CODEWITHSUNDEEP
javascriptregex
samiles
samiles

Reputation: 3900

Match regular expression to next special character while allowing spaces

I'm using Javascript to extract parts of a string using a regex.

I have a string like:

lorem ipsum !bang #hash #hash2 ^caret word @at sym

I am trying to pull out the words beginning with the various characters - sometimes they can have a space in them, and there can be multiple of each type. So I want to convert this string to a set of values such as:

text: "lorem ipsum"
!: "bang"
#: ["hash", "hash2"]
^: "caret word"
@: "at sym"

My current regex is / ([!#^@>\/*-]\w+)/gm. This sort of works, but it does not match spaces - so from the sample above it only produces caret for ^, and not the whole caret word.

My code to do this is:

var result = {};

var re = / ([!#^@>\/*-]\w+)/gm;
var m;

var firstSpecialCharIndex = inputString.search(/ [!#^@>\/*-]/);
result["text"] = inputString.substring(0, firstSpecialCharIndex);

while ((m = re.exec(inputString)) !== null) {
  if (m.index === re.lastIndex) {
    re.lastIndex++;
  }

  var index = m[1].substring(0,1);
  if(result[index] == null)
    result[index] = [];
  result[index].push(m[1].substring(1));
}

Does anyone know how I can match up to the next special character, including the space between multiple words (but not the space after it to the next special char)? Many thanks

Upvotes: 2

Views: 81

Answers (3)

Me.Name
Me.Name

Reputation: 12544

Just a suggestion for an alternative, but if a regex is used to split on the specialchars (preceding with a non capturing space and lookahead alphanumeric char), while keeping the captured special char, the logic could be rewritten to:

var inputString = 'lorem ipsum !bang #hash #hash2 ^caret word @at sym'

var rx = /(?:\s)([!#^@>\/*-](?=\w))/;
var arr = inputString.split(rx);
var result = {text:  arr[0]};
for(var i = 1; i < arr.length; i++){
		var ind = arr[i++], val = arr[i];
    var coll = (result[ind] = result[ind] || []);
    coll.push(val);    
}

console.log(JSON.stringify(result));

The main advantage is that the special chars are not repeated in the expression. A small secondary one is that the search is executed only once (the 'text' part is simply the first element in the results). It will also work with multiple words and/or special chars in the middle of a word such as 'lorem ipsum !bang #ha/sh adfa #ha3sh2 ^caret word asdf @at sym'

Upvotes: 0

rock321987
rock321987

Reputation: 11032

I am removing text part for simplicity. You can use lookahead

([!#^@>\/*-])(.*?)(?=\s[!#^@>\/*-]|$)

Regex Demo

Group 1 contains symbol and group 2 contains text and you can trim the result if you want.

JS Demo

var inputString = "lorem ipsum !bang #hash #hash2 ^caret word @at sym";

var result = {};

var re = /([!#^@>\/*-])(.*?)(?=\s[!#^@>\/*-]|$)/gm;
var m;

var firstSpecialCharIndex = inputString.search(/ [!#^@>\/*-]/);
result["text"] = inputString.substring(0, firstSpecialCharIndex);

while ((m = re.exec(inputString)) !== null) {
    var index = m[1];
    if(result[index] == null) {
        result[index] = [];
    }
    result[index].push(m[2].trim());
}
document.writeln("<pre>" + JSON.stringify(result) + "</pre>");

Upvotes: 1

spirit
spirit

Reputation: 3415

Try this:

/ ((?:[!#^@>\/*-]\w+)(?: [^!#^@>\/*-]\w+)*)/gm

regex101 example

?: in the parenthesis makes so subpattern does not do any capturing. remove it and look what's changed.

Tested this code in the Chrome:

var inputString = "lorem ipsum !bang #hash #hash2 ^long caret word @at sym";

var result = {};

//var re = / ([!#^@>\/*-]\w+)/gm;
var re = / ((?:[!#^@>\/*-]\w+)(?: [^!#^@>\/*-]\w+)*)/gm;
var m;

var firstSpecialCharIndex = inputString.search(/ [!#^@>\/*-]/);
result["text"] = inputString.substring(0, firstSpecialCharIndex);

while ((m = re.exec(inputString)) !== null) {
    if (m.index === re.lastIndex) {
        re.lastIndex++;
    }

    var index = m[1].substring(0,1);
    if(result[index] == null) {
        result[index] = [];
    }
    result[index].push(m[1].substring(1));
}
console.log(result);

it works well.

Upvotes: 0

Related Questions