Question

我有以下格式的 JSON 字符串。

{
"dt_total_score": [
{
  "score": 60.8332,
  "gText": "Total"
},
{
  "score": 66.6666,
  "gText": "Call Center"
},
{
  "score": 66.6666,
  "gText": "Restaurant reservation"
},
{
  "score": 50,
  "gText": "Arriving (Trams) - Experience or Observation"
},
{
  "score": 60,
  "gText": "Arrival & Parking Experience"
}
],
"dt_trend": [
{
  "score": 60,
  "gText": "Q1,2018",
  "seriesName": "Arrival & Parking Experience",
  "ONUM": 1
},
{
  "score": 50,
  "gText": "Q1,2018",
  "seriesName": "Arriving (Trams) - Experience or Observation",
  "ONUM": 1
},
{
  "score": 66.6666,
  "gText": "Q1,2018",
  "seriesName": "Call Center",
  "ONUM": 1
},
{
  "score": 66.6666,
  "gText": "Q1,2018",
  "seriesName": "Restaurant reservation",
  "ONUM": 1
},
{
  "score": 60.8333,
  "gText": "Q1,2018",
  "seriesName": "Total",
  "ONUM": 2
},
{
  "score": 60.8333,
  "gText": "Q4,2017",
  "seriesName": "Total",
  "ONUM": 2
},
{
  "score": 60.8333,
  "gText": "Q3,2017",
  "seriesName": "Total",
  "ONUM": 2
}
]
,
"dt_compartive":[  
  {  
     "gText":"Total",
     "percentDiff":6083.0000
  }
 ]
  }

在上面的JSON中：

dt_total_score具有特定类别的即时分数。
dt_trend在给定时间范围内有得分趋势（在上述情况下为季度，但也可以是周或月）。
dt_comparative在时间范围内得分的百分比差异

我必须以下列方式将基于类别的上述数据解析并合并到Javascript对象列表中。 请注意，此数据仅用于说明我希望输出的方式与上述数据无关

var dt_score_n_trend = [

{
gText:"Call Center",
score: 60.2223,
trendName:[ "Q1,2018", "Q2,2018"],
trendValue:[ 60, 70],
PercentageDiff:70
},
 {
gText:"Call Center",
score: 60.2223,
trendName: [ "Q1,2018", "Q2,2018"],
trendValue:[ 60, 70],
PercentageDiff:70
}
];

到目前为止，我已经无意中尝试将JSON解析为Javascript对象，然后尝试合并它们。

如何将它们合并为上述方式？

编辑：更新了问题并删除了格式错误

Answer 1

您可以使用array#reduce根据dt_trend对seriesName数组进行分组。创建一个累加器对象，并为每个新seriesName添加一个新对象，对于旧对象，更新趋势值。然后提取出所有的值。

＆＃13;

var data = { "dt_total_score": [ { "score": 60.8332, "gText": "Total" }, { "score": 66.6666, "gText": "Call Center" }, { "score": 66.6666, "gText": "Restaurant reservation" }, { "score": 50, "gText": "Arriving (Trams) - Experience or Observation" }, { "score": 60, "gText": "Arrival & Parking Experience" } ], "dt_trend": [ { "score": 60, "gText": "Q1,2018", "seriesName": "Arrival & Parking Experience", "ONUM": 1 }, { "score": 50, "gText": "Q1,2018", "seriesName": "Arriving (Trams) - Experience or Observation", "ONUM": 1 }, { "score": 66.6666, "gText": "Q1,2018", "seriesName": "Call Center", "ONUM": 1 }, { "score": 66.6666, "gText": "Q1,2018", "seriesName": "Restaurant reservation", "ONUM": 1 }, { "score": 60.8333, "gText": "Q1,2018", "seriesName": "Total", "ONUM": 2 }, { "score": 60.8333, "gText": "Q4,2017", "seriesName": "Total", "ONUM": 2 }, { "score": 60.8333, "gText": "Q3,2017", "seriesName": "Total", "ONUM": 2 } ], "dt_compartive":[{"gText":"Total","percentDiff":6083.0000}]},
    result = data.dt_trend.reduce((r,o) => {
      r[o.seriesName] = r[o.seriesName] || {gText: o.seriesName, score: o.score, trendName: [], trendValue: []};
      r[o.seriesName].trendName.push(o.gText);
      r[o.seriesName].trendValue.push(o.score);
      return r;
    },{});
data.dt_compartive.forEach(o => {
  result[o.gText].PercentageDiff = o.percentDiff;
});
var output = Object.values(result);
console.log(output);

＆＃13;

如何将JSON格式化为单个javascript对象

1 个答案: