CODEWITHSUNDEEP
javascriptregexstringreplace
paligap
paligap

Reputation: 942

Adding a condition to a regex

Given the Javascript below how can I add a condition to the clause? I would like to add a "space" character after a separator only if a space does not already exist. The current code will result in double-spaces if a space character already exists in spacedText.

var separators = ['.', ',', '?', '!'];
for (var i = 0; i < separators.length; i++) { 
      var rg = new RegExp("\\" + separators[i], "g"); 
      spacedText = spacedText.replace(rg, separators[i] + " "); 
}

Upvotes: 0

Views: 100

Answers (5)

flitig
flitig

Reputation: 531

I would suggest the following regexp to solve your problem:

"Test!Test! Test.Test 1,2,3,4 test".replace(/([!,.?])(?!\s)/g, "$1 ");
// "Test! Test! Test. Test 1, 2, 3, 4 test"

The regexp matches any character in the character class [!,.?] not followed by a space (?!\s). The parenthesis around the character class means that the matched separator will be contained in the first backreference $1, which is used in the replacement string. See this fiddle for working example.

Upvotes: 2

Lekhnath
Lekhnath

Reputation: 4625

Here is a working code : 

var nonSpaced= 'Hello World!Which is your favorite number? 10,20,25,30 or other.answer fast.';
var spaced;
var patt = /\b([!\.,\?])+\b/g;

spaced = nonSpaced.replace(patt, '$1 ');

If you console.log the value of spaced, It will be : Hello World! Which is your favorite number? 10, 20, 25, 30 or other. answer fast. Notice the number of space characters after the ? sign , it is only one, and there is not extra space after last full-stop.

Upvotes: 0

2cent
2cent

Reputation: 211

'. , ? ! .,?!foo'.replace(/([.,?!])(?! )/g, '$1 ');
//-> ". , ? ! . , ? ! foo"

Means replace every occurence of one of .,?! that is not followed by a space with itself and a space afterwards.

Upvotes: 2

pawel
pawel

Reputation: 36965

spacedText = spacedText.replace(/([\.,!\?])([^\s])/g,"$1 ")

This means: replace one of these characters ([\.,!\?]) followed by a non-whitespace character ([^\s]) with the match from first group and a space ("$1 ").

Upvotes: 0

Ben
Ben

Reputation: 13615

You could do a replace of all above characters including a space. In that way you will capture any punctuation and it's trailing space and replace both by a single space.

"H1a!. A ?. ".replace(/[.,?! ]+/g, " ")

[.,?! ] is a chararcter class. It will match either ., ,, ?, ! or and + makes it match atleast once (but if possible multiple times).

Upvotes: 0

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