JQuery - show only one if the same id and also add it on weeks [duplicate]











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  • Most efficient method to groupby on a array of objects

    33 answers




I have this sample data:



[{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}]


I wanted that I will only show one if id is redundant but add all of the weeks that the id's had



My Attempt:



var data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];

var weeks = 0;

for (var i=0; i<data.length; i++){

    weeks += data[i].weeks;

}


Desired output



[{"id":"2", "weeks":6}, {"id":3, "weeks":7}]





javascript jquery







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marked as duplicate by Andreas, Amit Joki javascript
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 Nov 20 at 6:27


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    up vote
    2
    down vote

    favorite
    1













    This question already has an answer here:




    • Most efficient method to groupby on a array of objects

      33 answers




    I have this sample data:



    [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}]


    I wanted that I will only show one if id is redundant but add all of the weeks that the id's had



    My Attempt:



    var data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
    
var weeks = 0;
    
for (var i=0; i<data.length; i++){
    
    weeks += data[i].weeks;
    
}


    Desired output



    [{"id":"2", "weeks":6}, {"id":3, "weeks":7}]





    javascript jquery







    share|improve this question













    marked as duplicate by Andreas, Amit Joki javascript
    Users with the  javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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     Nov 20 at 6:27


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

















      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1






      This question already has an answer here:




      • Most efficient method to groupby on a array of objects

        33 answers




      I have this sample data:



      [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}]


      I wanted that I will only show one if id is redundant but add all of the weeks that the id's had



      My Attempt:



      var data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
      
var weeks = 0;
      
for (var i=0; i<data.length; i++){
      
    weeks += data[i].weeks;
      
}


      Desired output



      [{"id":"2", "weeks":6}, {"id":3, "weeks":7}]





      javascript jquery







      share|improve this question














      This question already has an answer here:




      • Most efficient method to groupby on a array of objects

        33 answers




      I have this sample data:



      [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}]


      I wanted that I will only show one if id is redundant but add all of the weeks that the id's had



      My Attempt:



      var data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
      
var weeks = 0;
      
for (var i=0; i<data.length; i++){
      
    weeks += data[i].weeks;
      
}


      Desired output



      [{"id":"2", "weeks":6}, {"id":3, "weeks":7}]




      This question already has an answer here:




      • Most efficient method to groupby on a array of objects

        33 answers







      javascript jquery




      javascript jquery






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 20 at 6:24









      sad saddest

      338




      338




      marked as duplicate by Andreas, Amit Joki javascript
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       Nov 20 at 6:27


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by Andreas, Amit Joki javascript
      Users with the  javascript badge can single-handedly close javascript questions as duplicates and reopen them as needed.

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       Nov 20 at 6:27


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


























          2 Answers
          2






          active

          oldest

          votes

















          up vote
          3
          down vote



          accepted










          You can use array#reduce() with Object.values(). Group your data based on id and add the weeks for repeating id in an object. Take the values from the object using Object.values().






          const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}],
          
      result = Object.values(data.reduce((r,{id,weeks}) => {
          
        r[id] = r[id] || {id, weeks : 0};
          
        r[id].weeks += weeks;
          
        return r;
          
      },{}));
          
console.log(result);








          share|improve this answer




























            up vote
            1
            down vote













            You can Array.prototype.reduce() combined with Array.prototype.push() or Array.prototype.find() based on the number of repeated elements with same id



            Code:






            const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
            
const result = data
            
  .reduce((a, c) => {
            
    a.temp[c.id] = ++a.temp[c.id] || 1;
            
    a.temp[c.id] === 1
            
      ? a.array.push(c)
            
      : a.array.find(item => item.id == a.temp[c.id]).weeks += c.weeks;
            
    return a;
            
  }, {temp: {}, array: })
            
  .array;
            

            
console.log(result);








            share|improve this answer






























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              3
              down vote



              accepted










              You can use array#reduce() with Object.values(). Group your data based on id and add the weeks for repeating id in an object. Take the values from the object using Object.values().






              const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}],
              
      result = Object.values(data.reduce((r,{id,weeks}) => {
              
        r[id] = r[id] || {id, weeks : 0};
              
        r[id].weeks += weeks;
              
        return r;
              
      },{}));
              
console.log(result);








              share|improve this answer

























                up vote
                3
                down vote



                accepted










                You can use array#reduce() with Object.values(). Group your data based on id and add the weeks for repeating id in an object. Take the values from the object using Object.values().






                const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}],
                
      result = Object.values(data.reduce((r,{id,weeks}) => {
                
        r[id] = r[id] || {id, weeks : 0};
                
        r[id].weeks += weeks;
                
        return r;
                
      },{}));
                
console.log(result);








                share|improve this answer























                  up vote
                  3
                  down vote



                  accepted







                  up vote
                  3
                  down vote



                  accepted






                  You can use array#reduce() with Object.values(). Group your data based on id and add the weeks for repeating id in an object. Take the values from the object using Object.values().






