合并可选嵌套数组值以创建新对象

Question

像这样的对象中有多个字段。

{
  "image-processing": {
    "type": "lib",
    "data": {
      "files": [
        {
          "file": "libs/photography/processing/.babelrc",
          "hash": "62d50f586b2880f9d58ca4e9a84914c6a0d4936d"
        },
        {
          "file": "libs/photography/processing/src/lib/processing.ts",
          "hash": "a285d3554f264cc60f35bef6180298badb0478d1",
          "deps": [
            "npm:gm",
            "npm:mongodb"
          ]
        }
      ]
    }
  },
  "image-processing-2": {
    "type": "lib",
    "data": {
      "files": [
        {
          "file": "libs/photography/processing/.babelrc",
          "hash": "62d50f586b2880f9d58ca4e9a84914c6a0d4936d",
          "deps": [
            "npm:gm",
            "npm:faker"
          ]
        },
        {
          "file": "libs/photography/processing/src/lib/processing.ts",
          "hash": "a285d3554f264cc60f35bef6180298badb0478d1",
          "deps": [
            "npm:gm",
            "npm:mongodb"
          ]
        }
      ]
    }
  }
}

我只需要将data.files.deps的类型和内容作为键的值对象。如您所见，每个文件都有一个可选的deps数组。所有的deps值都应该合并。

因此，结果应该是：

{
    "image-processing": { type: "lib", packages: [ "npm:gm", "npm:mongodb" ] },
    "image-processing-2": { type: "lib", packages: [ "npm:faker", "npm:gm", "npm:mongodb" ] }
}

我试图遍历该对象，但这将覆盖该值，而不是向该值添加新的包名：

const result = {}
Object.entries(data).map(([key, value]) => {
    result[key].type = value.type
    result[key].packages = value?.data?.files.map(d => d.deps)
})

Answer 1

使用flatMap而不是map，这样就可以有一个扁平的依赖字符串数组，并与空数组交替使用，这样未定义的.deps属性就不会造成问题。

​

const data={"image-processing":{type:"lib",data:{files:[{file:"libs/photography/processing/.babelrc",hash:"62d50f586b2880f9d58ca4e9a84914c6a0d4936d"},{file:"libs/photography/processing/src/lib/processing.ts",hash:"a285d3554f264cc60f35bef6180298badb0478d1",deps:["npm:gm","npm:mongodb"]}]}},"image-processing-2":{type:"lib",data:{files:[{file:"libs/photography/processing/.babelrc",hash:"62d50f586b2880f9d58ca4e9a84914c6a0d4936d",deps:["npm:gm","npm:faker"]},{file:"libs/photography/processing/src/lib/processing.ts",hash:"a285d3554f264cc60f35bef6180298badb0478d1",deps:["npm:gm","npm:mongodb"]}]}}};

const output = Object.fromEntries(
  Object.entries(data).map(([key, value]) => [
    key,
    {
      type: 'lib',
      packages: [...new Set(value.data.files.flatMap(file => file.deps ?? []))]
    }
  ])
);
console.log(output);

​

Answer 2

根据您的新评论，只需将逻辑分开一点：

迭代在Object.entries上获得一个键/值对(值是对象)，然后遍历每个对象的文件。如果有任何filter出"npm“，那么map在该数组上并创建一个具有更新名称的新数组。

将该数组添加到packages数组中。

最后，压扁包数组，通过从它创建一套新的来简化它，然后将把它传开作为数组中的值，然后使用键添加到输出对象中。

​

const data={"image-processing":{type:"lib",data:{files:[{file:"libs/photography/processing/.babelrc",hash:"62d50f586b2880f9d58ca4e9a84914c6a0d4936d"},{file:"libs/photography/processing/src/lib/processing.ts",hash:"a285d3554f264cc60f35bef6180298badb0478d1",deps:["npm:gm","npm:mongodb"]}]}},"image-processing-2":{type:"lib",data:{files:[{file:"libs/photography/processing/.babelrc",hash:"62d50f586b2880f9d58ca4e9a84914c6a0d4936d",deps:["npm:gm","npm:faker"]},{file:"libs/photography/processing/src/lib/processing.ts",hash:"a285d3554f264cc60f35bef6180298badb0478d1",deps:["npm:gm","npm:mongodb"]}]}}};
// Output object
const out = {};

// Iterate over the object entries
for (const [key, obj] of Object.entries(data)) {
  
  // Destructure the type, and the files
  // array from each object
  const { type, data: { files } } = obj;
  
  const packages = [];

  // Iterate over each `files` array
  for (const file of files) {

    if (file.deps) {
      
      // `filter` out the "npm" names, and
      // then `map` over that array to update the names
      const list = file.deps
        .filter(dep => dep.startsWith('npm:'))
        .map(dep => dep.replace('npm:', ''));

      // Bang that completed array into `packages`
      packages.push(list);
    
    }

  }

  // Finally create a new key/value on the output
  // object by setting the value as an object
  // with `type`, and the flattened, deduped packages array
  out[key] = {
    type,
    packages: [...new Set(packages.flat())]
  };

}

console.log(out);

​