The $path always returns false.
My code
$path = Storage::disk('s3')->put($filePath, file_get_contents($request->file));
Please let me know what could have gone wrong.
Thanks
I realized that first I've to upload the files to the server and then to the s3.
Storage::disk('public')->put($filePath, file_get_contents($request->file));
Related
I am new to S3 and need to use it for image storage. I found a half dozen versions of an s2wrapper for cf but it appears that the only one set of for v4 is one modified by Leigh
https://gist.github.com/Leigh-/26993ed79c956c9309a9dfe40f1fce29
Dropped in the com directory and created a "test" page that contains the following code:
s3 = createObject('component','com.S3Wrapper').init(application.s3.AccessKeyId,application.s3.SecretAccessKey);
but got the following error :
So I changed the line 37 from
variables.Sv4Util = createObject('component', 'Sv4').init(arguments.S3AccessKey, arguments.S3SecretAccessKey);
to
variables.Sv4Util = createObject('component', 'Sv4Util').init(arguments.S3AccessKey, arguments.S3SecretAccessKey);
Now I am getting:
I feel like going through Leigh code and start changing things is a bad idea since I have lurked here for year an know Leigh's code is solid.
Does any know if there are any examples on how to use this anywhere? If not what I am doing wrong. If it makes a difference I am using Lucee 5 and not Adobe's CF engine.
UPDATE :
I followed Leigh's directions and the error is now gone. I am addedsome more code to my test page which now looks like this :
<cfscript>
s3 = createObject('component','com.S3v4').init(application.s3.AccessKeyId,application.s3.SecretAccessKey);
bucket = "imgbkt.domain.com";
obj = "fake.ping";
region = "s3-us-west-1"
test = s3.getObject(bucket,obj,region);
writeDump(test);
test2 = s3.getObjectLink(bucket,obj,region);
writeDump(test2);
writeDump(s3);
</cfscript>
Regardless of what I put in for bucket, obj or region I get :
JIC I did go to AWS and get new keys:
Leigh if you are still around or anyone how has used one of the s3Wrappers any suggestions or guidance?
UPDATE #2:
Even after Alex's help I am not able to get this to work. The Link I receive from getObjectLink is not valid and getObject never does download an object. I thought I would try the putObject method
test3 = s3.putObject(bucketName=bucket,regionName=region,keyName="favicon.ico");
writeDump(test3);
to see if there is any additional information, I received this :
I did find this article https://shlomoswidler.com/2009/08/amazon-s3-gotcha-using-virtual-host.html but it is pretty old and since S3 specifically suggests using dots in bucketnames I don't that it is relevant any longer. There is obviously something I am doing wrong but I have spent hours trying to resolve this and I can't seem to figure out what it might be.
I will give you a rundown of what the code does:
getObjectLink returns a HTTP URL for the file fake.ping that is found looking in the bucket imgbkt.domain.com of region s3-us-west-1. This link is temporary and expires after 60 seconds by default.
getObject invokes getObjectLink and immediately requests the URL using HTTP GET. The response is then saved to the directory of the S3v4.cfc with the filename fake.ping by default. Finally the function returns the full path of the downloaded file: E:\wwwDevRoot\taa\fake.ping
To save the file in a different location, you would invoke:
downloadPath = 'E:\';
test = s3.getObject(bucket,obj,region,downloadPath);
writeDump(test);
The HTTP request is synchronous, meaning the file will be downloaded completely when the functions returns the filepath.
If you want to access the actual content of the file, you can do this:
test = s3.getObject(bucket,obj,region);
contentAsString = fileRead(test); // returns the file content as string
// or
contentAsBinary = fileReadBinary(test); // returns the content as binary (byte array)
writeDump(contentAsString);
writeDump(contentAsBinary);
(You might want to stream the content if the file is large since fileRead/fileReadBinary reads the whole file into buffer. Use fileOpen to stream the content.
Does that help you?
Say for example I have the following bucket set up:
bucketone
…/folderone
…/text1.txt
…/text2.txt
…/foldertwo
…/file1.json
…/folderthree
…/folderthreesub
…/file2.json
…/file3.json
But it only goes down one level.
What’s the proper way of retrieving information under a bucket?
Will be sure to accept/upvote answer.
Whats wrong with just doing this from the CLI?
aws s3 cp s3://bucketing . --recursive
Contrary to the way you'd think it will work, rsplit() actually returns the splits from left-right, even though it applies it right-to-left.