                  const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}],
                  
      result = Object.values(data.reduce((r,{id,weeks}) => {
                  
        r[id] = r[id] || {id, weeks : 0};
                  
        r[id].weeks += weeks;
                  
        return r;
                  
      },{}));
                  
console.log(result);








                  share|improve this answer












                  You can use array#reduce() with Object.values(). Group your data based on id and add the weeks for repeating id in an object. Take the values from the object using Object.values().






                  const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}],
                  
      result = Object.values(data.reduce((r,{id,weeks}) => {
                  
        r[id] = r[id] || {id, weeks : 0};
                  
        r[id].weeks += weeks;
                  
        return r;
                  
      },{}));
                  
console.log(result);








                  const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}],
                  
      result = Object.values(data.reduce((r,{id,weeks}) => {
                  
        r[id] = r[id] || {id, weeks : 0};
                  
        r[id].weeks += weeks;
                  
        return r;
                  
      },{}));
                  
console.log(result);





                  const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}],
                  
      result = Object.values(data.reduce((r,{id,weeks}) => {
                  
        r[id] = r[id] || {id, weeks : 0};
                  
        r[id].weeks += weeks;
                  
        return r;
                  
      },{}));
                  
console.log(result);






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 20 at 6:29









                  Hassan Imam

                  11.8k31230




                  11.8k31230
























                      up vote
                      1
                      down vote













                      You can Array.prototype.reduce() combined with Array.prototype.push() or Array.prototype.find() based on the number of repeated elements with same id



                      Code:






                      const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
                      
const result = data
                      
  .reduce((a, c) => {
                      
    a.temp[c.id] = ++a.temp[c.id] || 1;
                      
    a.temp[c.id] === 1
                      
      ? a.array.push(c)
                      
      : a.array.find(item => item.id == a.temp[c.id]).weeks += c.weeks;
                      
    return a;
                      
  }, {temp: {}, array: })
                      
  .array;
                      

                      
console.log(result);








                      share|improve this answer



























                        up vote
                        1
                        down vote













                        You can Array.prototype.reduce() combined with Array.prototype.push() or Array.prototype.find() based on the number of repeated elements with same id



                        Code:






                        const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
                        
const result = data
                        
  .reduce((a, c) => {
                        
    a.temp[c.id] = ++a.temp[c.id] || 1;
                        
    a.temp[c.id] === 1
                        
      ? a.array.push(c)
                        
      : a.array.find(item => item.id == a.temp[c.id]).weeks += c.weeks;
                        
    return a;
                        
  }, {temp: {}, array: })
                        
  .array;
                        

                        
console.log(result);








                        share|improve this answer

























                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          You can Array.prototype.reduce() combined with Array.prototype.push() or Array.prototype.find() based on the number of repeated elements with same id



                          Code:






                          const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
                          
const result = data
                          
  .reduce((a, c) => {
                          
    a.temp[c.id] = ++a.temp[c.id] || 1;
                          
    a.temp[c.id] === 1
                          
      ? a.array.push(c)
                          
      : a.array.find(item => item.id == a.temp[c.id]).weeks += c.weeks;
                          
    return a;
                          
  }, {temp: {}, array: })
                          
  .array;
                          

                          
console.log(result);








                          share|improve this answer














                          You can Array.prototype.reduce() combined with Array.prototype.push() or Array.prototype.find() based on the number of repeated elements with same id



                          Code:






                          const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
                          
const result = data
                          
  .reduce((a, c) => {
                          
    a.temp[c.id] = ++a.temp[c.id] || 1;
                          
    a.temp[c.id] === 1
                          
      ? a.array.push(c)
                          
      : a.array.find(item => item.id == a.temp[c.id]).weeks += c.weeks;
                          
    return a;
                          
  }, {temp: {}, array: })
                          
  .array;
                          

                          
console.log(result);








                          const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
                          
const result = data
                          
  .reduce((a, c) => {
                          
    a.temp[c.id] = ++a.temp[c.id] || 1;
                          
    a.temp[c.id] === 1
                          
      ? a.array.push(c)
                          
      : a.array.find(item => item.id == a.temp[c.id]).weeks += c.weeks;
                          
    return a;
                          
  }, {temp: {}, array: })
                          
  .array;
                          

                          
console.log(result);





                          const data = [{"id":"2", "weeks":2}, {"id":"2", "weeks":4}, {"id":3, "weeks":7}];
                          
const result = data
                          
  .reduce((a, c) => {
                          
    a.temp[c.id] = ++a.temp[c.id] || 1;
                          
    a.temp[c.id] === 1
                          
      ? a.array.push(c)
                          
      : a.array.find(item => item.id == a.temp[c.id]).weeks += c.weeks;
                          
    return a;
                          
  }, {temp: {}, array: })
                          
  .array;
                          

                          
console.log(result);






                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited Nov 20 at 7:11

























                          answered Nov 20 at 6:29









                          Yosvel Quintero

                          10.8k42229




                          10.8k42229















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