Javascript string replace using regExp and replacer function

Question

How to use AND operator in regexp and string replacer function. Please see the below code sample. Here, since I am using a | (OR) operator I am getting an output like this

["I am number [1]", "I am number [1]2", "I am number [1]2[3]", "I am number [1]2[3]4", "I am number [1]2[3]45", "I am number [1]2[3]456"]

So, instead of OR how to use AND so that only strings containing 1 & 3 should be effected not others. I have tried using

var patt = /(?=.*1)(?=.*3)/g;

it did not work, although if I do patt.test("String") it works fine. But, with replacer function it is not. Can any one please help

var arr = [ "I am number 1", 
"I am number 12", "I am number 123", "I am number 1234",
"I am number 12345", "I am number 123456"];

var patt = /1|3/g;
for (var i = 0; i < arr.length; i++) {
    arr[i] = arr[i].replace(patt, function(match) {
        if (match.length) return '['+match+']';
        else return match;
    });
};
console.log(arr);

EDIT: An indirect way is to have two patterns

var pattToReplace = /1|3/g;

var pattToValidate = /(?=.*1)(?=.*3)/g;

var arr = [ "I am number 1", 
    "I am number 12", "I am number 123", "I am number 1234",
    "I am number 12345", "I am number 123456", "I am number 33234156"];

var pattToReplace = /1|3/g;
var pattToValidate = /(?=.*1)(?=.*3)/g;
for (var i = 0; i < arr.length; i++) {
	if (pattToValidate.test(arr[i])) {
		arr[i] = arr[i].replace(pattToReplace, function(match) {
			if (match.length) return '['+match+']';
			else return match;
		});
	}
};
document.write(JSON.stringify(arr, null, 2));

Use 1\d*3\d*. Or \d*1\d*3\d* if you expect [ "I am number 1", "I am number 12", "I am number [123]", "I am number [1234]", "I am number [12345]", "I am number [123456]" ]. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Feb 25, 2016 at 11:11
@gurvinder372 I have mentioned it in the post....." only strings containing 1 & 3 should be effected not others" — Roy
– Roy, Commented Feb 25, 2016 at 11:14
After affecting what should it do? What is replace function suppose to achieve? — gurvinder372
– gurvinder372, Commented Feb 25, 2016 at 11:15
why not arr[i] = arr[i].replace(/(?:(1)([0-9]*)(3))|(?:(3)([0-9]*)(1))/g, "[$1]$2[$3]"); — snoopen
– snoopen, Commented Feb 25, 2016 at 11:48

Wiktor Stribiżew · Accepted Answer · 2016-02-25 20:27:04Z

3

Update for a more generic solution

RegExp.escape = function(s) {
    return s.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');
};
var arr = [ "I am number (a)", 
"I am number -c-+b+", "I am number (a)+b+-c-", "I am number (a)+b+-c--d-", "I am number (a)+b+-c--d--e-", "I am number (a)+b+-c--d--e--f-", "I am number -c--c-+b+-c--d-(a)-e--f-"];
var key1 = "(a)";
var key2 = "+b+";
var patt = RegExp("(" + RegExp.escape(key1) + "|" + RegExp.escape(key2) + ")", "g");
for (var i = 0; i < arr.length; i++) {
    if (arr[i].indexOf(key1) > -1 && arr[i].indexOf(key2) > -1) {
    	arr[i] = arr[i].replace(patt, '[$1]');
    } 
};
document.body.innerHTML =  "<pre>" + JSON.stringify(arr, 0, 4) + "</pre>";

Code description:

You will have to get user input first (in the dem, key1 and key2). Then, you need to escape them (see RegExp.escape) for use in a regular expression and build the regexps dynamically(see how patt is built). Then, iterate through the input strings, check if each contains both key1 and key2 with indexOf method, and if yes, perform the replacements.

Original answer for numeric-only input

To match 1 or 3 in a chunk of consecutive digits that must contain 1 and then 3, use

/\b(?=[0-24-9]*3)(?=[02-9]*1)\d+\b/g

And then just replace all 1 and 3 with [1] and [3] with .replace(/([13])/g, '[$1]':

var arr = [ "I am number 1", 
"I am number 12", "I am number 123", "I am number 1234",
"I am number 12345", "I am number 123456", "I am number 33234156"];

var patt = /\b(?=[0-24-9]*3)(?=[02-9]*1)\d+\b/g;
for (var i = 0; i < arr.length; i++) {
    arr[i] = arr[i].replace(patt, function(match) {
        if (match.length) return match.replace(/([13])/g, '[$1]');
        else return match;
    });
};
document.body.innerHTML =  "<pre>" + JSON.stringify(arr, 0, 4) + "</pre>";

Regex explanation:

\b - leading word boundary (this allows matching whole words only, and makes it possible to execute the following lookahead checks only once)
(?=[0-24-9]*3) - a positive lookahead that will fail a match if there is no zero or more digits other than 3 followed by 3 (so, 3 must be inside a bigger chunk of digits)
(?=[02-9]*1) - a positive lookahead that will fail a match if there is no zero or more digits other than 1 followed by 1 (so, 1 must be inside a bigger chunk of digits)
\d+ - one or more digits (matching them, to return the result)
\b - trailing word boundary

edited Feb 25, 2016 at 20:27

answered Feb 25, 2016 at 11:17

Wiktor Stribiżew

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5 Comments

Roy Over a year ago

This is useful... but I wanted to know is there a generic way to handle such scenarios.. because this was a sample code which I posted and the solution provided by you is specific to digits only.

Wiktor Stribiżew Over a year ago

What do you consider a generic solution? Remember that regular expressions deal with patterns. What type of pattern do you need? Any non-whitespace? Any characters including a newline?

Roy Over a year ago

By generic I meant, what if from this array some user wants replace only those strings which contains both the inputted keywords like 1 & 3 (can be any two strings like for that matter). How to handle such scenario, is it possible using regEx. For this question and as a temporary solution I am going with an approach which I updated in post (please see edit). Looking forward to your suggestion.

Wiktor Stribiżew Over a year ago

I have posted a new suggestion.

Roy Over a year ago

Nice, before replacing we need to put an additional check.... so, we were on the same track... but your approach looks more generic, with the use of "escape" function.

Vince · Accepted Answer · 2016-02-25 11:44:56Z

I believe /.*(?=1).*(?=3).*/ will do most of what you want, but it doesn't account for strings which have the numbers in the wrong order. The difference is that the .* is outside of the parentheses.

var arr = [
  "I am number 1",
  "I am number 12",
  "I am number 123",
  "I am number 1234",
  "I am number 12345",
  "I am number 123456",
  "I am number 54321"
];

var result = "";
var patt = /.*(?=1).*(?=3).*/g;
for (var i = 0; i < arr.length; i++) {
    arr[i] = arr[i].replace(patt, function(match) {
        if (match.length) result += '['+match+']' + "\n";
    });
};
document.getElementById('result').innerHTML = result;

html, body, textarea {
  margin: 0;
  padding: 0;
  width: 100%;
  height: 100%;
  overflow: hidden;
}

<html>
  <body>
    <textarea id="result"></textarea>
  </body>
</html>

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Javascript string replace using regExp and replacer function

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Update for a more generic solution

Original answer for numeric-only input

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Original answer for numeric-only input

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