Question

我想合并两个对象数组，以使其更易于以HTML显示。该函数应在arr1中找到名为“ id”的键，在arr2中找到“ source”的键的匹配值。看起来是这样的：

   let arr1 = [
   {id = 1,
   name = "Anna"},

   {id = 2,
   name = "Chris"}
   ]

   let arr2 = [
   {childName = "Brian",
   {source = 1}},
   {childName = "Connie",
   {source = 2}}
   {childName = "Dory",
   {source = 1}}
   ]

我尝试了不同的方法，最好的方法是在数组上使用forEach和filter。我正在尝试在arr1对象中设置一个名为“儿童”的新属性。

  arr1.forEach(el => el.children = arr2.filter(checkMatch));

            function checkMatch(child){
                for(let i=0;i<arr1.length;i++){
                    child.childName.source === arr1[i].id
                }
            }

这会导致在第一个对象中添加适当的子代（Anna现在有Brian和Dory），这是正确的，但是它也将相同的子代添加到了第二个对象（因此Chris也有Brian和Dory）。我的错在哪里？我猜想循环无法按我希望的方式工作，但是我不知道哪个循环以及如何发生。

Answer 1

由于用于创建arr1和arr2对象的语法无效，因此我尝试猜测对象的结构。

let arr1 = [
  {
    id: 1,
    name: "Anna"
  },
  {
    id: 2,
    name: "Chris"
  }
];

let arr2 = [
  {
    childName: "Brian",
    source: 1
  },
  {
    childName: "Connie",
    source: 2
  },
  {
    childName: "Dory",
    source: 1
  }
];

arr2.map((child) => {
  for (let parent of arr1) {
    if (parent.id == child.source) {
      if (!parent.children) {
        parent.children = [];
      }
      parent.children.push(child);
    }
  }
});

console.log(arr1);

Answer 2

您的JSON出现问题，但我整理了一下，这里是使用map和filter的选项

const arr1 = [{
    id: 1,
    name: "Anna"
  },
  {
    id: 2,
    name: "Chris"
  }];

const arr2 = [{
    childName: "Brian",
    parent: {
      source: 1
    }
  },
  {
    childName: "Connie",
    parent: {
      source: 2
    }
  },
  {
    childName: "Dory", 
    parent: {
      source: 1
    }
  }];

let merge = arr1.map(p => {
  p.children = arr2.filter(c => c.parent.source === p.id).map(c => c.childName);
  return p;
});

console.log(merge);

Answer 3

您的json有一些您应该使用的问题

：

代替

=

还有一些花括号使结构不正确，但是我想在这里要做的是用主题的childNames填充children子数组，这是我的方法：

     var json = 
        [
           {
           "id" : 1,
           "name" : "Anna"
           },
           {
           "id" : 2,
           "name" : "Chris"
           }
          	 ];
             
         var subJson = [
           {
           "childName" : "Brian",
           "source" : 1
           },
           {
           "childName" : "Connie",
           "source" : 2
           },
           {"childName" : "Dory",
           "source" : 1
           }
           ];
        
        var newJson = [];
        $.each(json,function(index1,item){
        newJson.push({"id":item.id,"name":item.name, "children": []});
        	$.each(subJson,function(index2,subitem){
        	if(subitem.source == item.id){
          newJson[newJson.length - 1].children.push({"childName":subitem.childName}) ;
          }
          })
        })
        
        console.log(newJson);

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

希望有帮助

Answer 4

以下使用Map来存储和方便查找父母。

const parents = [
  {
    id: 1,
    name: "Anna"
  },
  {
    id: 2,
    name: "Chris"
  }
]

const children = [
  {
    childName: "Brian",
    source: 1
  },
  {
    childName: "Connie",
    source: 2
  },
  {
    childName: "Dory",
    source: 1
  }
]

// Create a map for easy lookup of parents.
const parentMap = new Map()

// Add parents to the map, based on their id.
parents.forEach(parent => parentMap.set(parent.id, parent))

// Add children to their parents.
children.forEach((child) => {
  // Get the parent from the map.
  const parent = parentMap.get(child.source)

  // Handle parent not found error.
  if (!parent) { return console.error('parent not found for', child.childName)}

  // Create the children array if it doesn't already exist.
  parent.children = parent.children || []

  // Add the child to the parent's children array.
  parent.children.push(child)
})

// Output the result.
Array.from(parentMap).forEach(parent => console.log(parent[1]))

结果：

{ 
  id: 1,
  name: 'Anna',
  children: [ 
    { childName: 'Brian', source: 1 },
    { childName: 'Dory', source: 1 } 
  ] 
}
{ 
  id: 2,
  name: 'Chris',
  children: [ 
    { childName: 'Connie', source: 2 } 
  ] 
}

通过匹配值组合对象数组

4 个答案: