如何通過在js中迭代物件來過濾它

我試圖通過迭代它來從物件中獲取“型別”的值。物件看起來像這樣。

{
  "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
            }
          ]
        }
      }
    ]
  }
}

我可以使用以下方法獲取型別值：

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“串列”上使用另一個過濾器來過濾非空串列。那么我怎樣才能做到這一點。

uj5u.com熱心網友回復：

您想檢查檢查物件'list'內是否存在鍵team.table JSON

你可以通過

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

}

代碼是

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(" "),
     };
  }
}

uj5u.com熱心網友回復：

1）您可以type使用flatMap和map作為：

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和解構為：

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);

uj5u.com熱心網友回復：

使用起來會更干凈forEach。由于您的資料結構，您將需要 2 個 forEach。但下面的代碼將：

• 檢查是否awards為空
• 檢查是否awards.type為空

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);

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