PHP 判断是否包含在某个字符串中

太过爱你忘了你带给我的痛 2022-03-28 03:47 559阅读 0赞

PHP 判断是否包含在某个字符串中

1.用strpos函数，查找字符首次出现的位置，如果不存在则会返回false  
$str= ‘abc’;  
$needle= ‘e’;  
 p o s = s t r p o s ( pos = strpos( pos=strpos(str, $needle);  
2.用strstr函数查找字符在字符串中首次出现，并返回字符出现开始的位置至结束的字符串  
$email = ‘user@example.com’;  
 d o m a i n = s t r s t r ( domain = strstr( domain=strstr(email, ‘@’);  
3.用explode分割字符串，如果返回的数组长度大于一则表示存在  
$str= ‘abc’;  
$needle = ‘E’;//判断是否包含a这个字符  
 t m p a r r a y = e x p l o d e ( tmparray = explode( tmparray=explode(needle,$str);  
4.用in\_array检查数组中是否存在某个值  
// bool in\_array ( mixed $needle , array $haystack \[, bool $strict = FALSE \] )  
//in\_array()例子  
 o s = a r r a y ( " M a c " , " N T " , " I r i x " , " L i n u x " ) ; i f ( i n a r r a y ( " I r i x " , os=array("Mac","NT","Irix","Linux"); if(in\_array("Irix", os=array("Mac","NT","Irix","Linux");if(inarray("Irix",os))\{  
echo “Got Irix”;  
\}  
if(in\_array(“mac”,$os))\{  
echo “Got mac”;  
\}  
//第二个条件失败，因为in\_array()是区分大小写的 所以只输出Got Irix  
//in\_array() 严格类型检查例子  
$a = array(‘1.10’, 12.4, 1.13);  
if (in\_array(‘12.4’, $a, true)) \{  
echo “‘12.4’ found with strict check\\n”;  
\}  
if (in\_array(1.13, $a, true)) \{  
echo “1.13 found with strict check\\n”;  
\}  
//程序会输出1.13 found with strict check  
//in\_array()中数组作为needle  
$a = array(array(‘p’, ‘h’), array(‘p’, ‘r’), ‘o’);  
if (in\_array(array(‘p’, ‘h’), $a)) \{  
echo “‘ph’ was found\\n”;  
\}  
if (in\_array(array(‘f’, ‘i’), $a)) \{  
echo “‘fi’ was found\\n”;  
\}  
if (in\_array(‘o’, $a)) \{  
echo “‘o’ was found\\n”;  
\}  
//程序会输出  
//‘ph’ was found  
//‘o’ was found