[英]Getting image URL from bucket Amazon Web Service iOS after uploading
I have an image that I'm uploading to my bucket in AWS this way: 我有一个图像,我以这种方式在AWS上传到我的存储桶:
BFTask *task = [BFTask taskWithResult:nil];
[[task continueWithBlock:^id(BFTask *task) {
self.URL = [NSURL fileURLWithPath:[NSTemporaryDirectory() stringByAppendingPathComponent:@"test"]];
NSData *data = UIImagePNGRepresentation(image);
//NSMutableString *dataString = [NSMutableString new];
[data writeToURL:self.URL atomically:YES];
return nil;
}]continueWithExecutor:[BFExecutor mainThreadExecutor] withBlock:^id(BFTask *task) {
self.uploadRequest1 = [AWSS3TransferManagerUploadRequest new];
self.uploadRequest1.bucket = S3BucketName;
self.uploadRequest1.key = S3KeyUploadName1;
self.uploadRequest1.body = self.URL;
return nil;
}];
AWSS3TransferManager *transferManager = [AWSS3TransferManager defaultS3TransferManager];
[[transferManager upload:self.uploadRequest1] continueWithExecutor:[BFExecutor mainThreadExecutor] withBlock:^id(BFTask *task) {
if (task.error != nil) {
if( task.error.code != AWSS3TransferManagerErrorCancelled
&&
task.error.code != AWSS3TransferManagerErrorPaused
)
{
NSLog(@"Upload Failed!");
}
} else {
self.uploadRequest1 = nil;
NSLog(@"Uploaded!");
}
return nil;
}];
The code for uploading the image works just fine. 上传图像的代码工作得很好。 When I open my bucket I see the image there. 当我打开我的水桶时,我看到那里的图像。
Now what I want to do is to get the URL of that image, is there a way to get the URL without getting the image again? 现在我想要做的是获取该图像的URL,有没有办法获取URL而不再获取图像?
You don't get it, you create it like: 你没有得到它,你创建它像:
https://s3.amazonaws.com/BUCKET_NAME/FILE_NAME.jpg https://s3.amazonaws.com/BUCKET_NAME/FILE_NAME.jpg
AWSS3TransferManagerUploadRequest *uploadRequest = [AWSS3TransferManagerUploadRequest new];
uploadRequest.body = [NSURL fileURLWithPath:filePath];
uploadRequest.key = fileName;
uploadReuest.bucket = S3BucketName;
[uploadRequest setACL:AWSS3ObjectCannedACLPublicRead]; [uploadRequest setACL:AWSS3ObjectCannedACLPublicRead];
Above Answer is right but you must have to set "ACL" to uploadRequest. 上面的答案是正确的,但您必须将“ACL”设置为uploadRequest。
In your Question,you are forget to set "ACL". 在您的问题中,您忘记设置“ACL”。
Thanks 谢谢
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