I want to add text in specific file content.
I used:
var file = await dataService.Files.Get(fileId).ExecuteAsync();
if (file == null)
{
throw new Exception(nameof(fileId));
}
using (var stream = new MemoryStream())
{
using (var sw = new StreamWriter(stream, Encoding.UTF8))
{
sw.Write(content); //there is a simple text like: bblablllbalblba
var request = dataService.Files.Update(file, fileId, stream, file.MimeType);
await request.UploadAsync();
}
}
but seems that content is not updated/written. I don't get any error.
Where I'm wrong ?
Related
I make a private class to get the bytes of the file and then return it into a zip file. Basically, I just want to pick specific files inside my folder and then zip it then download it. Here's my class:
private FileResult DownloadMultipleFiles(List<byte[]> byteArrayList)
{
var zipName = $"archive-EvidenceFiles-{DateTime.Now.ToString("yyyy_MM_dd-HH_mm_ss")}.zip";
using (MemoryStream ms = new MemoryStream())
{
using (var archive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
foreach (var file in byteArrayList)
{
string fPath = Encoding.ASCII.GetString(file);
var entry = archive.CreateEntry(fPath, CompressionLevel.Fastest);
using (var zipStream = entry.Open())
{
zipStream.Write(file, 0, file.Length);
}
}
}
return File(ms.ToArray(), "application/zip", zipName);
}
}
And then, here's my controller:
[HttpGet("GetBundleFiles/{rhaId}")]
public async Task<IActionResult> GetBundleFiles(string rhaId)
{
List<byte[]> filesPath = new List<byte[]>();
var results = await _rhaFileEvidence.GetByRhaID(rhaId);
var files = results.ToList();
if (files.Count == 0)
return Ok(new { status = "null", message = "Empty data" });
files.ForEach(file =>
{
var fPath = file.FilePath;
byte[] bytes = Encoding.ASCII.GetBytes(fPath);
filesPath.Add(bytes);
});
return DownloadMultipleFiles(filesPath);
}
The controller works well, I can download the zip but when I open it, I can't get the files instead I get the root directory of the project I saved, like D:. I think I make mistake when making the memory stream or something, is there any suggestion how can I fix this? (Paste some solution code in the answer please)
you are getting the directory root because when you use archive.CreateEntry you are passing the file full path in parameter, you should be using only the file name
var entry = archive.CreateEntry(System.IO.Path.GetFileName(fPath), CompressionLevel.Fastest);
a second issue is that you actually saving the file path to your files not the content of the original file. you can update your DownloadMultipleFiles like this
private FileResult DownloadMultipleFiles(List<byte[]> byteArrayList)
{
var zipName = $"archive-EvidenceFiles-{DateTime.Now.ToString("yyyy_MM_dd-HH_mm_ss")}.zip";
using (MemoryStream ms = new MemoryStream())
{
using (var archive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
foreach (var file in byteArrayList)
{
string fPath = Encoding.ASCII.GetString(file);
var entry = archive.CreateEntry(System.IO.Path.GetFileName(fPath), CompressionLevel.Fastest);
using (var zipStream = entry.Open())
{
var bytes = System.IO.File.ReadAllBytes(fPath);
zipStream.Write(bytes, 0, bytes.Length);
}
}
}
return File(ms.ToArray(), "application/zip", zipName);
}
}
I am creating a zip file that, appears, valid but is always invalid after I have put it to a Amazon S3 bucket. I am using System.IO.Compression for the task and AmazonS3Client for uploading:
private byte[] GenerateZipFile(string tenant)
{
byte[] zipData;
var results = QueryAggregateTable(tenant);
using (var memoryStream = new MemoryStream())
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true))
{
foreach (var item in results)
{
var archiveEntry = archive.CreateEntry($"{item.RowKey:D3}.json", CompressionLevel.Fastest);
using (var entryStream = archiveEntry.Open())
{
var entryBytes = Encoding.UTF8.GetBytes(item.Data);
entryStream.Write(entryBytes, 0, item.Data.Length);
}
}
zipData = memoryStream.ToArray();
}
return zipData;
}
and
private async Task UploadToAmazon(byte[] zipData, string tenant)
{
var bucketName = _config["egestionBucketName"];
var configCreds = _config["egestionAwsCredentials"].Split(":");
var awsCreds = new BasicAWSCredentials(configCreds[0], configCreds[1]);
var awsRegion = Amazon.RegionEndpoint.GetBySystemName(_config["egestionRegionEndpointSystemName"]);
var s3Client = new AmazonS3Client(awsCreds, awsRegion);
using (var stream = new MemoryStream(zipData))
{
var putRequest = new PutObjectRequest
{
BucketName = bucketName,
Key = $"{tenant}-{DateTime.UtcNow.ToString("s")}.zip",
InputStream = stream,
CannedACL = S3CannedACL.BucketOwnerFullControl
};
await s3Client.PutObjectAsync(putRequest);
}
}
The byte array looks good after returning from generation and the upload method does, in fact, load a file with the correct name to the bucket. When I attempt to download the file to check it I cannot open it with a message stating it is invalid.
I have had some problems with async/await and suspect it may be something related but there is no non async option for PutObject that I can find. Any help appreciated.
This is not an async-await issue.
The bytes from the memory stream are being collected before the archive has had a chance to write all the data to the stream. The uploaded archive is incomplete and therefore invalid when downloaded.
Move
zipData = memoryStream.ToArray();
to outside of the archive using block so that any buffered data is flushed to the backing stream when the archive is disposed.
//...
using (var memoryStream = new MemoryStream()) {
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
foreach (var item in results) {
var archiveEntry = archive.CreateEntry($"{item.RowKey:D3}.json", CompressionLevel.Fastest);
using (var entryStream = archiveEntry.Open()) {
var entryBytes = Encoding.UTF8.GetBytes(item.Data);
entryStream.Write(entryBytes, 0, entryBytes.Length);
}
}
}//Archive disposed and pushed any remaining buffered data to the stream.
zipData = memoryStream.ToArray();
}
//...
I have a project that requires pdf files to be zipped up from an URL link and then downloaded and clickable by the end users browser. So far, I was able to zip one pdf file which isn't nearly what I'm looking to for.
I'm not sure how to proceed from here. Below is the code. Any help would be very much appreciated.
ASP.NET Core
[HttpGet("zipFiles")]
public IActionResult ZipPDFFiles()
{
var fileNames = _repo.GetFileNames();
foreach (var filesName in fileNames)
{
var urlLink = "https://example.com/folder/" + $"{filesName.PdfFileName}";
var net = new System.Net.WebClient();
var data = net.DownloadData(urlLink);
var file = $"{filesName.PdfFileName}";
var contentType = "application/zip";
string zippedFolderName = "Archive.zip";
using (MemoryStream ms = new MemoryStream())
{
using (var archive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
var zipArchiveEntry = archive.CreateEntry($"{file}", System.IO.Compression.CompressionLevel.Fastest);
using (var zipStream = zipArchiveEntry.Open()) zipStream.Write(data, 0, data.Length);
}
return File(ms.ToArray(), contentType, $"{zippedFolderName}");
}
}
return NotFound();
}
Some improvements:
used HttpClient
files uploaded simultaneously
var fileUrls = new[]
{
new Uri("https://learn.microsoft.com/en-us/dotnet/standard/microservices-architecture/implement-resilient-applications/media/image3.5.png"),
new Uri("https://learn.microsoft.com/en-us/dotnet/standard/microservices-architecture/implement-resilient-applications/media/image4.png"),
new Uri("https://learn.microsoft.com/en-us/dotnet/standard/microservices-architecture/implement-resilient-applications/media/image6.png")
};
var downloadResults = fileUrls
.Select(uri => (uri: uri, response: HttpClientFactory.Create().SendAsync(new HttpRequestMessage(HttpMethod.Get, uri))))
.ToArray();
await Task.WhenAll(downloadResults.Select(v => v.response));
using (var ms = new MemoryStream())
{
using (var archive = new ZipArchive(ms, ZipArchiveMode.Create, true))
{
foreach (var download in downloadResults)
{
var entry = archive.CreateEntry(download.uri.Segments.Last(), CompressionLevel.Fastest);
using (var zipStream = entry.Open())
{
var data = await download.response.Result.Content.ReadAsByteArrayAsync();
zipStream.Write(data, 0, data.Length);
}
}
}
return File(ms.ToArray(), contentType, $"{zippedFolderName}");
}
I need to create a zip file in memory, then send the zip file to the client. However, there are cases where the created zip file will need to contain other zip files that were also generated in memory. For instance, the file structure might look like this:
SendToClient.zip
InnerZip1.zip
File1.xml
File2.xml
InnerZip2.zip
File3.xml
File4.xml
I've been attempting to use the System.IO.Compression.ZipArchive library. I cannot use the System.IO.Compression.ZipFile library because my project's version of .NET is not compatible with it.
Here's an example of what I've tried.
public Stream GetMemoryStream() {
var memoryStream = new MemoryStream();
string fileContents = "Lorem ipsum dolor sit amet";
string entryName = "Lorem.txt";
string innerZipName = "InnerZip.zip";
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
ZipArchiveEntry entry = archive.CreateEntry(Path.Combine(innerZipName, entryName), CompressionLevel.Optimal);
using (var writer = new StreamWriter(entry.Open())) {
writer.Write(fileContents);
}
}
return memoryStream
}
However, this just puts Lorem.txt in a folder called "Inner.zip" (instead of in an actual zip file).
I can create an empty inner zip file if I create an entry called "Inner.zip" without writing to it. I can't add anything to it, though, and writing to an entry called "Inner.zip\Lorem.txt" afterward just creates a folder again (alongside the identically named empty .zip file).
I've also tried creating a separate archive, serializing it with a memory stream, then writing that to the original archive as a .zip.
public Stream CreateOuterZip() {
var memoryStream = new MemoryStream();
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
ZipArchiveEntry entry = archive.CreateEntry("Outer.zip", CompressionLevel.NoCompression);
using (var writer = new BinaryWriter(entry.Open())) {
writer.Write(GetMemoryStream().ToArray());
}
}
return memoryStream;
}
This just creates an invalid .zip file that windows doesn't know how to open, though.
