CODEWITHSUNDEEP
javascriptregexnumbersreplace
stefan
stefan

Reputation: 3

Find numbers at a specific position

I'm trying to find an expression for JavaScript which gives me the two characters at a specific position.

It's always the same call so its may be not too complicated.

I have always a 10 char long number and i want to replace the first two, the two at place 3 and 4 or the two at place 5 and 6 and so on.

So far I've done this:

number.replace(/\d{2}/, index));

this replace my first 2 digits with 2 others digits. but now I want to include some variables at which position the digits should be replaced, something like:

number.replace(/\d{atposx,atpox+1}/, index));

that means:

01234567891

and I want sometimes to replace 01 with 02 and sometimes 23 with 56.

(or something like this with other numbers).

I hope I pointed out what I want.

Upvotes: 0

Views: 3800

Answers (3)

Stephen
Stephen

Reputation: 5470

Numbers are fairly easily interpreted as strings in JS. So, if you're working with an actual number (i.e. 9876543210) and not a number that's represented by a string (i.e. '987654321'), just turn the number into a string (''.concat(number); ) and don't limit yourself to the constraints of what you can do with just numbers.

Both of the above examples are fine (bah, they beat me to it), but you can even think about it like this:

var numberString = ''.concat(number);
var numberChunks = numberString.match(/(\d{2})/g);

You've now got an array of chunks that you can either walk through, switch through, or whatever other kind of flow you want to follow. When you're done, just say...

numberString = numberChunks.join('');
number = parseInt(numberString, 10);

You've got your number back as a native number (or skip the last part to just get the string back). And, aside from that, if you're doing multiple replacements.. the more replacements you do in the number, the more efficient breaking it up into chunks and dealing with the chunks are. I did a quick test, and running the 'replaceChars' function was faster on a single change, but will be slower than just splitting into an array if you're doing two or more changes to the data.

Hope that makes sense!

Upvotes: 1

Liviu T.
Liviu T.

Reputation: 23664

You can try this

function replaceAtIndex(str,value,index) {
    return str.substr(0,index)+value+str.substr(index+value.length);
}
replaceAtIndex('0123456789','X',3);  // returns "012X456789"
replaceAtIndex('0123456789','XY',3); // returns "012XY56789"

Upvotes: 0

Jamiec
Jamiec

Reputation: 136134

This function works fine:

function replaceChars(input, startPos, replacement){
   return input.substring(0,startPos) + 
        replacement + 
        input.substring(startPos+replacement.length)  
}

Usage:

replaceChars("0123456789",2,"55") // output: 0155456789

Live example: http://jsfiddle.net/FnkpT/

Upvotes: 1

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