更改陣列中特定物件的值-有解無憂

如何將每個條目的狀態屬性更改為 false 假設組“B”，而不更改其他條目？通過改變陣列或回傳一個新陣列

const personnel = [
  { id: 0, name: "Audrey Holcomb", group: "A", status: true },
  { id: 1, name: "Gwendolyn Heath", group: "B", status: true },
  { id: 2, name: "Rivas Mclean", group: "C", status: true },
  { id: 3, name: "Valeria Schneider", group: "A", status: true },
  { id: 4, name: "Leticia Keith", group: "C", status: true },
  { id: 5, name: "Nadine Woodward", group: "B", status: true },
  { id: 6, name: "Shaffer Oliver", group: "A", status: true },
  { id: 7, name: "Meghan Norman", group: "B", status: true },
  { id: 8, name: "Kyle Smith", group: "C", status: true },
]

uj5u.com熱心網友回復：

使用map和檢查組

const update = (arr, group = "B", status = false) =>
  arr.map((item) => ({
    ...item,
    status: item.group === group ? status : item.status,
  }));

const personnel = [
  { id: 0, name: "Audrey Holcomb", group: "A", status: true },
  { id: 1, name: "Gwendolyn Heath", group: "B", status: true },
  { id: 2, name: "Rivas Mclean", group: "C", status: true },
  { id: 3, name: "Valeria Schneider", group: "A", status: true },
  { id: 4, name: "Leticia Keith", group: "C", status: true },
  { id: 5, name: "Nadine Woodward", group: "B", status: true },
  { id: 6, name: "Shaffer Oliver", group: "A", status: true },
  { id: 7, name: "Meghan Norman", group: "B", status: true },
  { id: 8, name: "Kyle Smith", group: "C", status: true },
];

console.log(update(personnel))

console.log(update(personnel, "C", false))

uj5u.com熱心網友回復：

回圈遍歷陣列，檢查組是否為 B，然后將狀態更改為 false

const personnel = [
  { id: 0, name: "Audrey Holcomb", group: "A", status: true },
  { id: 1, name: "Gwendolyn Heath", group: "B", status: true },
  { id: 2, name: "Rivas Mclean", group: "C", status: true },
  { id: 3, name: "Valeria Schneider", group: "A", status: true },
  { id: 4, name: "Leticia Keith", group: "C", status: true },
  { id: 5, name: "Nadine Woodward", group: "B", status: true },
  { id: 6, name: "Shaffer Oliver", group: "A", status: true },
  { id: 7, name: "Meghan Norman", group: "B", status: true },
  { id: 8, name: "Kyle Smith", group: "C", status: true },
];

for (let i = 0; i < personnel.length; i  ) {
  if (personnel[i]["group"] == "B") {
    personnel[i]["status"] = false;
  }
}

console.log(personnel);

uj5u.com熱心網友回復：

您可以使用 map 回傳一個新陣列并隨時修改 B 組元素。在這個特定的例子中，這個方法沒有什么值得推薦的，我不認為，但在更復雜的情況下它可以簡單得多。

const personnel = [
  {id: 0, name: 'Audrey Holcomb', group: 'A', status: true},
  {id: 1, name: 'Gwendolyn Heath', group: 'B', status: true},
  {id: 2, name: 'Rivas Mclean', group: 'C', status: true},
  {id: 3, name: 'Valeria Schneider', group: 'A', status: true},
  {id: 4, name: 'Leticia Keith', group: 'C', status: true},
  {id: 5, name: 'Nadine Woodward', group: 'B', status: true},
  {id: 6, name: 'Shaffer Oliver', group: 'A', status: true},
  {id: 7, name: 'Meghan Norman', group: 'B', status: true},
  {id: 8, name: 'Kyle Smith', group: 'C', status: true},
];

let changed = personnel.map(m=>{
  m.status = m.group == 'B' ? false : m.status;
  return m;
});

console.log(changed);

