Question

我的目标是，如果digital_assets和产品中的id匹配，然后从products对象中获取digital_assets和ProductName的URL值。我能够遍历该对象并获取digital_assets和产品的值，但需要一些帮助来基于ID比较这两个对象以获得URL和ProductName的值。下面是我到目前为止所做的。

var data = [{
        "digital_assets": [{
            "id": "AA001",
            "url": "https://via.placeholder.com/150"
        },{
            "id": "AA002",
            "url": "https://via.placeholder.com/150"
        }]
    }, {
        "products": [{
            "id": ["BB001", "AA001"],
            "ProductName": "PROD 485"
        },{
            "id": ["BB002", "AA002"],
            "ProductName": "PROD 555"
        }]
    }
];

$.each(data, function () {
    var data = this;
    //console.log(data);

    $.each(data.digital_assets, function () {
        var dAssets = this,
            id = dAssets['id'];
        // console.log(id);
    });

    $.each(data.products, function () {
        var proData = this,
            prod_id = proData['id'];
        // console.log(prod_id);

        $.each(prod_id, function () {
            var arr_id = this;
            console.log(arr_id);
        });
    });
});

我需要创建新数组并将值推入新数组吗？然后将这些数组concat（）合并为1。？失去任何帮助将不胜感激。

Answer 1

这是您可以通过Array.reduce，Array.includes，Object.entries和Array.forEach进行此操作的一种方式：

var data = [{ "digital_assets": [{ "id": "AA001", "url": "https://via.placeholder.com/150" }, { "id": "AA002", "url": "https://via.placeholder.com/150" } ] }, { "products": [{ "id": ["BB001", "AA001"], "ProductName": "PROD 485" }, { "id": ["BB002", "AA002"], "ProductName": "PROD 555" } ] } ]

const result = data.reduce((r,c) => { 
  Object.entries(c).forEach(([k,v]) => 
    k == 'digital_assets' 
     ? v.forEach(({id, url}) => r[id] = ({ id, url }))
     : v.forEach(x => Object.keys(r).forEach(k => x.id.includes(k) 
       ? r[k].ProductName = x.ProductName 
       : null))
  )
  return r
}, {})

console.log(Object.values(result))

Answer 2

您可以使用Array.prototype.find，Array.prototype.includes和Array.prototype.map非常优雅地实现这一目标。

let data = [
   {
    "digital_assets": [
      {
        "id": "AA001",
        "url": "https://via.placeholder.com/150"
      },
      {
        "id": "AA002",
        "url": "https://via.placeholder.com/150"
      }
    ]
  },
  {
    "products": [
      {
        "id": ["BB001", "AA001"],
        "ProductName": "PROD 485"
      },
      {
        "id": ["BB002","AA002"],
        "ProductName": "PROD 555"
      }
    ]
  }
];

// Find the 'digital_assets' array
let assets = data.find(d => d['digital_assets'])['digital_assets'];

// Find the 'products' array
let products = data.find(d => d['products'])['products'];

// Return an array of composed asset objects
let details = assets.map(a => {
  return {
    id : a.id,
    url : a.url
    name : products.find(p => p.id.includes(a.id)).ProductName
  };
});

console.log(details);

Answer 3

已更改答案以适合您的需求：

var data = [
       {
        "digital_assets": [
          {
            "id": "AA001",
            "url": "https://via.placeholder.com/150"
          },
          {
            "id": "AA002",
            "url": "https://via.placeholder.com/150"
          }
        ]
      },
      {
        "products": [
          {
            "id": ["BB001", "AA001"],
            "ProductName": "PROD 485"
          },
          {
            "id": ["BB002","AA002"],
            "ProductName": "PROD 555"
          }
        ]
      }
    ]
let matchingIds = [];
let data_assetsObject = data.find(element => {
    return Object.keys(element).includes("digital_assets")
})
let productsObject = data.find(element => {
    return Object.keys(element).includes("products")
})
data_assetsObject["digital_assets"].forEach(da => {
  productsObject["products"].forEach(product => {
    if (product.id.includes(da.id)){
      matchingIds.push({
        url: da.url,
        productName: product.ProductName
      })
    }
  })
})
console.log(matchingIds);

有效的提琴：https://jsfiddle.net/z2ak1fvs/3/

希望能有所帮助。如果您不想使用新数组，也可以将各自的数据存储在要遍历的元素中。

编辑：我想我知道我为什么被拒绝。我的示例通过使数据成为对象而不是数组来工作。更改了代码段以更清楚地显示这一点。为什么数据仍然是数组？有什么原因吗？还是可以将其转换为对象？

编辑nr2：我根据您的评论理解了代码，从而更改了代码以满足期望。它现在使用您的数据结构，无论数据中有什么内容，您现在都可以搜索包含digital_assets / products属性的对象。

欢呼

Answer 4

https://jsfiddle.net/2b1zutvx/

使用地图。

 var myobj = data[0].digital_assets.map(function(x) {
    return {
        id: x.id,
        url: x.url,
        ProductName: data[1].products.filter(f => f.id.indexOf(x.id) > -1).map(m => m.ProductName)
    };
});

遍历对象的JSON数组，并根据对象中的匹配ID获取属性

4 个答案: