$a = 'How are you?';

if ($a contains ‘are’)
echo ‘true’;

假设我有上面的代码,编写语句 if ($a contains 'are') 的正确方法是什么?

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';


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;

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).

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

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.

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

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);


  • 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.


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)

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')) {

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.

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

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