正则表达式选择不包含某些字符的文本

Question

I have a text like this

MyText.xyz;
MyText.abc + "ss";

I want to capture text which comes after MyText. , but does not end with ; or empty space, ie I want the output like:

MyText.xyz
MyText.abc

I have created this regex: MyText.(.*(?=\\;)) in order to achieve my goal.

But, right now the text matching out from regex is MyText.xyz and MyText.abc + "ss" , ie the second result is incorrect.

Answer 1

You may fix it using a negated character class:

MyText\.[^\s;]+
        ^^^^^^^ 

See the regex demo

Regex details

• MyText\\. - a literal MyText. substring (note the . must be escaped to match a literal . char)
• [^\\s;]+ - a negated character class matching any 1+ chars other than whitespace ( \\s ) and ; char

Use it as var pattern = @"MyText\\.[^\\s;]+"; in C#.

Answer 2

MyText\..+?\b(?<!;)
   ^  ^ ^  ^  ^
   |__|_|__|__|______ MyText : "MyText" literal
      |_|__|__|______ \.     : "." literal, escaped by "\"
        |__|__|______ .      : any character
           |__|______ ?\b    : non-greedy search up to boundary (\b)
              |______ (?<!;) : not ended by ';'

Test:

$ cat sample.txt
MyText.xyz;
MyText.abc + "ss";
MyText.uuu+"yyy";

$ grep -Po 'MyText\..+?\b(?<!;)' <sample.txt
MyText.xyz
MyText.abc
MyText.uuu

Note: It's based on solution of @Wiktor Stribiżew, with looking behind added

Answer 3

You are using

MyText.(.*(?=\;))

The first mistake is the . after MyText , it should be \\. if you want to match a literal dot.

The second half is also incorrect, you're trying to match any number of non-linebreaking-characters, followed by a ; , which is why you got the results you did.

Try this regex instead:

MyText.[^ ;]*

The [^ ;]* matches any character that is not a space or a ; . If you also don't want tabs or linebreaks to match, you can use the following instead:

MyText.[^\s;]*

\\s matches any whitespace character.