Javascript replace every 6th colon in array

Question

I have some problems with replacing every 6th colon in my array. Have tried something with Regex, but that doesn't seem to work. I have red other questions were people are using nth and then set this variabele to the index you want to replace, but can't figure out why that isn't working. I used the join function to replace the ',' in my array with ':'.

arrayProducts[i] = arrayProducts[i].join(':');

When i use console.log(arrayProducts); this is my result:

F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3

This is what I want:

F200:0:0.0000:1:100:0:1,KPO2:0:0.0000:1:200:0:2,HGB1:0:0.0000:1:300:0:3

Thanks for reading!

Edit: F200, KP02 and HGB1, could also be numbers / digits like: 210, 89, 102 so the :[A-Z] method from regex doesn't work.

this may be helpful http://stackoverflow.com/questions/6843441/javascript-how-can-i-replace-only-nth-match-in-the-string — gurvinder372, Apr 03 '17 at 09:16
["F200:0:0.0000:1:100:0:1:KP02:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3"] — Koen van de Sande, Apr 03 '17 at 09:23

score 2 · Accepted Answer · answered Apr 03 '17 at 09:26

You can just count the number of colon occurences and replace every nth of them.

var str = 'F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3', counter = 0;
    res = str.replace(/:/g, function(v) {
      counter++;
      return !(counter % 7) ? ',' : v;
    });

    console.log(res);

score 1 · Answer 2 · edited Apr 03 '17 at 09:34

1

A regex solution is viable. You can use a function as the second parameter of the .replace method to make full use of backreferences.

var str = 'F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3';

str = str.replace(/((?:[^:]*:){6}(?:[^:]*)):/g, function() {
    var matches = arguments;
    return matches[1] + ',';
});

console.log(str);

edited Apr 03 '17 at 09:34

RomanPerekhrest

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answered Apr 03 '17 at 09:22

Scoots

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score 0 · Answer 3 · answered Apr 03 '17 at 09:22

0

What you are looking for is to split over the following expression :[A-Z] (assuming that your rows always start with this range)

a simple solution could be:

mystring.split(/:[A-Z]/).join(',')

/:[A-Z]/ matches any : followed by a uppercase letter

answered Apr 03 '17 at 09:22

amd

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Well ya, that was what i used before. The problem is that F200, KP02 and HGB1 could also be something like: 112, 141, 5 – Koen van de Sande Apr 03 '17 at 09:25
1

So you should mention that in your question :), to get better results – amd Apr 03 '17 at 09:26

NomNomNom · Answer 4 · 2017-04-03T09:55:14.197

maybe you can try this approach, loop your array and join it manually, something like :

     var strarr = "F200:0:00000:1:100:0:1:KPO2:0:00000:1:200:0:2:HGB1:0:00000:1:300:0:3";
     var arr = strarr.split(":")

      var resStr = "";

       for(var i = 0; i < arr.length; i++)
       {
           if(i > 0 && i%7 == 0)
               resStr = resStr + ","  + arr[i]
           else
               resStr =  resStr + ( resStr == "" ? "" : ":") + arr[i];
        }

        console.log(resStr);

Nina Scholz · Answer 5 · 2017-04-03T09:45:20.000

0

You could use replace with a look for six parts with colon and replace the seventh.

var string = 'F200:0:0.0000:1:100:0:1:KPO2:0:0.0000:1:200:0:2:HGB1:0:0.0000:1:300:0:3',
    result = string.replace(/(([^:]*:){6}[^:]*):/g, '$1,');

console.log(result);

edited Apr 03 '17 at 09:45

answered Apr 03 '17 at 09:25

Nina Scholz

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amd · Answer 6 · 2017-04-03T09:48:55.727

0

Another solution (based on the number of iteration)

using map method:

str.split(':').map((v, i) => (i % 7 === 0 ? ',' : ':') + v ).join('').slice(1)

using reduce method:

str.split(':').reduce((acc,v, i) =>  {    
    return acc +  (i % 7 === 0 ? ',' : ':' ) + v ;
}, '').slice(1)

Note: arrow expression does not work on old browsers

edited Apr 03 '17 at 09:48

answered Apr 03 '17 at 09:41

amd

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Javascript replace every 6th colon in array

Thanks for reading!

6 Answers6