Question

我很抱歉，如果以前已经解决了这个问题，但是我无法使它与我找到的任何内容一起使用。

假设我有2个数组-arr1，arr2。如果属性arr1在id和arr1中匹配，我想更新arr2中的对象。 arr2中存在但arr1中不存在的对象-表示属性id在arr1中不存在-应该被推到arr1。

示例：

let arr1 = [
  {id: 0, name: "John"},
  {id: 1, name: "Sara"},
  {id: 2, name: "Domnic"},
  {id: 3, name: "Bravo"}
]

let arr2 = [
  {id: 0, name: "Mark"},
  {id: 4, name: "Sara"}
] 

# Expected Outcome

let outcome = [
  {id: 0, name: "Mark"},
  {id: 1, name: "Sara"},
  {id: 2, name: "Domnic"},
  {id: 3, name: "Bravo"},
  {id: 4, name: "Sara"}
]

Answer 1

您可以为此使用reduce和find

const arr1 = [
  {id: 0, name: "John"},
  {id: 1, name: "Sara"},
  {id: 2, name: "Domnic"},
  {id: 3, name: "Bravo"}
];

const arr2 = [
  {id: 0, name: "Mark"},
  {id: 4, name: "Sara"}
];

arr2.reduce((res, item) => {
  const existingItem = res.find(x => x.id === item.id);
  if (existingItem) { existingItem.name = item.name; }
  else { res.push(item); }
  return res;
}, arr1);

console.log(arr1);

Answer 2

您可以这样做：

$END$

Answer 3

假设您要对If ThisWorkbook.Worksheets("Annexure").Range("G8").Value = 2 Then bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click ElseIf ThisWorkbook.Worksheets("Annexure").Range("G8").Value = 3 Then bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click ElseIf ThisWorkbook.Worksheets("Annexure").Range("G8").Value = 4 Then bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click ElseIf ThisWorkbook.Worksheets("Annexure").Range("G8").Value = 5 Then bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click bot.FindElementByXPath("//a[@class='add btn-add-raw-materials action']//i[@class='fa fa-plus-circle']").Click End If bot.FindElementById("export-HS-code").SendKeys ThisWorkbook.Worksheets("Annexure").Range("B11") bot.FindElementById("raw-material-description-1").SendKeys ThisWorkbook.Worksheets("Annexure").Range("C11") bot.FindElementById("raw-material-quantity-unit-1").SendKeys ThisWorkbook.Worksheets("Annexure").Range("D11") bot.FindElementByName("raw_material_unit[1]").SendKeys ThisWorkbook.Worksheets("Annexure").Range("E11") bot.FindElementById("raw-material-unit-cif-value-1").SendKeys ThisWorkbook.Worksheets("Annexure").Range("F11") bot.FindElementById("raw-material-total-cif-value-1").SendKeys ThisWorkbook.Worksheets("Annexure").Range("G11") bot.FindElementByName("raw_country_id[1]").SendKeys ThisWorkbook.Worksheets("Annexure").Range("H11") bot.FindElementById("bill-invoice-no-1").SendKeys ThisWorkbook.Worksheets("Annexure").Range("I11") bot.FindElementById("bill-invoice-date-1").SendKeys ThisWorkbook.Worksheets("Annexure").Range("J11") bot.FindElementById("supplier-details-1").SendKeys ThisWorkbook.Worksheets("Annexure").Range("K11")中的对象进行突变而不是创建新对象，一种方法是使用for...of对arr1中的对象进行迭代，然后检查是否存在已使用Array.prototype.find()在arr2中具有相同id的对象：

如果有一个，则用Object.assign对其进行突变。
否则，将新对象推送到arr1：

arr1

const arr1 = [
  { id: 0, name: 'John' },
  { id: 1, name: 'Sara' },
  { id: 2, name: 'Domnic' },
  { id: 3, name: 'Bravo' },
];

const arr2 = [
  { id: 0, name: 'Mark', sometingElse: 123 },
  { id: 2, foo: 'bar' },
  { id: 4, name: 'Sara' },
];

for (const currentElement of arr2) {
  let previousElement = arr1.find(el => el.id === currentElement.id);
  
  if (previousElement) {
    Object.assign(previousElement, currentElement);
  } else {
    arr1.push(currentElement);
  }
}

console.log(arr1);

Answer 4

如果您想尝试其他方法，可以使用foreach和filter来实现

let arr1 = [
  {id: 0, name: "John"},
  {id: 1, name: "Sara"},
  {id: 2, name: "Domnic"},
  {id: 3, name: "Bravo"}
]

let arr2 = [
  {id: 0, name: "Mark"},
  {id: 4, name: "Sara"}]


  arr1.forEach(x=>{
    arr2.forEach(y=>{
      if(x.id==y.id){
        x.name=y.name
      }
    })
  })
 arr2.filter((a)=>{if(!arr1.some(b=>a.id==b.id)) arr1.push(a)})


  console.log(arr1)

Answer 5

您应该可以使用Array.prototype.find来解决这个问题！

let arr1 = [
  {id: 0, name: "John"},
  {id: 1, name: "Sara"},
  {id: 2, name: "Domnic"},
  {id: 3, name: "Bravo"}
];

let arr2 = [
  {id: 0, name: "Mark"},
  {id: 4, name: "Sara"}
];

let updateArrayOfObjects = (arr1, arr2) => {
  for (let obj of arr2) {
    let item = arr1.find(v => v.id === obj.id);
    if (item) item.name = obj.name;
    else      arr1.push({ ...obj });
  }
  return arr1;
};

console.log(updateArrayOfObjects(arr1, arr2));

用对象数组更新对象数组

5 个答案: