Question

我有一个对象，我想将类型的值作为属性访问另一个对象，例如a.Fax，a.Email，a.Call等。但是我在下面写下了一个函数，它将值推入attt。因此，我不想访问ttt，而是要访问ttt中存储的类型（即传真，电子邮件，电话，文本）中的每个值。

export const ActionTypes2 = [
    { value: "OtherLawyer", label: "Other Lawyer", types: ["Fax", "Email", "Call", "Text"] },
    { value: "OtherClerk", label: "Other Clerk", types: ["Fax", "Email", "Call", "Text"]  },
    { value: "BrokerageAgent", label: "Brokerage Agent", types: ["Fax", "Email", "Call", "Text"]  },
    { value: "Lawyer", label: "Lawyer", types: ["Email", "Call", "Text"]  },
    { value: "Clerk", label: "Clerk", types: ["Email", "Call", "Text"] },
    { value: "Client", label: "Client", types: ["Email", "Call", "Text"] },
    { value: "ListingAgent", label: "Listing Agent", types: ["Email", "Call", "Text"] },
    { value: "CooperatingAgent", label: "Cooperating Agent", types: ["Email", "Call", "Text"] }
]


getActionTypes() {
    let t = ActionTypes2;
    let f = <any>{};
    t.map((tt: any) => {
      tt.types.map((ttt: any) => {
        if (!f.ttt) {
          f.ttt = [];
        }
        f.ttt.push(t);
        // console.log(ttt);
      })
    })
    console.log(f);
  }

Answer 1

下面怎么样，

getActionTypes() {
   let t = ActionTypes2;
   let f = <any>{};
   t.map((tt: any) => {
     if (!f.ttt) {
       f.ttt = [];
     }
     f.ttt.push(tt.types);
   })
   console.log(f.ttt);  }

Answer 2

您应按如下所示将f.ttt替换为f [ttt]

getActionTypes() {
    let t = ActionTypes2;
    let f = <any>{};
    t.map((tt: any) => {
        tt.types.map((ttt: any) => {
            if (!f[ttt]) {
                f[ttt] = [];
            }
            f[ttt].push(t);
            // console.log(ttt);
        })
    })
    console.log(f);
}

Answer 3

您可以将数组here {{}}放置到一个对象上（请注意，SO打印控制台有些奇怪，请按F12键并查看控制台以获取实际结果）：

const data = [{"value":"OtherLawyer","label":"Other Lawyer","types":["Fax","Email","Call","Text"]},{"value":"OtherClerk","label":"Other Clerk","types":["Fax","Email","Call","Text"]},{"value":"BrokerageAgent","label":"Brokerage Agent","types":["Fax","Email","Call","Text"]},{"value":"Lawyer","label":"Lawyer","types":["Email","Call","Text"]},{"value":"Clerk","label":"Clerk","types":["Email","Call","Text"]},{"value":"Client","label":"Client","types":["Email","Call","Text"]},{"value":"ListingAgent","label":"Listing Agent","types":["Email","Call","Text"]},{"value":"CooperatingAgent","label":"Cooperating Agent","types":["Email","Call","Text"]}];

const groupByType = (data) =>
  data.reduce(
    (result, { types, ...rest }) =>
      types.reduce((result, type) => {
        result[type] = result[type] || [];
        result[type].push(rest);
        return result;
      }, result),
    {},
  );
console.log(groupByType(data));

reduce用于将'T数组转换为'U数组，甚至产生相同类型但翻倍的新数组（例如[1,2,3].map(x=>x*2)）。传递给map的函数应该是纯函数，因此它不应使用或变异该函数外部定义的内容或更改要映射的数组。

访问数组内的对象

3 个答案: