如何使用node.js将多个json对象合并到一个json文件中

Question

我有多个json文件，它们只是想从中合并特定对象，然后将其保存到新的json文件中：例如：

one.json
two.json
...

每个json文件都是这样的：

一个.json

{
  "passed": 1,
  "fixtures": [
    {
      "name": "Getting Started one",
      "path": "testOne.ts"
    }
  ]
}

two.json

{
  "passed": 1,
  "fixtures": [
    {
      "name": "Getting Started two",
      "path": "testTwo.ts"
    }
  ]
}

所以我的最终json文件应该是这样的：

final.json

{
    "passed": 2,
    "fixtures": [{
            "name": "Getting Started one",
            "path": "testOne.ts"
        },
        {
            "name": "Getting Started two",
            "path": "testTwo.ts"
        }
    ]
}

您知道这样合并的简单方法是什么吗？

节点：json文件不仅是两个，而且很多，因此最好找到每个扩展名为json的文件。最好在shell脚本或node.js中执行此操作。

谢谢。

Answer 1

Can do something like this using Promise:

const fs = require('fs');
const data = {
    "passed": 0,
    "fixtures": []
};
const dir = `${__dirname}/data/`;
fs.readdir(dir, (err, files) => {
    return new Promise((resolve, reject) => {
        if (err) reject(err);
        files.forEach(file => {
           console.log(file);
           let content = require(`${dir}${file}`);
           data['passed'] += content.passed;
           data['fixtures'] = data['fixtures'].concat(content['fixtures']);
        });
        resolve(data);
    }).then(data => {
        fs.writeFileSync('./final.json',JSON.stringify(data));
    });
})

Using Async/Await it can done like this:

const fs = require('fs');
const path = require('path');
const dir = path.join(__dirname, 'data');

let finalContent = { "fixtures": [], "passed": 0 };
const read_directory = async dir =>
    fs.readdirSync(dir).reduce((finalContent, file) => {
        filePath = path.join(dir, file);
        console.log(filePath);
        let content = require(filePath);
        finalContent.passed += content.passed;
        finalContent.fixtures = finalContent.fixtures.concat(content.fixtures);
        return finalContent;
    }, { "passed": 0, "fixtures": [] });

read_directory(dir).then(data => {
    fs.writeFileSync('./final.json', JSON.stringify(data));
});