Node.js 保存JSON数据时出错-NodeJS Express_Node.js_Arrays_Json_Express

Node.js 保存JSON数据时出错-NodeJS Express

node.js arrays json express

Node.js 保存JSON数据时出错-NodeJS Express,node.js,arrays,json,express,Node.js,Arrays,Json,Express,我在保存和访问JSON数据时遇到问题，JSON文件中的所有对象都被转换为字符串，偶数这是我从HTML表单 control.post('/like/:id',getuserdata, async (req, res, next) => { try{ let likedata = './bin/like/likedata.json' let title = res.accountuser.title; let pass = res.ac

我在保存和访问JSON数据时遇到问题，JSON文件中的所有对象都被转换为字符串，偶数

这是我从

HTML表单

control.post('/like/:id',getuserdata, async (req, res, next) => {
    try{
        let likedata = './bin/like/likedata.json'
        let title = res.accountuser.title;
        let pass = res.accountuser.password;
        let tags = req.body.likes;
        let actions = req.body.action;
        let like =  {
            username: title,
            password: pass,
            tag: tags,
            times: actions,
        }
        
        let jsonlike = JSON.stringify(like)
        fs.appendFileSync(likedata, jsonlike)
        next();
        res.redirect('/dashboard')
        } catch (err){
            console.log(err)
        }
});

这就是我的JSON文件是如何产生的当我添加两次或更多值时，JSON文件显示错误

预期的文件结尾。

{"username":"account1","password":"somepass1","tag":"hello","times":"5"}{"username":"account2","password":"somepass2","tag":"helllo","times":"444"}

我想单独使用这个JSON数据，这里：

const ig = require('./like');
var cron = require('node-cron');

const iglike = async() => {

   await ig.initialize();

   await ig.login(username, password); //here

   await ig.liketagsprocess(tag, times); //here

};
cron.schedule('* */4 */12 * * *', function() {
   iglike();
});

module.exports = iglike;

如何执行此操作？

您在JSON文件中写入的内容根本不是有效的JSON

// Contents of your file

{"username":"account1","password":"somepass1","tag":"hello","times":"5"}{"username":"account2","password":"somepass2","tag":"helllo","times":"444"}

请尝试此结构，它是一个JSON对象数组：

[{"username":"account1","password":"somepass1","tag":"hello","times":"5"}, {"username":"account2","password":"somepass2","tag":"helllo","times":"444"}]

如果您生成此数据结构并以与现在相同的方式将其写入一个

JSON

文件中，那么您就不会出现错误

另外，您可能应该尽量减少读/写操作

appendFileSync（）

是一个阻塞操作，您可以使用

require

加载一个有效的

JSON

文件，并对其进行处理，定期尝试通过覆盖对象JSON数组的当前状态来更新文件中的内容

将下面的代码复制粘贴到名为

app.js

的文件中，并通过键入

node app.js

运行，检查

console.log（）

输出和生成的

json

文件。结果不言自明

const like = require('./like.json');
const fs = require('fs');

const init = () => {
    console.log(like);
  
    let likedata = [
        {
            "hello": "world",
            "one": 1
        }, 
        {
            "hi": "there",
            "two": 2
        }
    ];

    let likedatastr = [
        {
            "hey": "whatsup",
            "three_str": "3"
        }
    ];

    let likedatamod = [
        {
            "hey": likedatastr[0].hey,
            "three_int": parseInt(likedatastr[0].three_str)
        }
    ]
    fs.writeFileSync('./likedata1.json', likedata);
    fs.writeFileSync('./likedata2.json', JSON.stringify(likedata));
    fs.writeFileSync('./likedatastr.json', JSON.stringify(likedatastr));
    fs.writeFileSync('./likedatamod.json', JSON.stringify(likedatamod));

};

init();

而

like.json

如下所示：

[{"username":"account1","password":"somepass1","tag":"hello","times":"5"}, {"username":"account2","password":"somepass2","tag":"helllo","times":"444"]

同样，确保这些数字实际上是从请求主体中提取的数据中的数字。理想情况下，您应该在请求正文数据上添加一个

parseInt（）

（假设您需要整数），并处理

NaN

，以防万一

祝您好运。

您正在写入JSON文件的内容根本不是有效的JSON

// Contents of your file

{"username":"account1","password":"somepass1","tag":"hello","times":"5"}{"username":"account2","password":"somepass2","tag":"helllo","times":"444"}