如何删除点后跟非数字或任何特殊字符？

Question

Example: ABCD, NC exg. 58/2095, s.2.7 A. 2A. ABCD, NC exg. 58/2095, s.2.7 A. 2A. is the text and am using Regexp

`String.replace(/([az]+).|.([ ?!\\d])/ig, '$1')

O/p : Missing some Text from the Text.

Expected is: ABCD, NC exg 58/2095, s.2.7 A2A

--dot followed by non digit(special Char or Alphabets) should replace with null.

Answer 1

You can use alternation and grouping.

([az]+)\\. - Matches one or more character (captures as group) followed by dot.

\\.([az]+) - Matches dot followed by one or more character (captures chracters as group)

And in replace by the matched group.

 let str = `ABCD, NC exg. 58/2095, s. 2.7 ` let op = str.replace(/([az]+)\\.|\\.([az]+)/ig, '$1') console.log(op)

Answer 2

If your input is only ASCII, just replace every letter followed by a dot with the character itself:

 console.log( "ABCD, NC exg. 58/2095, s. 2.7".replace(/([az])\\./ig, '$1') );

Removing every special character.

Because that's what you are telling it to do. [^\\w.\\s] matches every character that is not a letter, number, _ , . or white space.