合併多個JSON對象iniside數組到期望的格式

我有出外下面合併多個JSON對象iniside數組到期望的格式

[{ 
    "data": { 
     "aggregations": { 
      "openissue": { 
       "buckets": [{ 
         "doc_count": 5, 
         "date": "25 aug" 
        }, 
        { 
         "doc_count": 6, 
         "date": "26 aug" 
        }, 
        { 
         "doc_count": 7, 
         "date": "27 aug" 
        }, 
        { 
         "doc_count": 8, 
         "date": "28 aug" 
        } 
       ] 
      } 
     } 

    }, 
    "a": "not needed", 
    "b": "not needed" 
}, 
{ 
    "data": { 
     "aggregations": { 
      "closeissue": { 
       "buckets": [{ 
         "doc_count": 9, 
         "date": "29 aug" 
        }, 
        { 
         "doc_count": 10, 
         "date": "30 aug" 
        }, 
        { 
         "doc_count": 11, 
         "date": "31 aug" 
        }, 
        { 
         "doc_count": 12, 
         "date": "1 sept" 
        } 
       ] 
      } 
     } 

    }, 
    "a": "not needed", 
    "b": "not needed" 
}, { 
    "data": { 
     "aggregations": { 
      "pendingissue": { 
       "buckets": [{ 
         "doc_count": 13, 
         "date": "10 sept" 
        }, 
        { 
         "doc_count": 14, 
         "date": "13 sept" 
        }, 
        { 
         "doc_count": 15, 
         "date": "18 sept" 
        }, 
        { 
         "doc_count": 16, 
         "date": "19 sept" 
        } 
       ] 
      } 
     } 

    }, 
    "a": "not needed", 
    "b": "not needed" 
}

]

對象的數目是該我被AJAX調用獲得動態的，這是一個JSON文件json是在ajax調用之後創建的。我試圖過濾掉json，並將其轉換爲所需的格式，其中對象的數量取決於日期的數量。

所需的輸出應該像

[ 
    { 
    "Key_as_string": "25aug", 
    "values": [ 
     { 
     "label": "open-issues", 
     "doc_count": 10 
     }, 
     { 
     "label": "closed-issues", 
     "doc_count": 10 
     }, 
     { 
     "label": "exempted-issues", 
     "doc_count": 10 
     } 
    ] 
    }, 
    { 
    "Key_as_string": "26aug", 
    "values": [ 
     { 
     "label": "open-issues", 
     "doc_count": 10 
     }, 
     { 
     "label": "closed-issues", 
     "doc_count": 10 
     }, 
     { 
     "label": "pending-issues", 
     "doc_count": 10 
     } 
    ] 
    } 
]

使用JavaScript或jQuery的

來源

2017-09-14 Trinanjan Saha

的想法就是擊敗所有的問題納入一個單一的陣列，然後按日期，最後構建它我該怎麼辦它到所需的格式。

const payload = [ 
 
{ 
 
    "data": { 
 
     "aggregations": { 
 
      "openissue": { 
 
       "buckets": [{ 
 
         "doc_count": 5, 
 
         "date": "25 aug" 
 
        }, 
 
        { 
 
         "doc_count": 6, 
 
         "date": "26 aug" 
 
        }, 
 
        { 
 
         "doc_count": 7, 
 
         "date": "27 aug" 
 
        }, 
 
        { 
 
         "doc_count": 8, 
 
         "date": "28 aug" 
 
        } 
 
       ] 
 
      } 
 
     } 
 

 
    }, 
 
    "a": "not needed", 
 
    "b": "not needed" 
 
}, 
 
{ 
 
    "data": { 
 
     "aggregations": { 
 
      "closeissue": { 
 
       "buckets": [{ 
 
         "doc_count": 9, 
 
         "date": "29 aug" 
 
        }, 
 
        { 
 
         "doc_count": 10, 
 
         "date": "30 aug" 
 
        }, 
 
        { 
 
         "doc_count": 11, 
 
         "date": "31 aug" 
 
        }, 
 
        { 
 
         "doc_count": 12, 
 
         "date": "1 sept" 
 
        } 
 
       ] 
 
      } 
 
     } 
 

 
    }, 
 
    "a": "not needed", 
 
    "b": "not needed" 
 
}, { 
 
    "data": { 
 
     "aggregations": { 
 
      "pendingissue": { 
 
       "buckets": [{ 
 
         "doc_count": 13, 
 
         "date": "10 sept" 
 
        }, 
 
        { 
 
         "doc_count": 14, 
 
         "date": "13 sept" 
 
        }, 
 
        { 
 
         "doc_count": 15, 
 
         "date": "18 sept" 
 
        }, 
 
        { 
 
         "doc_count": 16, 
 
         "date": "19 sept" 
 
        } 
 
       ] 
 
      } 
 
     } 
 

 
    }, 
 
    "a": "not needed", 
 
    "b": "not needed" 
 
} 
 
] 
 
// 
 
let flat = [] 
 
payload.forEach(item => { 
 
\t const preIssueType = Object.keys(item.data.aggregations)[0] 
 
    item.data.aggregations[preIssueType].buckets.forEach(issue => { 
 
\t \t \t flat.push({label: preIssueType, doc_count: issue.doc_count, date: issue.date}) \t 
 
    }) 
 
}) 
 
const grouped = _.groupBy(flat, 'date') 
 
let result = [] 
 
for(const key in grouped){ 
 
\t result.push({'Key_as_string': key, values: grouped[key]}) 
 
} 
 

 
console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>

忘了說了，lodash是不是強制性的，但能節省一些編碼工作。

來源

2017-09-14 01:58:31 spiritwalker

非常感謝您的回覆。我不熟悉loadsh。所以我會嘗試使用普通的javascript。在這種情況下，代碼不會改變很多嗎？對不起，我在JavaScript中很新。 –

@TrinanjanSaha _.groupBy是唯一使用lodash的地方。所以，如果你想使用原生的JavaScript，那麼在純JavaScript中有如何做到這一點。 https://stackoverflow.com/questions/14446511/what-is-the-most-efficient-method-to-groupby-on-a-javascript-array-of-objects。 – spiritwalker

我正在嘗試，因爲你提到。我得到一些error.this是小提琴附加http://jsfiddle.net/y45Lfupw/62/ –

合併多個JSON對象iniside數組到期望的格式

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