合并 Javascript 数组而不进行连接

Question

我需要组合两个数组。我能找到的大多数关于组合数组的建议都使用concat .但是我不想添加到数组的末尾，我需要从array1向array2中的每个对象添加一个键/值对。

我需要合并这个数组1：

[
    "Basket Abandonment",
    "Downloads",
    "App Version"    
]

使用此数组2：

[
    {
        bottom: {
            comp : "",
            details, : "3.1.39 22nd Jul 2015",
            status : "",
            title : "Previous Version",
            value : "8.7%"
        },
        top: {
            details: "3.1.40 25th August 2015", 
            status: "", 
            comp: "", 
            title: "Latest Version", 
            value: "86%",
        }
    },
    {
        bottom: {
            value: "469", 
            title: "Total Reviews", 
            status: "neutral", 
            comp: "same", 
            details: "2 New This Week"
        },
        top:  {
            details: "Version 3.1.40", 
            status: "neutral", 
            comp: "same", 
            title: "Average Rating", 
            value: "4.0"
        }
    },  
    {
        bottom: {
            value: "469", 
            title: "Total Reviews", 
            status: "neutral", 
            comp: "same", 
            details: "2 New This Week"
        },
        top:  {
            details: "Version 3.1.40", 
            status: "neutral", 
            comp: "same", 
            title: "Average Rating", 
            value: "4.0"
        }
    }       
]

在一个新的组合数组中，我需要为每个对象添加一个 title 键使用第一个数组中的值，以便生成的数组如下所示：

[
    {
        title: "Basket Abandonment",
        bottom: {
            comp : "",
            details, : "3.1.39 22nd Jul 2015",
            status : "",
            title : "Previous Version",
            value : "8.7%"
        },
        top: {
            details: "3.1.40 25th August 2015", 
            status: "", 
            comp: "", 
            title: "Latest Version", 
            value: "86%",
        }
    },
    {
        title: "Downloads",
        bottom: {
            value: "469", 
            title: "Total Reviews", 
            status: "neutral", 
            comp: "same", 
            details: "2 New This Week"
        },
        top:  {
            details: "Version 3.1.40", 
            status: "neutral", 
            comp: "same", 
            title: "Average Rating", 
            value: "4.0"
        }
    },  
    {
        title: "App Version",
        bottom: {
            value: "469", 
            title: "Total Reviews", 
            status: "neutral", 
            comp: "same", 
            details: "2 New This Week"
        },
        top:  {
            details: "Version 3.1.40", 
            status: "neutral", 
            comp: "same", 
            title: "Average Rating", 
            value: "4.0"
        }
    }       
]

Answer 1

您可以在

第二个数组中执行一个简单的for-loop，插入从第一个数组中获取的新title属性。 但是，如果你想要一个函数，在不修改源的情况下给你一个新数组，那么一种解决方案是创建一个新数组，将第二个数组中的对象的克隆与第一个数组中的字符串映射，如下所示：

const mixed = objects.map((obj, index) => (clone = {...obj}, clone.title = titles[index], clone));

这里有一个将函数与数组一起使用的示例：

const titles = [
  "Basket Abandonment",
  "Downloads",
  "App Version"
];
const objects = [
  {
    bottom: {
      comp: "",
      details : "3.1.39 22nd Jul 2015",
      status: "",
      title: "Previous Version",
      value: "8.7%"
    },
    top: {
      details: "3.1.40 25th August 2015",
      status: "",
      comp: "",
      title: "Latest Version",
      value: "86%",
    }
  },
  {
    bottom: {
      value: "469",
      title: "Total Reviews",
      status: "neutral",
      comp: "same",
      details: "2 New This Week"
    },
    top: {
      details: "Version 3.1.40",
      status: "neutral",
      comp: "same",
      title: "Average Rating",
      value: "4.0"
    }
  },
  {
    bottom: {
      value: "469",
      title: "Total Reviews",
      status: "neutral",
      comp: "same",
      details: "2 New This Week"
    },
    top: {
      details: "Version 3.1.40",
      status: "neutral",
      comp: "same",
      title: "Average Rating",
      value: "4.0"
    }
  }
];
const mix = (o, t) => o.map((m, i) => (c = {...m}, c.title = t[i], c));
const mixed = mix(objects, titles);
console.log(mixed);