AWSS3TransferManagerUploadRequest得到

Question

我使用V2 AWS iOS的SDK，以我的圖片上傳到服務器 我的代碼兩岸向前

NSString * uniqueIdentifier = [[NSUUID UUID]UUIDString]; 

AWSS3TransferManagerUploadRequest *uploadRequest = [AWSS3TransferManagerUploadRequest new]; 
uploadRequest.bucket = @"my bucket"; 
uploadRequest.key = [NSString stringWithFormat:@"%@/images/%@.jpeg",objectId,uniqueIdentifier]; 
uploadRequest.ACL = AWSS3ObjectCannedACLPublicRead; 

NSData * data = UIImageJPEGRepresentation(newImage, 0.8); 

NSString *tmpDirectory = NSTemporaryDirectory(); 

NSString *tmpFile = [tmpDirectory stringByAppendingPathComponent:[NSString stringWithFormat:@"%@.jpeg",uniqueIdentifier]]; 
NSLog(@"Temp File:%@", tmpFile); 

[data writeToFile:tmpFile atomically:YES]; 

uploadRequest.body = [NSURL fileURLWithPath:tmpFile]; 

AWSS3TransferManager *transferManager = [AWSS3TransferManager defaultS3TransferManager]; 

[[transferManager upload:uploadRequest] continueWithExecutor:[BFExecutor defaultExecutor] 
                withBlock:^id(BFTask *task) { 
                 if (task.error != nil) { 
                  NSLog(@"%s %@","Error uploading :", uploadRequest.key); 
                 }else { 

                  AWSS3TransferManagerUploadOutput * output = task.result; 


                  NSLog(@"Upload completed %@",[output description]); 


                 } 
                 return nil; 
                }]; 

後，我將文件上傳到S3服務這裏是響應URL響應URL ？ 我如何知道該圖像的網址？ 我想通過數據庫把圖像放在一個對象中。

感謝

Answer 1

公共可讀亞馬遜S3對象按照此URL模式：

https://S3Endpoint/BucketName/ObjectName 

找出對象的URL最簡單的方法是去AWS Management Console和look在「鏈接：「爲對象。