Therefore, you actually want to obtain the last element of the split:
filename = obj['Key'].rsplit('/', 1)[-1]
See: Python rsplit() documentation
Also, be careful of 'pretend directories' that might be created via the console. They are actually zero-length files the make the folder appear in the UI. Therefore, skip files with no name after the final slash.
Make those fixes and it works as desired:
import boto3
import os
s3client = boto3.client('s3')
for obj in s3client.list_objects_v2(Bucket='my-bucket')['Contents']:
filename = obj['Key'].rsplit('/', 1)[-1]
localfiledir = os.path.join('/tmp', filename)
if filename != '':
s3client.download_file('my-bucket', obj['Key'], localfiledir)
I'm trying to build a proper file upload handling via Pear quickform 2.
My serverside approach would be:
$submitValues = $editForm->getValue();
$filename = submitValues['uploaded_image']['name'];
$move_file = move_uploaded_file(.....)
Is there still a function like in quickform1: isUploadedFile() to make sure its an uploaded file?
unfortunately searching the documentary didn't give me the hints I needed.
Any advice regarding this issue is very much appreciated.
You can use the php function directly :
is_uploaded_file($submitValues['uploaded_image']['tmp_name']);
See http://php.net/is_uploaded_file
Can anyone help me to implement how to move uploaded file from one server to another
I am not talking about the move_uploaded_file() function.
for example,
If the image is uploaded from http://example.com
How can I move it to http://image.example.com
It is possible right? Not by sending another post or put request?
Take the Uploaded file, move it to a temporary location and push it then to any FTP-Acount you like.
$tempName = tempnam(sys_get_temp_dir(), 'upload');
move_uploaded_file($_FILES["file"]["tmpname"], $tempName);
$handle = fopen("ftp://user:password#example.com/somefile.txt", "w");
fwrite($handle, file_get_contents($uploadedFile));
fclose($handle);
unlink($tempName);
Actually you don't even need the part with the move_uploaded_file. It is totally sufficent to take the uploaded file and write it's content to the file opened with fopen. For more informations on opening URLs with fopenhave a look at the php-documentation. For more information on uploading files have a look at the File-Upload-Section of the PHP-Manual
[Edit] Added file_get_contents to the code-example
[Edit] Shorter Example
$handle = fopen("ftp://user:password#example.com/somefile.txt", "w");
fwrite($handle, file_get_contents($_FILES["file"]["tmpname"]);
fclose($handle);
// As the uploaded file has not been moved from the temporary folder
// it will be deleted from the server the moment the script is finished.
// So no cleaning up is required here.
I have just moved a site from a dedicated server to a GoDaddy shared hosting account, and have just encountered loads of problems! One being with ASP Upload.
In Classic ASP, this is what I would normally do to upload files to my folder:
Set upload = Server.CreateObject("Persits.Upload")
uploadPath = Server.MapPath("../../files/photos/"&token_&"/")
upload.IgnoreNoPost = True
upload.Save(uploadPath)
Set upload = Nothing
But since moving to GoDaddy, I get this nasty message:
This feature has been disabled by the system administrator. Use SaveVirtual instead.
I went on to ASP Upload's website documentation and I could not find SaveVirtual, only SaveAsVirtual which is where I have become unstuck.
I tried using SaveAsVirtual but it threw an error stating that I was using a physical path and I should be using a virtual path! I really don't understand this and was hoping that somebody could put me straight. My website is now deemed broken and is offline, please help.
This is what I tried before the physical/virtual path error:
Set upload = Server.CreateObject("Persits.Upload")
uploadPath = Server.MapPath("../../files/photos/"&token_&"/")
upload.IgnoreNoPost = True
upload.SaveAsVirtual(uploadPath)
Set upload = Nothing
According to the docs, the method is named SaveVirtual. It does the Server.MapPath conversion for you.
So, try:
Set upload = Server.CreateObject("Persits.Upload")
uploadPath = "../../files/photos/"&token_&"/"
upload.IgnoreNoPost = True
upload.SaveVirtual(uploadPath)
Set upload = Nothing
You already set a folder path to upload not need use like this "server.mappath"
please use below of code.
Set Upload = Server.CreateObject("Persits.Upload" )
Upload.SetMaxSize 10000000 ' Maksimum dosya büyüklüğü Byte cinsinden
Upload.OverwriteFiles = True
Path = "../../duyurular/"
Count = Upload.SaveAsVirtual(Path)
for godaddy you have to go in and change the server folder permission to read/write. Otherwise you get that error message.
well I also get encounter with this problem, I just changed the folder privileges from Go-daddy Hosting server so Any one can read or Write file from specific location
Many Thanks
Ahsan Aziz Abbasi