javascript - 如何修改这个函数来合并键的值？

Question

我希望将“year”键的值合并到“userName”对象中。

我有以下对象数组

userName=[{
            loginName:'jimmy',
            deviation:[{dates:[20150222,20150223,20150224,20150225], type:'vacation'}],
            year:'2015'
          },
         {
           loginName:'jimmy',
           deviation:[{dates:[20150322,20150323,20150324,20150325], type:'sick'}],
           year:'2016'
         },
         {
          loginName:'chloe',
          deviation:[{dates:[20150222,20150221,20150224,20150225], type:'national free day'}],
          year:'2016'
         },
        ];      

通过使用此函数，我可以根据登录名对偏差日期进行排序。因此，在这种情况下，用户“jimmy”将仅显示一次，但所有偏差日期和类型都会合并。然而，关键“年份”的值仅继承第一个值并保持不变。如何通过修改我当前的函数来合并键“年份”，以便用户“吉米”年份将显示“2015，2016”:

var mergedUsername = [];

    userName.forEach(function(deviation) { 
       var existing = mergedUsername.filter(function(v, i) { 
       return v.loginName == deviation.loginName; 
       }); 
    if(existing.length) {
      var existingIndex = mergedUsername.indexOf(existing[0]);
      mergedUsername[existingIndex].deviation = mergedUsername[existingIndex].deviation.concat(deviation.deviation); 
    }  else {
        if(typeof deviation.deviation == 'string')
        deviation.deviation = [deviation.deviation];
        mergedUsername.push(deviation);  
     }
    });

提前致谢!

Answer 1

您可以使用reduce()来获得所需的结果。

var data = [{
  loginName: 'jimmy',
  deviation: [{
    dates: [20150222, 20150223, 20150224, 20150225],
    type: 'vacation'
  }],
  year: '2015'
}, {
  loginName: 'jimmy',
  deviation: [{
    dates: [20150322, 20150323, 20150324, 20150325],
    type: 'sick'
  }],
  year: '2016'
}, {
  loginName: 'chloe',
  deviation: [{
    dates: [20150222, 20150221, 20150224, 20150225],
    type: 'national free day'
  }],
  year: '2016'
}];

var o = {}
var result = data.reduce(function(r, e) {
  if (!o[e.loginName]) {
    o[e.loginName] = e;
    r.push(o[e.loginName]);
  } else {
    o[e.loginName].deviation[0].dates.push(...e.deviation[0].dates)
    o[e.loginName].deviation[0].type += ' ' + e.deviation[0].type
    o[e.loginName].year += ' ' + e.year
  }
  return r;
}, [])

console.log(result)