如果在正则表达式中存在特定的域，如何跳过并说不匹配？

Question

我想跳过和preg_match() 0返回一个字符串或文本包含例如

$var="for see our post visit phyteachers.com |click now"; 
if(preg_match('(((https?|ftps?)\:\/\/)?((www([0-9]+)?)\.)?.*\.[a-zA-Z0-9](\/)?((?!.*phyteachers\b).*))','$var')==0){ 
    echo "ok"; 
} 

如果http://phyteachers.com或http://www.phyteachers.com或https://www.phyteachers.com或https://phyteachers.com或www.phyteachers.com或phyteachers.com然后返回0特定领域。

Answer 1

我的下列模式可能会稍微改进一点，但它应该可以满足您的目的。

~(?:(?:ht|f)tps?:/{2})?(?:w{3}\d*\.)?(?:phyteachers\.com(*SKIP)(*FAIL)|\S+\.[a-z\d]+/?\S*)~ 

https://regex101.com/r/iZ9HRQ/6

有关我的模式的组件正式用语，看到的演示链接。否则，我会提供一个非正式的解释...

(?:     # start optional non-capturing group 
    (?:ht|f)tps? # match http, https, ftp, ftps 
    :/{2}   # match colon then two slashes 
)?     # end optional group (none of this HAS to exist) 
(?:     # start optional non-capturing group 
    w{3}\d*\.  # match www then zero or more digits, then optiona 
)?     # end optional group 
(?:        # start non-capturing group 
    phyteachers\.com(*SKIP)(*FAIL) # abort match if found 
    |        # or 
    \S+\.[a-z\d]+/?\S*    # see the demo, too involved to explain in limited space 
)         # end non-capturing group