【发布时间】：2018-03-29 06:07:24
【问题描述】：
我想合并两个包含电影信息的 JSON 文件。这些文件有一些共同点。我想编写第三个文件，其中包含所有电影而不重复它们。
这是我目前所拥有的：
const fs = require('fs');
const path = require('path');
const readMovies1 = () => {
return new Promise((resolve, reject) => {
fs.readFile(path.join(__dirname,'../models/movies.json'), 'utf8', (err, data1) => {
if(err) reject(err);
let data = JSON.parse(data1);
resolve(data);
});
});
};
const readMovies2 = (data1) => {
return new Promise((resolve, reject) =>{
fs.readFile(path.join(__dirname,'../models/movies2.json'), 'utf8', (err, data) => {
if(err) reject(err);
data = JSON.parse(data);
resolve([data1,data]);
});
});
};
const merging = (data1, data2) => {
var args = arguments;
var hash = {};
var arr = [];
for (var i = 0; i < args.length; i++) {
for (var j = 0; j < args[i].length; j++) {
if (hash[args[i][j]] !== true) {
arr[arr.length] = args[i][j];
hash[args[i][j]] = true;
}
}
}
return arr;
};
readMovies1()
.then(readMovies2)
.catch((err) => console.error(err))
.then((data) => console.log(merging(data[0],data[1])))
.catch((err) => console.error(err));
但是控制台输出给了我这个：
[ undefined,
'/',
'U',
's',
'e',
'r',
'g',
'o',
'n',
'z',
'P',
'j',
'c',
't',
'J',
'S',
'a',
'd',
'm',
'-',
'v',
'i',
'l',
'b',
'.' ]
我在这里的堆栈溢出问题中找到了合并函数，但它正在合并一个数字数组，我正在合并对象数组。我不知道这是否与我的问题有关。
这是我的一个源文件（另一个遵循相同的模式，但有一些其他电影）：
[{
"title": "Spider-Man: Homecoming",
"usersScore": "92%",
"criticsScore": "89%"
}, {
"title": "Girls Trip",
"usersScore": "89%",
"criticsScore": "83%"
}, {
"title": "Captain Underpants: The First Epic Movie (Captain Underpants)",
"usersScore": "87%",
"criticsScore": "62%"
}, {
"title": "Guardians of the Galaxy Vol. 2",
"usersScore": "82%",
"criticsScore": "88%"
}, {
"title": "Wonder Woman",
"usersScore": "92%",
"criticsScore": "89%"
}, {
"title": "First They Killed My Father",
"usersScore": "88%",
"criticsScore": "83%"
}, {
"title": "Baby Driver",
"usersScore": "93%",
"criticsScore": "87%"
}, {
"title": "Demon",
"usersScore": "91%",
"criticsScore": "56%"
}, {
"title": "The Music of Strangers: Yo-Yo Ma and the Silk Road Ensemble",
"usersScore": "84%",
"criticsScore": "85%"
}, {
"title": "Colossal",
"usersScore": "80%",
"criticsScore": "59%"
}, {
"title": "Certain Women",
"usersScore": "92%",
"criticsScore": "51%"
}, {
"title": "Godzilla Resurgence (Shin Godzilla)",
"usersScore": "84%",
"criticsScore": "73%"
}, {
"title": "My Cousin Rachel",
"usersScore": "76%",
"criticsScore": "47%"
}, {
"title": "The Meyerowitz Stories (New and Selected)",
"usersScore": "93%",
"criticsScore": "84%"
}, {
"title": "Raw",
"usersScore": "90%",
"criticsScore": "77%"
}, {
"title": "The Wedding Plan",
"usersScore": "86%",
"criticsScore": "65%"
}, {
"title": "Maudie",
"usersScore": "88%",
"criticsScore": "92%"
}, {
"title": "Heal the Living (Réparer les vivants)",
"usersScore": "90%",
"criticsScore": "70%"
}, {
"title": "Lady Macbeth",
"usersScore": "89%",
"criticsScore": "72%"
}, {
"title": "The Exception (The Kaiser's Last Kiss)",
"usersScore": "76%",
"criticsScore": "67%"
}, {
"title": "Citizen Jane: Battle for the City",
"usersScore": "94%",
"criticsScore": "61%"
}, {
"title": "The Beguiled",
"usersScore": "78%",
"criticsScore": "50%"
}, {
"title": "The Big Sick",
"usersScore": "98%",
"criticsScore": "89%"
}, {
"title": "The Little Hours",
"usersScore": "77%",
"criticsScore": "53%"
}, {
"title": "A Ghost Story",
"usersScore": "91%",
"criticsScore": "66%"
}, {
"title": "The Hero",
"usersScore": "77%",
"criticsScore": "64%"
}, {
"title": "Megan Leavey",
"usersScore": "84%",
"criticsScore": "83%"
}, {
"title": "Band Aid",
"usersScore": "85%",
"criticsScore": "73%"
}, {
"title": "It Comes At Night",
"usersScore": "89%",
"criticsScore": "43%"
}, {
"title": "The Midwife (Sage femme)",
"usersScore": "86%",
"criticsScore": "82%"
}, {
"title": "Brawl in Cell Block 99",
"usersScore": "93%",
"criticsScore": "75%"
}, {
"title": "Gerald's Game",
"usersScore": "89%",
"criticsScore": "78%"
}]
【问题讨论】：
-
您的源数据是什么样的？
-
一种更简单的思考方式是连接两个数组，然后只过滤掉唯一的（或有时称为不同的）值。合并通常被理解为一个不同的问题，并且比仅仅选择独特的属性要复杂得多。搜索“合并”将产生与您的案例几乎不相关的结果。
-
@BCartolo OP 想要“合并”数组，而不是对象。
-
@AndrewLohr OP 想要“合并”数组，而不是对象。此外，
Object.assign并没有真正合并，它只会覆盖属性并且它的工作很浅，即仅在顶层。任何深度嵌套的值都不会正确合并。如果打算合并对象，则必须为深度嵌套的对象实现自定义处理或使用执行此操作的库，例如lodash.merge.
标签： javascript arrays json node.js merge