outputs: bool(true)
var_dump(contains('stare', 'Are'));          // Outputs: bool(true)
var_dump(contains('stare', 'Are', true));    // Outputs: bool(false)
var_dump(contains('hair', 'are'));           // Outputs: bool(false)
var_dump(contains('aren\'t', 'are'));        // Outputs: bool(true)
var_dump(contains('Aren\'t', 'are'));        // Outputs: bool(true)
var_dump(contains('Aren\'t', 'are', true));  // Outputs: bool(false)
var_dump(contains('aren\'t', 'Are'));        // Outputs: bool(true)
var_dump(contains('aren\'t', 'Are', true));  // Outputs: bool(false)
var_dump(contains('broad', 'are'));          // Outputs: bool(false)
var_dump(contains('border', 'are'));         // Outputs: bool(false)

Answer 15

You can use the strstr function:

$haystack = "I know programming";
$needle   = "know";
$flag = strstr($haystack, $needle);

if ($flag){

    echo "true";
}

Without using an inbuilt function:

$haystack  = "hello world";
$needle = "llo";

$i = $j = 0;

while (isset($needle[$i])) {
    while (isset($haystack[$j]) && ($needle[$i] != $haystack[$j])) {
        $j++;
        $i = 0;
    }
    if (!isset($haystack[$j])) {
        break;
    }
    $i++;
    $j++;

}
if (!isset($needle[$i])) {
    echo "YES";
}
else{
    echo "NO ";
}

Answer 16

The function below also works and does not depend on any other function; it uses only native PHP string manipulation. Personally, I do not recommend this, but you can see how it works:

<?php

if (!function_exists('is_str_contain')) {
  function is_str_contain($string, $keyword)
  {
    if (empty($string) || empty($keyword)) return false;
    $keyword_first_char = $keyword[0];
    $keyword_length = strlen($keyword);
    $string_length = strlen($string);

    // case 1
    if ($string_length < $keyword_length) return false;

    // case 2
    if ($string_length == $keyword_length) {
      if ($string == $keyword) return true;
      else return false;
    }

    // case 3
    if ($keyword_length == 1) {
      for ($i = 0; $i < $string_length; $i++) {

        // Check if keyword's first char == string's first char
        if ($keyword_first_char == $string[$i]) {
          return true;
        }
      }
    }

    // case 4
    if ($keyword_length > 1) {
      for ($i = 0; $i < $string_length; $i++) {
        /*
        the remaining part of the string is equal or greater than the keyword
        */
        if (($string_length + 1 - $i) >= $keyword_length) {

          // Check if keyword's first char == string's first char
          if ($keyword_first_char == $string[$i]) {
            $match = 1;
            for ($j = 1; $j < $keyword_length; $j++) {
              if (($i + $j < $string_length) && $keyword[$j] == $string[$i + $j]) {
                $match++;
              }
              else {
                return false;
              }
            }

            if ($match == $keyword_length) {
              return true;
            }

            // end if first match found
          }

          // end if remaining part
        }
        else {
          return false;
        }

        // end for loop
      }

      // end case4
    }

    return false;
  }
}

Test:

var_dump(is_str_contain("test", "t")); //true
var_dump(is_str_contain("test", "")); //false
var_dump(is_str_contain("test", "test")); //true
var_dump(is_str_contain("test", "testa")); //flase
var_dump(is_str_contain("a----z", "a")); //true
var_dump(is_str_contain("a----z", "z")); //true 
var_dump(is_str_contain("mystringss", "strings")); //true 

Answer 17

I had some trouble with this, and finally I chose to create my own solution. Without using regular expression engine:

function contains($text, $word)
{
    $found = false;
    $spaceArray = explode(' ', $text);

    $nonBreakingSpaceArray = explode(chr(160), $text);

    if (in_array($word, $spaceArray) ||
        in_array($word, $nonBreakingSpaceArray)
       ) {

        $found = true;
    }
    return $found;
 }

You may notice that the previous solutions are not an answer for the word being used as a prefix for another. In order to use your example:

$a = 'How are you?';
$b = "a skirt that flares from the waist";
$c = "are";

With the samples above, both $a and $b contains $c, but you may want your function to tell you that only $a contains $c.

Answer 18

Lot of answers that use substr_count checks if the result is >0. But since the if statement considers zero the same as false, you can avoid that check and write directly:

if (substr_count($a, 'are')) {

To check if not present, add the ! operator:

if (!substr_count($a, 'are')) {

Answer 19

Another option to finding the occurrence of a word from a string using strstr() and stristr() is like the following:

<?php
    $a = 'How are you?';
    if (strstr($a,'are'))  // Case sensitive
        echo 'true';
    if (stristr($a,'are'))  // Case insensitive
        echo 'true';
?>

Answer 20

It can be done in three different ways:

 $a = 'How are you?';

1- stristr()

 if (strlen(stristr($a,"are"))>0) {
    echo "true"; // are Found
 } 

2- strpos()

 if (strpos($a, "are") !== false) {
   echo "true"; // are Found
 }

3- preg_match()

 if( preg_match("are",$a) === 1) {
   echo "true"; // are Found
 }

Answer 21

The short-hand version

$result = false!==strpos($a, 'are');

Answer 22

In order to find a 'word', rather than the occurrence of a series of letters that could in fact be a part of another word, the following would be a good solution.

$string = 'How are you?';
$array = explode(" ", $string);

if (in_array('are', $array) ) {
    echo 'Found the word';
}

Answer 23

You should use case Insensitive format,so if the entered value is in small or caps it wont matter.

