Question

var databaseRef = firebase.database().ref("Test_Applicants/");

        databaseRef.push({
            firstname: first_name,
            middlename: middle_name,
            lastname: last_name,
            university: university_full_name,
            email: email_address,
            phone: phone_number,
            date: getDate(),
            docs: docURL
        }, function(error){
            //Callback failed
            console.error(error);
        });

这是正确的做法吗？如何通过回调或通过任何其他方式从Firebase接收变量或参数以确认数据是否已成功写入？我希望在请求成功时上传文件，或者如果写入失败则向用户返回错误消息。

Answer 1

This should work

var databaseRef = firebase.database().ref("Test_Applicants/");
        databaseRef.push({
            firstname: first_name,
            middlename: middle_name,
            lastname: last_name,
            university: university_full_name,
            email: email_address,
            phone: phone_number,
            date: getDate(),
            docs: docURL
        }, function(error){
            if (error) {
             console.error(error)
             return
            }
            console.log('Push successful')
            //add upload function here
        });

Answer 2

You can listen child events in this case using following

databaseRef.on('child_added', function(data) {
  console.log(data);
});

More info: https://firebase.google.com/docs/database/web/lists-of-data#listen_for_child_events

Answer 3

databaseRef.push().set({name:"john doe",telephone:"25400000000"}).catch((error)=>console.log(error))

如何在Firebase Web中确认推送请求的成功？

3 个答案: