从 javascript 文件调用时，函数不会在剃刀页面上执行

Question

我有两个 JS 数组。他们是arr1和arr2。我想创建另一个名为arr3. 现在它想要比较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。所以它应该包含在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必须包含 中的元素arr1。中不能有附加元素arr2。预期输出：const arr3 = [{  des: "cont3",  note: "cont3",  pro_code: "QAS"}我试过的代码，不起作用。请帮我解决这个问题。

Answer 1

Set您可以对所有值取 apro_code并通过检查该值是否不在集合中来过滤第二个数组。

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);

Answer 2

这是另一个优化的解决方案

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);

Answer 3

简而言之，您的要求是过滤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);