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($a包含“are”),写语句的正确方法是什么?
当前回答
如果只想检查一个字符串是否包含在另一个字符串中,请不要使用preg_match()。改用strpos()或strstr(),因为它们会更快。(http://in2.php.net/preg_match)
if (strpos($text, 'string_name') !== false){
echo 'get the string';
}
其他回答
这里有一个小的实用函数,在这种情况下很有用
// returns true if $needle is a substring of $haystack
function contains($needle, $haystack)
{
return strpos($haystack, $needle) !== false;
}
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($text, 'is') > 0) {
echo "is exists";
}
许多使用subst_count的答案会检查结果是否大于0。但由于if语句将零视为false,因此可以避免该检查并直接写入:
if (substr_count($a, 'are')) {
若要检查是否不存在,请添加!操作员:
if (!substr_count($a, 'are')) {
检查字符串是否包含特定单词?
这意味着字符串必须解析为单词(请参见下面的注释)。
执行此操作并指定分隔符的一种方法是使用preg_split(doc):
<?php
function contains_word($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');
?>
跑一次就可以
$ 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'
注意:这里我们不是指每个符号序列的单词。
单词的实际定义是PCRE正则表达式引擎,其中单词是仅由单词字符组成的子字符串,由非单词字符分隔。
“单词”字符是任何字母、数字或下划线字符,也就是说,可以是Perl“单词”一部分的任何字符。这个字母和数字的定义由PCRE的字符控制表,如果正在进行特定于区域设置的匹配,则可能会有所不同(..)
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.";
}