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從 javascript 文件調(diào)用時(shí)，函數(shù)不會(huì)在剃刀頁(yè)面上執(zhí)行

從 javascript 文件調(diào)用時(shí)，函數(shù)不會(huì)在剃刀頁(yè)面上執(zhí)行

搖曳的薔薇 2023-05-19 19:44:47
我有兩個(gè) JS 數(shù)組。他們是arr1和arr2。我想創(chuàng)建另一個(gè)名為arr3. 現(xiàn)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(yīng)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(yù)期輸出：const arr3 = [{  des: "cont3",  note: "cont3",  pro_code: "QAS"}我試過(guò)的代碼，不起作用。請(qǐng)幫我解決這個(gè)問(wèn)題。
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3 回答

?
慕哥9229398

TA貢獻(xiàn)1877條經(jīng)驗(yàn) 獲得超6個(gè)贊

Set您可以對(duì)所有值取 apro_code并通過(guò)檢查該值是否不在集合中來(lái)過(guò)濾第二個(gè)數(shù)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);


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?
滄海一幻覺(jué)

TA貢獻(xiàn)1824條經(jīng)驗(yàn) 獲得超5個(gè)贊

這是另一個(gè)優(yōu)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);



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?
波斯汪

TA貢獻(xiàn)1811條經(jīng)驗(yàn) 獲得超4個(gè)贊

簡(jiǎn)而言之，您的要求是過(guò)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);


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