javascript - How to return other URL requests in node?

Question

I want to encapsulate the token method. Why can’t this call be returned correctly?

app.js

app.use('/getUserInfo',function(req,res,next){
    console.log("进入getUserInfo")
    utils.getToken(appid,appsecret)
})

utils.js

utils.getToken=function(appid,appsecret){
    console.log("我是utils里面的")
    let tokenUrl = "https://api.weixin.qq.com/cgi-bin/token?grant_type=client_credential&appid=" + appid + "&secret=" + appsecret;
    let jsonResult=''
    return function(){        
        request(tokenUrl,(err,response,body)=>{
            let jsonResult = JSON.parse(body);
            return jsonResult
        })
    }
}

I want to call getUserInfo to obtain user information, how to change it?

Answer 1

utils.getToken(appid,appsecret)() call?

Answer 2

Node is asynchronous. If you can return directly, what else should you do with async/await?

You can also use callback directly:

app.use('/getUserInfo',function(req,res,next){
    console.log("进入getUserInfo")
    utils.getToken(appid,appsecret,function(res){
      console.log(res);
    })
})

........................

utils.getToken=function(appid,appsecret,callback){
    console.log("我是utils里面的")
    let tokenUrl = "https://api.weixin.qq.com/cgi-bin/token?grant_type=client_credential&appid=" + appid + "&secret=" + appsecret;  
    request(tokenUrl,(err,response,body)=>{
        let jsonResult = JSON.parse(body);
        callback(sonResult);
    })
}