使用Typescript中的conition对象过滤嵌套数组的数组

Question

我有如下所示的对象数组（Json格式）

var datas = [
  {
    "Id": "1",
    // Here 10 fields
    "tests": [
            {
                "id":"1-1",
                "isSelected": true,
            },
            {
                "id":"1-2",
                "isSelected": false,
            },
        ]
  },
  {
    "Id": "2",
    // Here 10 fields
    "tests": [
            {
                "id":"2-1",
                "isSelected": true,
            },
            {
                "id":"2-2",
                "isSelected": true,
            },
        ]
  },
  {
    "Id": "3",
    // Here 10 fields
    "tests": [
            {
                "id":"3-1",
                "isSelected": false,
            },
            {
                "id":"3-2",
                "isSelected": false,
            },
        ]
  }
]

我尝试如下

var filteredData = datas.filter( t => t.tests.filter(o => o.isSelected));

获取所有具有测试值为"isSelected"的数据是true，它不起作用。反正有没有获取以下格式的数据。

[
  {
    "Id": "1",
    // Here 10 fields
    "t": [
            {
                "id":"1-1",
                "isSelected": true,
            }
        ]
  },
  {
    "Id": "2",
    // Here 10 fields
    "tests": [
            {
                "id":"2-1",
                "isSelected": true,
            },
            {
                "id":"2-2",
                "isSelected": true,
            },
        ]
  }
]

Answer 1

您在正确的轨道上，但是您正在从外部filter回调中返回内部filter的结果（数组）。数组是真实的，因此您要保留所有内容。

您需要分两个步骤处理每个外部对象：

1. 过滤其tests和

2. 如果tests为空，则将其完全过滤掉

所以：

var filteredData = datas.filter(t => {
  t.tests = t.tests.filter(o => o.isSelected);
  return t.tests.length !== 0;
});

var datas = [
  {
    "Id": "1",
    // Here 10 fields
    "tests": [
            {
                "id":"1-1",
                "isSelected": true,
            },
            {
                "id":"1-2",
                "isSelected": false,
            },
        ]
  },
  {
    "Id": "2",
    // Here 10 fields
    "tests": [
            {
                "id":"2-1",
                "isSelected": true,
            },
            {
                "id":"2-2",
                "isSelected": true,
            },
        ]
  },
  {
    "Id": "3",
    // Here 10 fields
    "tests": [
            {
                "id":"3-1",
                "isSelected": false,
            },
            {
                "id":"3-2",
                "isSelected": false,
            },
        ]
  }
]

var filteredData = datas.filter(t => {
  t.tests = t.tests.filter(o => o.isSelected);
  return t.tests.length !== 0;
});

console.log(filteredData);

.as-console-wrapper {
  max-height: 100% !important;
}

或者如果您真的想要简洁的箭头功能：

所以：

var filteredData = datas.filter(t =>
  (t.tests = t.tests.filter(o => o.isSelected)).length !== 0
);

var datas = [
  {
    "Id": "1",
    // Here 10 fields
    "tests": [
            {
                "id":"1-1",
                "isSelected": true,
            },
            {
                "id":"1-2",
                "isSelected": false,
            },
        ]
  },
  {
    "Id": "2",
    // Here 10 fields
    "tests": [
            {
                "id":"2-1",
                "isSelected": true,
            },
            {
                "id":"2-2",
                "isSelected": true,
            },
        ]
  },
  {
    "Id": "3",
    // Here 10 fields
    "tests": [
            {
                "id":"3-1",
                "isSelected": false,
            },
            {
                "id":"3-2",
                "isSelected": false,
            },
        ]
  }
]

var filteredData = datas.filter(t =>
  (t.tests = t.tests.filter(o => o.isSelected)).length !== 0
);

console.log(filteredData);

.as-console-wrapper {
  max-height: 100% !important;
}