通过从Firebase存储访问来更新用户的photoURL

Question

我正在使用此javascript代码将图像上传到Firebase存储，然后我希望将此图像URL更新为用户的photoURL（firebase身份验证）

const fileButton = document.querySelector('#fileButton');
var imageURL;

fileButton.addEventListener('change', function(e) {
    //get file 
    var file = e.target.files[0];
    //create a storage reference
    var storageRef = storage.ref('profilePictures/' + userUID + '.jpg');
    //upload file
    var task = storageRef.put(file);
    //update progress bar
    task.on('state_changed',
        function progress(snapshot){

        },
        function error(err) {

        },
        function complete(){
            storageRef.getDownloadURL().then(function(url) {
                console.log(url);
                imageURL = url;
            })
            .catch(function(error) {
                // Handle any errors
                console.log(error);
            });
            // window.location.replace('/profile');
            var user = auth.currentUser;
            user.updateProfile({
                photoURL: imageURL
            })
            .then(function() {
                // Update successful.
                console.log(user);
            })
            .catch(function(error) {
                // An error happened.
                console.log(error);
            });
        }
    );
});

存储上的图像上传成功已满但用户数据中的photoURL为NOT SUCCESSFULL

uid: "t2TbJS6ut7NZH4UU8HeAVarGPOw2"
displayName: "Sachin Kumar"
photoURL: null
email: "[email protected]"
emailVerified: false

我做错了什么

Answer 1

[您需要等待storageRef.getDownloadURL()完成，然后才能将其传递给user.updateProfile：

function complete() {
  storageRef.getDownloadURL().then(function(url) {
      console.log(url);
      imageURL = url;

      // Now you have valid `imageURL` from async call
      var user = auth.currentUser;
      user.updateProfile({
          photoURL: imageURL
        })
        .then(function() {
          console.log(user);
        })
        .catch(function(error) {
          console.log(error);
        });

    })
    .catch(function(error) {
      console.log(error);
    });    
}