如何检查字符串是否包含数组中的单词然后回显它？

Question

我有一个包含位置的句子 例如 纽约开发商活动

在我的数组中我有一个位置列表

我想检查字符串是否包含数组中的位置，如果是，则将其回显

$string = 'Activities in New York for developers';
$array = array("New York","Seattle", "San Francisco");
if(0 == count(array_intersect(array_map('strtolower', explode(' ', $string)), $array))){

echo $location /*echo out location that is in the string on its own*/

} else {

/*do nothing*/

}

Answer 1

像这样遍历你的数组：

$string = 'Activities in New York for developers';
$array = array("New York","Seattle", "San Francisco");
//loop over locations
foreach($array as $location) {
    //strpos will return false if the needle ($location) is not found in the haystack ($string)
    if(strpos($string, $location) !== FALSE) {
        echo $string;
        break;
    }
}

修改

如果您想回显该位置，只需使用$string更改$location：

$string = 'Activities in New York for developers';
$array = array("New York","Seattle", "San Francisco");
//loop over locations
foreach($array as $location) {
    //strpos will return false if the needle ($location) is not found in the haystack ($string)
    if(strpos($string, $location) !== FALSE) {
        echo $location;
        break;
    }
}

Answer 2

使用stripos

$string = 'Activities in New York for developers';
$array = array("New York","Seattle", "San Francisco");
foreach($array as $location) {
    if(stripos($string, $location) !== FALSE) {
        echo $string;
        break;
    }
}