如何通过在js中迭代对象来过滤对象答案

【问题标题】：How to filter from an object by iterating over it in js如何通过在js中迭代对象来过滤对象
【发布时间】：2021-12-25 20:26:18
【问题描述】：

我试图通过迭代从对象中获取“类型”的值。对象看起来像这样。

{
  "team": {
    "table": [
      {
        "cityCode": 123,
        "list": {
          "players": [
            {
              "name": "peter",
              "school": "x",
              "awards": {
                "type": "gold"
              },
              "year": 2019
            }
          ]
        }
      },
      {
        "cityCode": 456,
        "list": {
          "players": [
            {
              "name": "Dave",
              "school": "y",
              "awards": {
                "type": "silver"
              },
              "year": 2018
            }
          ]
        }
      }
    ]
  }
}

我可以使用这个来获取类型值：

const table = team.table;
for (let i = 0; i < table.length; i++) {
  const values = {
    type: table[i].list.players
      .filter((a) => a.awards != null)
      .map((a) => a.awards.type)
      .join(" "),
  };
}

但是，我想在“列表”上使用另一个过滤器来过滤非空列表。那么我该如何实现呢。

【问题讨论】：

标签： javascript arrays json sorting object

【解决方案1】：

您要检查 'list' 键是否存在于 team.table JSON 对象中

你可以检查

if(table[i].hasOwnProperty('list')){

}

代码是

const table = team.table;
for (let i = 0; i < table.length; i++) {
if(table[i].hasOwnProperty('list')){
     const values = {
       type: table[i].list.players
         .filter((a) => a.awards != null)
         .map((a) => a.awards.type)
         .join(" "),
     };
  }
}

【讨论】：

【解决方案2】：

1)您可以使用flatMap 和map 获取所有type：

obj.team.table.flatMap((o) => o.list.players.map((o) => o.awards.type))

const obj = {
  team: {
    table: [
      {
        cityCode: 123,
        list: {
          players: [
            {
              name: "peter",
              school: "x",
              awards: {
                type: "gold",
              },
              year: 2019,
            },
          ],
        },
      },
      {
        cityCode: 456,
        list: {
          players: [
            {
              name: "Dave",
              school: "y",
              awards: {
                type: "silver",
              },
              year: 2018,
            },
          ],
        },
      },
    ],
  },
};

const types = obj.team.table.flatMap((o) => o.list.players.map((o) => o.awards.type));
console.log(types);

2) 使用forEach 并解构为：

const obj = {
  team: {
    table: [
      {
        cityCode: 123,
        list: {
          players: [
            {
              name: "peter",
              school: "x",
              awards: {
                type: "gold",
              },
              year: 2019,
            },
          ],
        },
      },
      {
        cityCode: 456,
        list: {
          players: [
            {
              name: "Dave",
              school: "y",
              awards: {
                type: "silver",
              },
              year: 2018,
            },
          ],
        },
      },
    ],
  },
};

const table = obj.team.table;
const types = [];
for (let i = 0; i < table.length; i++) {
  const { list: { players } } = table[i]
  players.forEach(({ awards: { type }}) => types.push(type))
}

console.log(types);

【讨论】：

【解决方案3】：

使用forEach 会更简洁。由于您的数据结构，您将需要 2 个 forEach。但下面的代码会：

检查awards是否为空
检查awards.type是否为空

const data = {
  "team": {
    "table": [
      {
        "cityCode": 123,
        "list": {
          "players": [
            {
              "name": "peter",
              "school": "x",
              "awards": {
                "type": "gold"
              },
              "year": 2019
            }
          ]
        }
      },
      {
        "cityCode": 456,
        "list": {
          "players": [
            {
              "name": "Dave",
              "school": "y",
              "awards": {
                "type": "silver"
              },
              "year": 2018
            },
            {
              "name": "Dave",
              "school": "y",
              "awards": {
                "type": "gold"
              },
              "year": 2016
            }
          ]
        }
      },
      {
        "cityCode": 444,
        "list": {
          "players": [
            {
              "name": "James",
              "school": "y",
              "awards": {
                "type": null
              },
              "year": 2016
            }
          ]
        }
      },
      
      {
        "cityCode": 555,
        "list": {
          "players": [
            {
              "name": "Name 101",
              "school": "y",
              "awards": {
                "type": "platinum"
              },
              "year": 2016
            },
            {
              "name": "Name 102",
              "school": "y",
              "awards": {
                "type": null
              },
              "year": 2016
            },
            {
              "name": "Name 103",
              "school": "y",
              "awards": null,
              "year": 2016
            },
            
          ]
        }
      }
    ]
  }
}

// Expanded your data with more items
const data1 = data.team.table;

let types = []
data1.forEach((item, index)  => {
  
  item.list.players.forEach((player) => {
    const awards = player.awards;
    if (awards !== null && awards.type !== null) {
        types = [...types, awards.type];
    }
  })
  
})

// Get the list of types
console.log(types);

// Get unique list of types
let unique_types = [...new Set(types)]
console.log(unique_types);

【讨论】：