<?php
$grass = "This is pratik joshi";
$needle = "pratik";
if (stripos($grass,$needle) !== false) { 

 /*If i EXCLUDE : !== false then if string is found at 0th location, 
   still it will say STRING NOT FOUND as it will return '0' and it      
   will goto else and will say NOT Found though it is found at 0th location.*/
    echo 'Contains word';
}else{
    echo "does NOT contain word";
}
?>

Here stripos finds needle in heystack without considering case (small/caps).

PHPCode Sample with output

Answer 24

Maybe you could use something like this:

<?php
    findWord('Test all OK');

    function findWord($text) {
        if (strstr($text, 'ok')) {
            echo 'Found a word';
        }
        else
        {
            echo 'Did not find a word';
        }
    }
?>

Answer 25

Do not use preg_match() if you only want to check if one string is contained in another string. Use strpos() or strstr() instead as they will be faster. (http://in2.php.net/preg_match)

if (strpos($text, 'string_name') !== false){
   echo 'get the string';
}

Answer 26

If you want to check if the string contains several specifics words, you can do:

$badWords = array("dette", "capitale", "rembourser", "ivoire", "mandat");

$string = "a string with the word ivoire";

$matchFound = preg_match_all("/\b(" . implode($badWords,"|") . ")\b/i", $string, $matches);

if ($matchFound) {
    echo "a bad word has been found";
}
else {
    echo "your string is okay";
}

This is useful to avoid spam when sending emails for example.

Answer 27

The strpos function works fine, but if you want to do case-insensitive checking for a word in a paragraph then you can make use of the stripos function of PHP.

For example,

$result = stripos("I love PHP, I love PHP too!", "php");
if ($result === false) {
    // Word does not exist
}
else {
    // Word exists
}

Find the position of the first occurrence of a case-insensitive substring in a string.

If the word doesn't exist in the string then it will return false else it will return the position of the word.

Answer 28

You need to use identical/not identical operators because strpos can return 0 as it's index value. If you like ternary operators, consider using the following (seems a little backwards I'll admit):

echo FALSE === strpos($a,'are') ? 'false': 'true';

Answer 29

Check if string contains specific words?

This means the string has to be resolved into words (see note below).

One way to do this and to specify the separators is using preg_split (doc):

<?php

function contains_word($str, $word) {
  // split string into words
  // separators are substrings of at least one non-word character
  $arr = preg_split('/\W+/', $str, NULL, PREG_SPLIT_NO_EMPTY);

  // now the words can be examined each
  foreach ($arr as $value) {
    if ($value === $word) {
      return true;
    }
  }
  return false;
}

function test($str, $word) {
  if (contains_word($str, $word)) {
    echo "string '" . $str . "' contains word '" . $word . "'\n";
  } else {
    echo "string '" . $str . "' does not contain word '" . $word . "'\n" ;
  }
}

$a = 'How are you?';

test($a, 'are');
test($a, 'ar');
test($a, 'hare');

?>

A run gives

$ php -f test.php                   
string 'How are you?' contains word 'are' 
string 'How are you?' does not contain word 'ar'
string 'How are you?' does not contain word 'hare'

Note: Here we do not mean word for every sequence of symbols.

A practical definition of word is in the sense the PCRE regular expression engine, where words are substrings consisting of word characters only, being separated by non-word characters.

A "word" character is any letter or digit or the underscore character, that is, any character which can be part of a Perl " word ". The definition of letters and digits is controlled by PCRE's character tables, and may vary if locale-specific matching is taking place (..)

Answer 30

Use:

$text = 'This is a test';
echo substr_count($text, 'is'); // 2

// So if you want to check if is exists in the text just put
// in a condition like this:
if (substr_count($text, 'is') > 0) {
    echo "is exists";
}

Answer 31

Another solution for a specific string:

$subject = 'How are you?';
$pattern = '/are/';
preg_match($pattern, $subject, $match);
if ($match[0] == 'are') {
    echo true;
}

You can also use strpos() function.

Answer 32

You can also use built-in functions strchr() and strrchr() and extensions for multibyte strings mb_strchr() and mb_strrchr(). These functions return parts of strings, and FALSE if nothing is found.

• strchr() - Find the first occurrence of a string (is an alias of strstr()).
• strrchr() - Find the last occurrence of a character in a string.

Answer 33

A string can be checked with the below function:

function either_String_existor_not($str, $character) {
    return strpos($str, $character) !== false;
}

Answer 34

I think that a good idea is to use mb_stpos:

$haystack = 'How are you?';
$needle = 'are';

if (mb_strpos($haystack, $needle) !== false) {

    echo 'true';
}

Because this solution is case sensitive and safe for all Unicode characters.

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But you can also do it like this (sauch response was not yet):

if (count(explode($needle, $haystack)) > 1) {

    echo 'true';
}

This solution is also case sensitive and safe for Unicode characters.

In addition you do not use the negation in the expression, which increases the readability of the code.

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Here is other solution using function:

function isContainsStr($haystack, $needle) {

    return count(explode($needle, $haystack)) > 1;
}

if (isContainsStr($haystack, $needle)) {

    echo 'true';
}

Answer 35

Use:

$a = 'How are you?';
if (mb_strpos($a, 'are')) {
    echo 'true';
}

It performs a multi-byte safe strpos() operation.

Answer 36

A simpler option:

return ( ! empty($a) && strpos($a, 'are'))? true : false;