I have an image that I'm uploading to my bucket in AWS this way:
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. When I open my bucket I see the image there.
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?
You don't get it, you create it like:
AWSS3TransferManagerUploadRequest *uploadRequest = [AWSS3TransferManagerUploadRequest new];
uploadRequest.body = [NSURL fileURLWithPath:filePath];
uploadRequest.key = fileName;
uploadReuest.bucket = S3BucketName;
[uploadRequest setACL:AWSS3ObjectCannedACLPublicRead];
Above Answer is right but you must have to set "ACL" to uploadRequest.
In your Question,you are forget to set "ACL".
Thanks
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