JavaScript循环和对象创建

问题描述 投票:1回答:1

我有以下JSON:

[
  {
    "name": "sp5",
    "damage": "68",
    "penetration": "35",
    "class1": "6",
    "class2": "6",https://stackoverflow.com/editing-help
    "class3": "6",
    "class4": "5",
    "class5": "3",
    "class6": "2"
  },
  {
    "name": "sp6",
    "damage": "58",
    "penetration": "43",
    "class1": "6",
    "class2": "6",
    "class3": "6",
    "class4": "6",
    "class5": "5",
    "class6": "4"
  }
]

我有一个函数,它遍历数组中的对象,然后遍历该对象的属性/键,并尝试通过每个对象的class6对其属性进行破坏,并将其推到其自己的新对象上,并转移到chart.data.datasets上数组。我在每个对象数据数组中得到32个值,而不是所需的8。

function createObjectsForChart(data) {
        console.log(`Data: ${data}`);
        const chart = {
          type: 'bar',
            data: {
                labels: ["Damage", "Penetration", "Class1", "Class 2", "Class 3", "Class 4", "Class 5", "Class 6"],
                datasets: [],
            }
          }
          const dataset = {
            label: "",
            fillColor: "rgba(220,220,220,0.5)",
            strokeColor: "rgba(220,220,220,0.8)",
            highlightFill: "rgba(220,220,220,0.75)",
            highlightStroke: "rgba(220,220,220,1)",
            data: []
          }

          let myChart = Object.create(chart);

          data.forEach((item, i) => {
            console.log(`Item: ${item}, Index: ${i}`);


            console.log(`Data length: ${data.length}`);
            //data[i]

            let myData = Object.create(dataset);
            count = 0;
            for (const property in item) {
              if(count >= 1) {

                //console.log(`Value: ${data[i][property]}`);
                myData.data.push(item[property]);
              }
              console.log(`Property: ${property}, Value: ${item[property]}`);
              count++;
            }
            myData.label = data[i].name;
            myChart.data.datasets.push(myData);
          });






          //myChart.data.datasets[0].highlightStroke = "";
          console.log(myChart.data.datasets);
      }
javascript arrays loops object prototype
1个回答
0
投票

这里的问题是您试图使用dataset创建Object.create变量的深层副本。它不会创建深层副本,并且所有副本将共享一个data数组的单个实例。作为快速的肮脏修复程序,您可以使用Object.assign({}, dataset, {data: []})

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