Angular: sorting software versions alphabetically

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I need to alphabetize a list of software versions in Angular from latest to oldest (i.e. 1.10.3.e, 1.9.5.b, 1.7.3, 1.5.1.c). I have been using .sort() which sorts alphabetically but treats the values as a string resulting in any 1.10 version to be sorted after a 1.5 version as it looks at the 1 and 0 independently.I tried doing .sort((a, b) => b - a) and it completely randomized the list rather than sorting it. I have also tried using .slice('.') which does not seem to work.

Any ideas?

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

asked Jan 3 at 13:42

Cole Henderson

155

This is not really an angular question, any typescript/javascript solution will work. You can check this question: stackoverflow.com/questions/6832596/…

– LambdaCruiser
Jan 3 at 14:54

add a comment |

Any ideas?

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

asked Jan 3 at 13:42

Cole Henderson

155

This is not really an angular question, any typescript/javascript solution will work. You can check this question: stackoverflow.com/questions/6832596/…

– LambdaCruiser
Jan 3 at 14:54

add a comment |

Any ideas?

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

asked Jan 3 at 13:42

Cole Henderson

155

Any ideas?

sorting version alphabetical-sort

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

asked Jan 3 at 13:42

Cole Henderson

155

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

asked Jan 3 at 13:42

Cole Henderson

155

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

edited Jan 3 at 16:25

peinearydevelopment

5,06642548

asked Jan 3 at 13:42

Cole Henderson

155

asked Jan 3 at 13:42

Cole Henderson

155

asked Jan 3 at 13:42

Cole Henderson

155

This is not really an angular question, any typescript/javascript solution will work. You can check this question: stackoverflow.com/questions/6832596/…

– LambdaCruiser
Jan 3 at 14:54

add a comment |

This is not really an angular question, any typescript/javascript solution will work. You can check this question: stackoverflow.com/questions/6832596/…

– LambdaCruiser
Jan 3 at 14:54

This is not really an angular question, any typescript/javascript solution will work. You can check this question: stackoverflow.com/questions/6832596/…

– LambdaCruiser
Jan 3 at 14:54

add a comment |

2 Answers
2

active

oldest

votes

Solution 1): temporary pad with '0':

(If a part of the version number could become more than 2 digits, padStart could be changed to 3, 4, ...)

var vers = [ '1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c' ];

console.log( vers );



var temp = vers.map( function( s ){

    return s.split('.').map( function( n ){ return n.padStart(2,'0'); } ).join('.');

} )



temp.sort();



vers = temp.map( function( s ){

    return s.split('.').map( function( n ){ return n.replace(/^0+/, ''); } ).join('.');

} )



console.log( temp );

console.log( vers );

Solution 2): sort seperately by integers and strings.

I would call this the more elegant solution, even if it is a bit more complicated.

(Such an answer, with integers, has been given already)

edited Jan 3 at 14:24

answered Jan 3 at 14:15

kicia

1214

Worked like a charm. Some of my versions have the number 0 at the end so I replaced .padStart(2, '0') with .padStart(2. '!'). Thank you!

– Cole Henderson
Jan 3 at 17:19

add a comment |

This isn't an easy question to answer as there are a few variables to consider. Assuming that you will never have versions like 1.1, 1.a, then the following should do the trick.

var arr = ['1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c', '1.5.1.a', '1.5.1.d', '1.10', '1.5']

          .map(a => a.split('.') // split by decimals

                     .map(i => Number.isNaN(parseInt(i, 10)) ? i : parseInt(i, 10))) // make numbers numbers, leave chars as strings

          .sort((a,b) => {

              // sort by multiple indexes and if one array is shorter, it's i item will be undefined

              var longerIndex = a.length > b.length ? a.length : b.length;

              for(var i = 0, l = longerIndex; i < l; i++) {

                if (a[i] > b[i] || b[i] === undefined) return 1;

                else if (a[i] < b[i] || a[i] === undefined) return -1;

              }



              return 0;

          })

          .map(a => a.join('.')) // join the version parts back into a single string

console.log(arr);

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Solution 1): temporary pad with '0':

(If a part of the version number could become more than 2 digits, padStart could be changed to 3, 4, ...)

var vers = [ '1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c' ];

console.log( vers );



var temp = vers.map( function( s ){

    return s.split('.').map( function( n ){ return n.padStart(2,'0'); } ).join('.');

} )



temp.sort();



vers = temp.map( function( s ){

    return s.split('.').map( function( n ){ return n.replace(/^0+/, ''); } ).join('.');

} )



console.log( temp );

console.log( vers );

Solution 2): sort seperately by integers and strings.

I would call this the more elegant solution, even if it is a bit more complicated.

(Such an answer, with integers, has been given already)

edited Jan 3 at 14:24

answered Jan 3 at 14:15

kicia

1214

Worked like a charm. Some of my versions have the number 0 at the end so I replaced .padStart(2, '0') with .padStart(2. '!'). Thank you!

– Cole Henderson
Jan 3 at 17:19

add a comment |

Solution 1): temporary pad with '0':

(If a part of the version number could become more than 2 digits, padStart could be changed to 3, 4, ...)

var vers = [ '1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c' ];

console.log( vers );



var temp = vers.map( function( s ){

    return s.split('.').map( function( n ){ return n.padStart(2,'0'); } ).join('.');

} )



temp.sort();



vers = temp.map( function( s ){

    return s.split('.').map( function( n ){ return n.replace(/^0+/, ''); } ).join('.');

} )



console.log( temp );

console.log( vers );

Solution 2): sort seperately by integers and strings.

I would call this the more elegant solution, even if it is a bit more complicated.

(Such an answer, with integers, has been given already)

edited Jan 3 at 14:24

answered Jan 3 at 14:15

kicia

1214

Worked like a charm. Some of my versions have the number 0 at the end so I replaced .padStart(2, '0') with .padStart(2. '!'). Thank you!

– Cole Henderson
Jan 3 at 17:19

add a comment |

Solution 1): temporary pad with '0':

(If a part of the version number could become more than 2 digits, padStart could be changed to 3, 4, ...)

var vers = [ '1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c' ];

console.log( vers );



var temp = vers.map( function( s ){

    return s.split('.').map( function( n ){ return n.padStart(2,'0'); } ).join('.');

} )



temp.sort();



vers = temp.map( function( s ){

    return s.split('.').map( function( n ){ return n.replace(/^0+/, ''); } ).join('.');

} )



console.log( temp );

console.log( vers );

Solution 2): sort seperately by integers and strings.

I would call this the more elegant solution, even if it is a bit more complicated.

(Such an answer, with integers, has been given already)

edited Jan 3 at 14:24

answered Jan 3 at 14:15

kicia

1214

Solution 1): temporary pad with '0':

(If a part of the version number could become more than 2 digits, padStart could be changed to 3, 4, ...)

var vers = [ '1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c' ];

console.log( vers );



var temp = vers.map( function( s ){

    return s.split('.').map( function( n ){ return n.padStart(2,'0'); } ).join('.');

} )



temp.sort();



vers = temp.map( function( s ){

    return s.split('.').map( function( n ){ return n.replace(/^0+/, ''); } ).join('.');

} )



console.log( temp );

console.log( vers );

Solution 2): sort seperately by integers and strings.

I would call this the more elegant solution, even if it is a bit more complicated.

(Such an answer, with integers, has been given already)

edited Jan 3 at 14:24

answered Jan 3 at 14:15

kicia

1214

edited Jan 3 at 14:24

answered Jan 3 at 14:15

kicia

1214

answered Jan 3 at 14:15

kicia

1214

answered Jan 3 at 14:15

kicia

1214

Worked like a charm. Some of my versions have the number 0 at the end so I replaced .padStart(2, '0') with .padStart(2. '!'). Thank you!

– Cole Henderson
Jan 3 at 17:19

add a comment |

Worked like a charm. Some of my versions have the number 0 at the end so I replaced .padStart(2, '0') with .padStart(2. '!'). Thank you!

– Cole Henderson
Jan 3 at 17:19

Worked like a charm. Some of my versions have the number 0 at the end so I replaced .padStart(2, '0') with .padStart(2. '!'). Thank you!

– Cole Henderson
Jan 3 at 17:19

add a comment |

This isn't an easy question to answer as there are a few variables to consider. Assuming that you will never have versions like 1.1, 1.a, then the following should do the trick.

var arr = ['1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c', '1.5.1.a', '1.5.1.d', '1.10', '1.5']

          .map(a => a.split('.') // split by decimals

                     .map(i => Number.isNaN(parseInt(i, 10)) ? i : parseInt(i, 10))) // make numbers numbers, leave chars as strings

          .sort((a,b) => {

              // sort by multiple indexes and if one array is shorter, it's i item will be undefined

              var longerIndex = a.length > b.length ? a.length : b.length;

              for(var i = 0, l = longerIndex; i < l; i++) {

                if (a[i] > b[i] || b[i] === undefined) return 1;

                else if (a[i] < b[i] || a[i] === undefined) return -1;

              }



              return 0;

          })

