如何從陣列物件中洗掉元素-有解無憂

我使用陣列物件，我想洗掉 1 個陣列中的特定值。

let medici= ["Person1","Person2", "Person3", "Person4", "Person5",  "Person6" ];
let giorni = ["Lun", "Mar", "Mer","Gio","Ven"]; let presenti ={};
for  (let giorno of giorni){ presenti[giorno] = medici; }

我得到：

Gio: ["Person1", "Person2", "Person3", "Person4", "Person5", "Person6"],
  Lun: ["Person1", "Person2", "Person3", "Person4", "Person5", "Person6"],
  Mar: ["Person1", "Person2", "Person3", "Person4", "Person5", "Person6"],
  Mer: ["Person1", "Person2", "Person3", "Person4", "Person5", "Person6"],
  Ven: ["Person1", "Person2", "Person3", "Person4", "Person5", "Person6"]

現在我想洗掉特定陣列中的特定值：

giorno="Ven";
nome="Person1";
presenti[giorno].splice(presenti[giorno].indexOf(nome), 1);

它洗掉了所有陣列中的值！！！不僅在“Ven”中...如何洗掉特定陣列中的單個元素而不影響物件的其他陣列？

謝謝

uj5u.com熱心網友回復：

const medici = ["Person1","Person2", "Person3", "Person4", "Person5",  "Person6" ];
const gironi = ["Lun", "Mar", "Mer","Gio","Ven"];
const presenti = {};

for (const giorno of gironi) {
    presenti[giorno] = [...medici];
}

presenti["Ven"].splice(presenti["Ven"].indexOf("Person1"), 1);

console.log(presenti);

您的問題與記憶體參考有關，當您將兩個變數等同起來時，您不會創建另一個相同的物件，而是將物件參考從一個變數復制到另一個變數，因此如果您修改Xchange Y，更正您可以使用擴展運算子創建另一個陣列[...medici]

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax

uj5u.com熱心網友回復：

let medici= ["Person1","Person2", "Person3", "Person4", "Person5",  "Person6" ];
let giorni = ["Lun", "Mar", "Mer","Gio","Ven"]; let presenti ={};
for  (let giorno of giorni){ presenti[giorno] = medici; }

giorno="Ven";
nome="Person1";

const filteredData = presenti[giorno].filter((p)=>p!==nome)

presenti[giorno] = filteredData

console.log(presenti)

這樣做是因為在 JavaScriptArray中Objects被稱為參考型別。[在此處了解更多資訊]。1

要解決您的問題，您可以這樣做：

giorno="Ven";

nome="Person1";

Person1首先從陣列中過濾掉。const filteredData = presenti[giorno].filter((p)=>p!==nome)

然后用Ven更新的物件替換filteredData

presenti[giorno] = filteredData

uj5u.com熱心網友回復：

構建物件時，所有陣列都具有相同的參考，因此您需要slice medici對每個物件值進行參考。看看按值復制陣列

現在你可以使用方法了。

let medici= ["Person1","Person2", "Person3", "Person4", "Person5",  "Person6" ];
let giorni = ["Lun", "Mar", "Mer","Gio","Ven"]; let presenti ={};
for  (let giorno of giorni){ presenti[giorno] = medici.slice(); }

let giorno="Ven";
let nome="Person1";

presenti[giorno].splice(presenti[giorno].indexOf(nome), 1);
console.log(presenti)

uj5u.com熱心網友回復：

使用這個功能：

const obj= {
  "Lun": [
    "Person1",
    "Person2",
    "Person3",
    "Person4",
    "Person5",
    "Person6"
  ],
  "Mar": [
    "Person1",
    "Person2",
    "Person3",
    "Person4",
    "Person5",
    "Person6"
  ],
  "Mer": [
    "Person1",
    "Person2",
    "Person3",
    "Person4",
    "Person5",
    "Person6"
  ],
  "Gio": [
    "Person1",
    "Person2",
    "Person3",
    "Person4",
    "Person5",
    "Person6"
  ],
  "Ven": [
    "Person2",
    "Person3",
    "Person4",
    "Person5",
    "Person6"
  ]
}
function changeObject(mainObject, {georno, nome}) {
  //action -> {georno : "Ven", nome : "Person1"}
  let newValue = mainObject[georno].filter(person => person !== nome)
  let changedObject = Object.create(null)
  changedObject[georno] = newValue
  return Object.assign({}, mainObject, changedObject)
}



console.log(changeObject(obj, {georno:  "Ven", name : "Person1"}))

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