我正在创建一个应用程序,您可以在其中将照片和其他数据上传到Firebase。上载部分与一张图片完美配合。但是,我现在添加了多张图片(选择1至5张图片),并且希望我的图片上传功能可以上传5张图片而不是1张图片。 图像上传可使用提供的1张图像,因此如何重新排列代码以上传阵列中x量的照片?
这些图片将与以下数据一起添加到photos数组中(下面显示的输出是从获取的图片中获取的console.log);
Array [
Object {
"exists": true,
"file": "ph://8905951D-1D94-483A-8864-BBFDC4FAD202/L0/001",
"isDirectory": false,
"md5": "f9ebcab5aa0706847235887c1a7e4740",
"modificationTime": 1574493667.505371,
"size": 104533,
"uri": "ph://8905951D-1D94-483A-8864-BBFDC4FAD202/L0/001",
},
有了这个didFocus,我检查是否设置了fethedImages参数,并将photos数组设置为获取的图像(所以上面显示的所有数据)
const didFocusSubscription = props.navigation.addListener(
'didFocus', () => {
let fetchedImages = props.navigation.getParam('fetchedImages')
console.log(fetchedImages)
setPhotos(fetchedImages)
setImageValid(true)
calculateImageDimensions()
}
);
当我保存页面并开始调度数据时,运行以下命令,运行uploadImage函数并返回一个uploadurl,然后将其保存在调度函数中,稍后存储到Firebase数据库中;
uploadurl = await uploadImageAsync(photos)
因此,uploadImageAsync从转发的photos数组开始。如何确保阵列中的每个photo.uri都启动了以下功能?我可以为此使用.map,在什么情况下应该使用.map? 另外,我不太确定如何发送回URL数组以与其余信息一起保存。
async function uploadImageAsync(photos) {
console.log('uploadImage is gestart')
// Why are we using XMLHttpRequest? See:
// https://github.com/expo/expo/issues/2402#issuecomment-443726662
const blob = await new Promise((resolve, reject) => {
const xhr = new XMLHttpRequest();
xhr.onload = function () {
resolve(xhr.response);
};
xhr.onerror = function (e) {
console.log(e);
reject(new TypeError('Network request failed'));
};
xhr.responseType = 'blob';
xhr.open('GET', photos, true);
xhr.send(null);
});
const ref = firebase
.storage()
.ref()
.child(uuid.v4());
const snapshot = await ref.put(blob);
// We're done with the blob, close and release it
blob.close();
return await snapshot.ref.getDownloadURL();
}
=============由于上载进度进行了编辑===================
再一次,我又走了一点。但是,图像上传功能现在正在运行,由于存在多个图像,因此我想在继续之前等待所有图像的响应。
try {
uploadurl = await uploadImageAsync()
address = await getAddress(selectedLocation)
console.log(uploadurl)
if (!uploadurl.lenght) {
Alert.alert('Upload error', 'Something went wrong uploading the photo, plase try again', [
{ text: 'Okay' }
]);
return;
}
dispatch(
此刻,当我启动uploadImageAsync函数时。在console.log的帮助下,我看到它上传了图像,它们也在线显示。但是,在上传图片时,上传网址已经返回0,并显示警告并停止了该功能。
uploadImageAsync = async () => {
const provider = firebase.database().ref(`providers/${uid}`);
let imagesArray = [];
try {
await photos.map(img => {
let file = img.data;
const path = "Img_" + uuid.v4();
const ref = firebase
.storage()
.ref(`/${uid}/${path}`);
ref.putString(file).then(() => {
ref
.getDownloadURL()
.then(images => {
imagesArray.push({
uri: images
});
console.log("Out-imgArray", imagesArray);
})
return imagesArray <== this return imagesArray is fired to early and starts the rest of my upload function.
} catch (e) {
console.error(e);
}
};
答案 0 :(得分:0)
因此,Discord聊天为我指明了promise.all函数的所有功能。我试过了,但是打开了另一个堆栈溢出主题以使它起作用。
await response of image upload before continue function
我的图像上传功能的解决方案在上面的主题中;
uploadImages = () => {
const provider = firebase.database().ref(`providers/${uid}`);
// CHANGED: removed 'let imagesArray = [];', no longer needed
return Promise.all(photos) // CHANGED: return the promise chain
.then(photoarray => {
console.log('all responses are resolved successfully');
// take each photo, upload it and then return it's download URL
return Promise.all(photoarray.map((photo) => { // CHANGED: used Promise.all(someArray.map(...)) idiom
let file = photo.data;
const path = "Img_" + uuid.v4();
const storageRef = firebase // CHANGED: renamed 'ref' to 'storageRef'
.storage()
.ref(`/${uid}/${path}`);
let metadata = {
contentType: 'image/jpeg',
};
// upload current photo and get it's download URL
return storageRef.putString(file, 'base64', metadata) // CHANGED: return the promise chain
.then(() => {
console.log(`${path} was uploaded successfully.`);
return storageRef.getDownloadURL() // CHANGED: return the promise chain
.then(fileUrl => ({uri: fileUrl}));
});
}));
})
.then((imagesArray) => { // These lines can
console.log("Out-imgArray: ", imagesArray) // safely be removed.
return imagesArray; // They are just
}) // for logging.
.catch((err) => {
console.error(err);
});
};