How To Compare Strings In Which Appears Similar Characters But Different Char Codes?

I have the problem with comparing strings with different char codes but similar characters like the following: console.log('³' === '3') // false; False value from the code above

Solution 1:

Look into ASCII folding, which is primarily used to convert accented characters to unaccented ones. There's a JS library for it here.

For your provided example, it will work - for other examples, it might not. It depends on how the equivalence is defined (nobody but you knows what you mean by "similar" - different characters are different characters).

If you know all of the characters that you want to map already, the easiest way will simply be to define a mapping yourself:

var eqls = function(first, second) {
    var mappings = { '³': '3', '3': '3' };

    if (mappings[first]) {
        return mappings[first] == mappings[second];
    }

    returnfalse;
}

if (eqls('³', '3')) { ... }

Solution 2:

There is no "universal solution"

If you've only to deal with digits you may build up your "equivalence table" where for each supported character you define a "canonical" character.

For example

var eqTable = []; // the table is just an array

eqTable[179] = 51; // ³ --> 3/* ... */

Then build a simple algorythm to turn a string into its canonical form

var original,         // the source string - let's assume original=="³3"var canonical = "";   // the canonical resulting stringvar i,
    n,
    c;

n = original.length;
for( i = 0; i < n; i++ )
{
    c = eqTable[ original.charCodeAt( i ) ];
    if( typeof( c ) != 'undefined' )
    {
        canonical += String.fromCharCode( c );
    }
    else
    {
        canonical += original[ i ]; // you *may* leave the original character if no match is found
    }
}

// RESULT: canonical == "33"