如何在PHP中使用正则表达式来获得双反斜杠之后的东西

Question

I have this sentence

C:\\wamp\\www\\callCenter\\joomlatools-files\\docman-files\\test.pdf

And I want to get as a result of my regex exression test but without using another word from the sentence.

My regex is ~\\\(.+)\.pdf~ but I have wamp\www\portail-callcenter\joomlatools-files\docman-files\test as a result. Can someone please explain how to get it done?

Answer 1

I'm not sure what are you trying to get. But I guess you want to get the word of test from your first sentence. So use this pattern:

~\\\\(\w+)\.pdf$~

And then $1 containing what you need.

Online Demo

• ~ delimiter
• \\\\ matches two backslashes literally
• ( capturing group $1 (which is containing what you need to get)
• \w+ matches one or more letter, number or underscore
• \. matches a dot literally
• $ end of string

---

Here is your pattern in the question ~\\\(.+)\.pdf~. The problem of your pattern is .+. Because it means one or more character(s) (every character even \). So your pattern matches that two backslashes which are in the beginning of your string, and the rest of backslashes will be match as every character (.+).

To avoid this, you have to use \w which means any letter, number or underscore instead. In this case the matched part will be just test.