您還可以創建一個箭頭函式來使事情更簡潔，然后由于您只在地圖中使用一行，您可以省略return和花括號：

const personnel = [
  {id: 0, name: 'Audrey Holcomb', group: 'A', status: true},
  {id: 1, name: 'Gwendolyn Heath', group: 'B', status: true},
  {id: 2, name: 'Rivas Mclean', group: 'C', status: true},
  {id: 3, name: 'Valeria Schneider', group: 'A', status: true},
  {id: 4, name: 'Leticia Keith', group: 'C', status: true},
  {id: 5, name: 'Nadine Woodward', group: 'B', status: true},
  {id: 6, name: 'Shaffer Oliver', group: 'A', status: true},
  {id: 7, name: 'Meghan Norman', group: 'B', status: true},
  {id: 8, name: 'Kyle Smith', group: 'C', status: true},
];
const checkGroup=(a)=>{
  a.status = a.group=='B' ? false : a.status;
  return a;
};    

let changed = personnel.map(m=>checkGroup(m));

console.log(changed);

為了獲得更多的可擴展性，您可以設定輔助函式接受一個引數，比如要更改的目標：

const personnel = [
  {id: 0, name: 'Audrey Holcomb', group: 'A', status: true},
  {id: 1, name: 'Gwendolyn Heath', group: 'B', status: true},
  {id: 2, name: 'Rivas Mclean', group: 'C', status: true},
  {id: 3, name: 'Valeria Schneider', group: 'A', status: true},
  {id: 4, name: 'Leticia Keith', group: 'C', status: true},
  {id: 5, name: 'Nadine Woodward', group: 'B', status: true},
  {id: 6, name: 'Shaffer Oliver', group: 'A', status: true},
  {id: 7, name: 'Meghan Norman', group: 'B', status: true},
  {id: 8, name: 'Kyle Smith', group: 'C', status: true},
];
const checkGroup=(a,b)=>{
  a.status = a.group==b ? false : a.status;
  return a;
};    

let changed = personnel.map(m=>checkGroup(m, 'B'));

console.log(changed);

你也可以收緊布爾邏輯以避免使用三元運算子，盡管在這種特殊情況下，它可能比其他方式更難閱讀和維護，而且它只是稍微短了一點：

const personnel = [
  {id: 0, name: 'Audrey Holcomb', group: 'A', status: true},
  {id: 1, name: 'Gwendolyn Heath', group: 'B', status: true},
  {id: 2, name: 'Rivas Mclean', group: 'C', status: true},
  {id: 3, name: 'Valeria Schneider', group: 'A', status: true},
  {id: 4, name: 'Leticia Keith', group: 'C', status: true},
  {id: 5, name: 'Nadine Woodward', group: 'B', status: true},
  {id: 6, name: 'Shaffer Oliver', group: 'A', status: true},
  {id: 7, name: 'Meghan Norman', group: 'B', status: true},
  {id: 8, name: 'Kyle Smith', group: 'C', status: true},
];
const checkGroup=(a,b)=>{
  a.status = !(a.group==b || !a.status)
  return a;
};    

let changed = personnel.map(m=>checkGroup(m, 'B'));

console.log(changed);

但是，如果您的用例真的像您所描述的一樣簡單，那么您將很難擊敗 forEach，您可以在一行中完成：

const personnel = [
  {id: 0, name: 'Audrey Holcomb', group: 'A', status: true},
  {id: 1, name: 'Gwendolyn Heath', group: 'B', status: true},
  {id: 2, name: 'Rivas Mclean', group: 'C', status: true},
  {id: 3, name: 'Valeria Schneider', group: 'A', status: true},
  {id: 4, name: 'Leticia Keith', group: 'C', status: true},
  {id: 5, name: 'Nadine Woodward', group: 'B', status: true},
  {id: 6, name: 'Shaffer Oliver', group: 'A', status: true},
  {id: 7, name: 'Meghan Norman', group: 'B', status: true},
  {id: 8, name: 'Kyle Smith', group: 'C', status: true},
];

personnel.forEach(m=>{m.status = m.group == 'B' ? false : m.status});

console.log(personnel);

uj5u.com熱心網友回復：

personnel.map(item => {
      if (item?.group === 'B') {
        item.status = false;
      }
    });

轉載請註明出處，本文鏈接：https://www.uj5u.com/qiye/365174.html

標籤：javascript

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