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
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 $pos.";
}
其他回答
另一个选项是使用strstr()函数。类似于:
if (strlen(strstr($haystack,$needle))>0) {
// Needle Found
}
注意:strstr()函数区分大小写。对于不区分大小写的搜索,请使用stristr()函数。
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.";
}
如果搜索不区分大小写,则使用strstr()或stristr(。
短手版本
$result = false!==strpos($a, 'are');
现在,使用PHP 8,您可以使用str_contains执行此操作:
if (str_contains('How are you', 'are')) {
echo 'true';
}
RFC
PHP 8之前
您可以使用strpos()函数,该函数用于查找一个字符串在另一个字符串中的出现:
$haystack = 'How are you?';
$needle = 'are';
if (strpos($haystack, $needle) !== false) {
echo 'true';
}
请注意,使用!==false是故意的(!=false或==true都不会返回期望的结果);strpos()返回大海捞针字符串开始时的偏移量,如果找不到针,则返回布尔值false。由于0是有效的偏移量,0是“false”,因此我们不能使用更简单的构造,如!strpos($a,'are')。