Question

我想将图像上传到Firebase（正在运行），然后返回图像的下载URL并将其存储为字符串。

这是我的代码：

uploadImage = async (uri, imageName) => {
    const response = await fetch(uri);
    const blob = await response.blob();

    firebase.storage().ref().child(imageName).put(blob)
      .then(snap => {
        return snap.ref.getDownloadURL();
      })
      .then(downloadURL => {
        return downloadURL;
      })
      .catch(error => {
        console.log(`An error occurred while uploading the file.\n\n${error}`);
      });
  }

图像可以很好地上传到Firebase存储。目前，当我尝试将上传的图像的URL写入数据库时，它仅显示此内容： https://ibb.co/WHHHxBY

这是我创建用户记录的代码块：

  firebase
    .auth()
    .createUserWithEmailAndPassword(this.state.email, this.state.password)
    .then(userCredentials => {
      let imageUrl = '';

      let db = firebase.database().ref('users/' + userCredentials.user.uid);

      if (this.state.image) {
        imageUrl = this.uploadImage(this.state.image.uri, `images/user-${userCredentials.user.uid}`);
      }

      db.set({
        email: this.state.email,
        imageUrl: imageUrl,
        username: this.state.username
      });

      return userCredentials.user.updateProfile({
        displayName: this.state.username
      });
    })
    .catch(error => this.setState({ errorMessage: error.message }));

Answer 1

在... foreach ($content_types as $content_type) { dispatch(new CreateContent($content_type)); }函数中，您正在链接承诺，但没有返回链接。您应该按照以下方式对其进行修改：

uploadImage

但是，您可以按uploadImage = async (uri, imageName) => { const response = await fetch(uri); const blob = await response.blob(); return firebase.storage().ref().child(imageName).put(blob) // <-- Here return the chain .then(snap => { return snap.ref.getDownloadURL(); }) .then(downloadURL => { return downloadURL; }) .catch(error => { console.log(`An error occurred while uploading the file.\n\n${error}`); }); }“样式”转换此代码，如下所示：

async/await

然后，由于此uploadImage = async (uri, imageName) => { try { const response = await fetch(uri); const blob = await response.blob(); const snap = await firebase.storage().ref().child(imageName).put(blob); const downloadURL = await snap.ref.getDownloadURL(); return downloadURL; } catch (e) { console.error(e); throw e; } }函数是异步的，因此您应该调整调用它的方式。我建议如下修改代码的另一部分：

uploadImage

如何返回上传到Firebase存储的图像的下载URL

1 个答案: