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從 javascript 文件調用時，函數不會在剃刀頁面上執行

JavaScript

搖曳的薔薇 2023-05-19 19:44:47

我有兩個 JS 數組。他們是arr1和arr2。我想創建另一個名為arr3. 現在它想要比較arr1并插入所有包含和不包含的arr2元素。arr2arr1例子：arr1: 0: des: "cont1" note: "cont1" pro_code: "XXY" 1: des: "cont2" note: "cont2" pro_code: "NNB" 2: des: "cont4" note: "cont4" pro_code: "QQA" 3: des: "cont5" note: "cont5" pro_code: "GFD"arr2： 0: des: "cont1" note: "cont1" pro_code: "XXY" 1: des: "cont2" note: "cont2" pro_code: "NNB" 2: des: "cont3" note: "cont3" pro_code: "QAS"在arr2包含pro_code: QAS. 但它不在arr1。所以它應該包含在arr3.在arr1包含arr1[4] pro_code: GFD. 它should not包含在arr3. Becaseaar1可以包含其他元素。Herearr2必須包含中的元素arr1。中不能有附加元素arr2。預期輸出：const arr3 = [{ des: "cont3", note: "cont3", pro_code: "QAS"}我試過的代碼，不起作用。請幫我解決這個問題。

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3 回答

慕哥9229398

TA貢獻1877條經驗獲得超6個贊

Set您可以對所有值取 apro_code并通過檢查該值是否不在集合中來過濾第二個數組。

const

? ? array1 = [{ des: "cont1", note: "cont1", pro_code: "XXY" }, { des: "cont2", note: "cont2", pro_code: "NNB" }, { des: "cont4", note: "cont4", pro_code: "QQA" }, { des: "cont5", note: "cont5", pro_code: "GFD" }],

? ? array2 = [{ des: "cont1", note: "cont1", pro_code: "XXY" }, { des: "cont2", note: "cont2", pro_code: "NNB" }, { des: "cont3", note: "cont3", pro_code: "QAS" }],

? ? pro_codes = new Set(array1.map(({ pro_code }) => pro_code)),

? ? result = array2.filter(({ pro_code }) => !pro_codes.has(pro_code));

console.log(result);

反對回復 2023-05-19

滄海一幻覺

TA貢獻1824條經驗獲得超5個贊

這是另一個優化的解決方案

const arr1 = [{

des: "cont1",

note: "cont1",

pro_code: "XXY"

}, {

des: "cont2",

note: "cont2",

pro_code: "NNB"

}, {

des: "cont4",

note: "cont4",

pro_code: "QQA"

}, {

des: "cont5",

note: "cont5",

pro_code: "GFD"

}];

const arr2 = [{

des: "cont1",

note: "cont1",

pro_code: "XXY"

}, {

des: "cont2",

note: "cont2",

pro_code: "NNB"

}, {

des: "cont3",

note: "cont3",

pro_code: "QAS"

}];

const results = arr2.filter(({ pro_code: id1 }) => !arr1.some(({ pro_code: id2 }) => id2 === id1));

console.log(results);

反對回復 2023-05-19

波斯汪

TA貢獻1811條經驗獲得超4個贊

簡而言之，您的要求是過濾arr2并保留不在arr1.

const arr3 = arr2.filter(function(arr2item) {

// only keep this item if it is not in arr1

return !arr1.some(function(arr1item) {

return arr1item.pro_code === arr2item.pro_code;

})

});

const arr1 = [{des: "cont1", note: "cont1", pro_code: "XXY"}, {des: "cont2", note: "cont2", pro_code: "NNB"}, {des: "cont4", note: "cont4", pro_code: "QQA"}, {des: "cont5", note: "cont5", pro_code: "GFD"}];

const arr2 = [{des: "cont1", note: "cont1", pro_code: "XXY"}, {des: "cont2", note: "cont2", pro_code: "NNB"}, {des: "cont3", note: "cont3", pro_code: "QAS"}]

const arr3 = arr2.filter(function(arr2item) {

return !arr1.some(function(arr1item) {

return arr1item.pro_code === arr2item.pro_code;

})

});

console.log(arr3);

反對回復 2023-05-19

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從 javascript 文件調用時，函數不會在剃刀頁面上執行

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從 javascript 文件調用時，函數不會在剃刀頁面上執行