CODEWITHSUNDEEP
javascriptnode.js
Cain Nuke
Cain Nuke

Reputation: 3103

How can I find an exact substring using String.prototype.search?

I have this code:

const ids = '[2][13]';
const myid = '1';

const result = ids.search('[' + myid + ']') != -1 ? 'found' : 'not found';
console.log(result);

this is returning found whereas it should be returning not found because id #1 is not present. However since [13] has a 1 in it then it misinterprets it. How can I fix this so it searches for the EXACT value match?

Upvotes: 0

Views: 188

Answers (5)

T.J. Crowder
T.J. Crowder

Reputation: 1075755

search converts any string you give it to a regular expression. [] is special in regular expressions, it defines a character class.

I suspect you want includes, not search, which looks for exactly what you pass it.

const result = ids.includes("[" + myid + "]") ? "found" : "not found";

(Or with a template literal:

const result = ids.includes(`[${myid}]`) ? "found" : "not found";

)

Upvotes: 2

Amir Rahman
Amir Rahman

Reputation: 1119

what your are trying to do can be achieved with .indexOf this returns index of exact match occurance or -1 in not found

you can check details about indexOf here indexOf 

var ids = '[2][13]';
var myid = '1';

var r = ids.indexOf('['+myid+']') != -1 ? 'found' : 'not found';
console.log(r)

Upvotes: 2

ProfDFrancis
ProfDFrancis

Reputation: 9441

The nasty trap is that you need to double-escape the square bracket

Whatever you pass into the .search() method is assumed to be a regular expression. In a regular expression, the "[" and "]" are the delimiters for a set of characters, allowing a match with any one of those characters.

For example the regular expression [A-Z34] will match with any capital letter or the digits 3 or 4.

Unfortunately your data contains [ and ] as characters within it.

You need one \ for Javascript and one \ for the regular expression parser, so that your code correctly looks for exactly the string [1].

const ids = '[2][13]';
const myid = '1';

const result = ids.search('['+myid+']') !== -1 ? 'found' : 'not found';
console.log(result) // found


const result2 = ids.search('\['+myid+'\]') !== -1 ? 'found' : 'not found';
console.log(result2) // found


const result3 = ids.search('\\['+myid+'\\]') !== -1 ? 'found' : 'not found';
console.log(result3) // not found

Upvotes: 0

Peter Badida
Peter Badida

Reputation: 12199

For the String.prototype.search() you are using a RegExp which treats some characters as reserved for the pattern matching - angle brackets for your case.

ids.search(/\[1\]/);

If you are assembling the expression manually, the backslash is also treated differently because it's parsed twice. The first parsing is for a conversion of string into a RegExp object (which uses \ for special characters such as \n, \t, ...) and the second parsing is when the regular expression is utilized.

If you escape it only once (\[) it'll create a regular expression of [ which is a reserved character, therefore you need to use \\[ so that the first parsing converts it to \[ and then a regular expression \[ i.e. the literal value of [ is used for pattern matching.

var myid = 1;
'[2][13]'.search(new RegExp(`\\[${myid}\\]`));
// or as pointed out in the comments:
'[2][13]'.search(`\\[${myid}\\]`);

Upvotes: 2

Andreas Bonini
Andreas Bonini

Reputation: 44832

var ids = '[2][13]';
let isMatch = function(n)
{
    return ids.match(/\[(\d+)\]/g).includes(`[${n}]`);
}

console.log(isMatch(1), isMatch(2), isMatch(3), isMatch(10), isMatch(13));
false true false false true

Upvotes: 1

Related Questions