          .map(a => a.join('.')) // join the version parts back into a single string

console.log(arr);

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

add a comment |

This isn't an easy question to answer as there are a few variables to consider. Assuming that you will never have versions like 1.1, 1.a, then the following should do the trick.

var arr = ['1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c', '1.5.1.a', '1.5.1.d', '1.10', '1.5']

          .map(a => a.split('.') // split by decimals

                     .map(i => Number.isNaN(parseInt(i, 10)) ? i : parseInt(i, 10))) // make numbers numbers, leave chars as strings

          .sort((a,b) => {

              // sort by multiple indexes and if one array is shorter, it's i item will be undefined

              var longerIndex = a.length > b.length ? a.length : b.length;

              for(var i = 0, l = longerIndex; i < l; i++) {

                if (a[i] > b[i] || b[i] === undefined) return 1;

                else if (a[i] < b[i] || a[i] === undefined) return -1;

              }



              return 0;

          })

          .map(a => a.join('.')) // join the version parts back into a single string

console.log(arr);

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

add a comment |

This isn't an easy question to answer as there are a few variables to consider. Assuming that you will never have versions like 1.1, 1.a, then the following should do the trick.

var arr = ['1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c', '1.5.1.a', '1.5.1.d', '1.10', '1.5']

          .map(a => a.split('.') // split by decimals

                     .map(i => Number.isNaN(parseInt(i, 10)) ? i : parseInt(i, 10))) // make numbers numbers, leave chars as strings

          .sort((a,b) => {

              // sort by multiple indexes and if one array is shorter, it's i item will be undefined

              var longerIndex = a.length > b.length ? a.length : b.length;

              for(var i = 0, l = longerIndex; i < l; i++) {

                if (a[i] > b[i] || b[i] === undefined) return 1;

                else if (a[i] < b[i] || a[i] === undefined) return -1;

              }



              return 0;

          })

          .map(a => a.join('.')) // join the version parts back into a single string

console.log(arr);

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

This isn't an easy question to answer as there are a few variables to consider. Assuming that you will never have versions like 1.1, 1.a, then the following should do the trick.

var arr = ['1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c', '1.5.1.a', '1.5.1.d', '1.10', '1.5']

          .map(a => a.split('.') // split by decimals

                     .map(i => Number.isNaN(parseInt(i, 10)) ? i : parseInt(i, 10))) // make numbers numbers, leave chars as strings

          .sort((a,b) => {

              // sort by multiple indexes and if one array is shorter, it's i item will be undefined

              var longerIndex = a.length > b.length ? a.length : b.length;

              for(var i = 0, l = longerIndex; i < l; i++) {

                if (a[i] > b[i] || b[i] === undefined) return 1;

                else if (a[i] < b[i] || a[i] === undefined) return -1;

              }



              return 0;

          })

          .map(a => a.join('.')) // join the version parts back into a single string

console.log(arr);

var arr = ['1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c', '1.5.1.a', '1.5.1.d', '1.10', '1.5']

          .map(a => a.split('.') // split by decimals

                     .map(i => Number.isNaN(parseInt(i, 10)) ? i : parseInt(i, 10))) // make numbers numbers, leave chars as strings

          .sort((a,b) => {

              // sort by multiple indexes and if one array is shorter, it's i item will be undefined

              var longerIndex = a.length > b.length ? a.length : b.length;

              for(var i = 0, l = longerIndex; i < l; i++) {

                if (a[i] > b[i] || b[i] === undefined) return 1;

                else if (a[i] < b[i] || a[i] === undefined) return -1;

              }



              return 0;

          })

          .map(a => a.join('.')) // join the version parts back into a single string

console.log(arr);

var arr = ['1.10.3.e', '1.9.5.b', '1.7.3', '1.5.1.c', '1.5.1.a', '1.5.1.d', '1.10', '1.5']

          .map(a => a.split('.') // split by decimals

                     .map(i => Number.isNaN(parseInt(i, 10)) ? i : parseInt(i, 10))) // make numbers numbers, leave chars as strings

          .sort((a,b) => {

              // sort by multiple indexes and if one array is shorter, it's i item will be undefined

              var longerIndex = a.length > b.length ? a.length : b.length;

              for(var i = 0, l = longerIndex; i < l; i++) {

                if (a[i] > b[i] || b[i] === undefined) return 1;

                else if (a[i] < b[i] || a[i] === undefined) return -1;

              }



              return 0;

          })

          .map(a => a.join('.')) // join the version parts back into a single string

console.log(arr);

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

answered Jan 3 at 14:06

peinearydevelopment

5,06642548

add a comment |

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