考虑:
$a = 'How are you?';
if ($a contains 'are')
echo 'true';
假设我有上面的代码,如果($a包含“are”),写语句的正确方法是什么?
考虑:
$a = 'How are you?';
if ($a contains 'are')
echo 'true';
假设我有上面的代码,如果($a包含“are”),写语句的正确方法是什么?
当前回答
if (preg_match('/(are)/', $a)) {
echo 'true';
}
其他回答
使用strstr()和stristr(()从字符串中查找单词出现的另一个选项如下:
<?php
$a = 'How are you?';
if (strstr($a,'are')) // Case sensitive
echo 'true';
if (stristr($a,'are')) // Case insensitive
echo 'true';
?>
Use:
$a = 'How are you?';
if (mb_strpos($a, 'are')) {
echo 'true';
}
它执行多字节安全strpos()操作。
查看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 $pos.";
}
我在这方面遇到了一些麻烦,最后我选择了自己的解决方案。不使用正则表达式引擎:
function contains($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;
}
您可能会注意到,前面的解决方案并不是对另一个用作前缀的单词的答案。为了使用您的示例:
$a = 'How are you?';
$b = "a skirt that flares from the waist";
$c = "are";
对于上面的示例,$a和$b都包含$c,但您可能希望函数告诉您,只有$a包含$c。
可以通过三种不同的方式完成:
$a = 'How are you?';
1-stristr()
if (strlen(stristr($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
}