Thanks in advance!
So I created a FileStream instead of a MemoryStream so the code can be tested easier
public static Stream CreateOuterZip()
{
string fileContents = "Lorem ipsum dolor sit amet";
// Final zip file
var fs = new FileStream(
Path.Combine(AppDomain.CurrentDomain.BaseDirectory, "SendToClient.zip"), FileMode.OpenOrCreate);
// Create inner zip 1
var innerZip1 = new MemoryStream();
using (var archive = new ZipArchive(innerZip1, ZipArchiveMode.Create, true))
{
var file1 = archive.CreateEntry("File1.xml");
using (var writer = new BinaryWriter(file1.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
var file2 = archive.CreateEntry("File2.xml");
using (var writer = new BinaryWriter(file2.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
}
// Create inner zip 2
var innerZip2 = new MemoryStream();
using (var archive = new ZipArchive(innerZip2, ZipArchiveMode.Create, true))
{
var file3 = archive.CreateEntry("File3.xml");
using (var writer = new BinaryWriter(file3.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
var file4 = archive.CreateEntry("File4.xml");
using (var writer = new BinaryWriter(file4.Open()))
{
writer.Write(fileContents); // Change fileContents to real XML content
}
}
using (var archive = new ZipArchive(fs, ZipArchiveMode.Create, true))
{
// Create inner zip 1
var innerZipEntry = archive.CreateEntry("InnerZip1.zip");
innerZip1.Position = 0;
using (var s = innerZipEntry.Open())
{
innerZip1.WriteTo(s);
}
// Create inner zip 2
var innerZipEntry2 = archive.CreateEntry("InnerZip2.zip");
innerZip2.Position = 0;
using (var s = innerZipEntry2.Open())
{
innerZip2.WriteTo(s);
}
}
fs.Close();
return fs; // The file is written, can probably just close this
}
You can obviously modify this method to return a MemoryStream, or change the method to Void to just have the zip file written out to disk
You should create ZipArchive for internal zip file also. Write it to stream (memorystream). And after write this stream as general stream into main zip.
static Stream Inner() {
var memoryStream = new MemoryStream();
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
var demoFile = archive.CreateEntry("foo2.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream)) {
streamWriter.Write("Bar2!");
}
}
return memoryStream;
}
static void Main(string[] args) {
using (var memoryStream = new MemoryStream()) {
using (var archive = new ZipArchive(memoryStream, ZipArchiveMode.Create, true)) {
var demoFile = archive.CreateEntry("foo.txt");
using (var entryStream = demoFile.Open())
using (var streamWriter = new StreamWriter(entryStream)) {
streamWriter.Write("Bar!");
}
var zip = archive.CreateEntry("inner.zip");
using (var entryStream = zip.Open()) {
var inner = Inner();
inner.Seek(0, SeekOrigin.Begin);
inner.CopyTo(entryStream);
}
}
using (var fileStream = new FileStream(#"d:\test.zip", FileMode.Create)) {
memoryStream.Seek(0, SeekOrigin.Begin);
memoryStream.CopyTo(fileStream);
}
}
Thanks to this answer.
I am required to read the contents of an .xml file using the Stream (Here the xml file is existing with in the zip package). Here in the below code, I need to get the file path at runtime (here I have hardcoded the path for reference). Please let me know how to read the file path at run time.
I have tried to use string s =entry.FullName.ToString(); but get the error "Could not find the Path". I have also tried to hard code the path as shown below. however get the same FileNotFound error.
string metaDataContents;
using (var zipStream = new FileStream(#"C:\OB10LinuxShare\TEST1\Temp" + "\\"+zipFileName+".zip", FileMode.Open))
using (var archive = new ZipArchive(zipStream, ZipArchiveMode.Read))
{
foreach (var entry in archive.Entries)
{
if (entry.Name.EndsWith(".xml"))
{
FileInfo metadataFileInfo = new FileInfo(entry.Name);
string metadataFileName = metadataFileInfo.Name.Replace(metadataFileInfo.Extension, String.Empty);
if (String.Compare(zipFileName, metadataFileName, true) == 0)
{
using (var stream = entry.Open())
using (var reader = new StreamReader(stream))
{
metaDataContents = reader.ReadToEnd();
clientProcessLogWriter.WriteToLog(LogWriter.LogLevel.DEBUG, "metaDataContents : " + metaDataContents);
}
}
}
}
}
I have also tried to get the contents of the .xml file using the Stream object as shown below. But here I get the error "Stream was not readable".
Stream metaDataStream = null;
string metaDataContent = string.Empty;
using (Stream stream = entry.Open())
{
metaDataStream = stream;
}
using (var reader = new StreamReader(metaDataStream))
{
metaDataContent = reader.ReadToEnd();
}
Kindly suggest, how to read the contents of the xml with in a zip file using Stream and StreamReader by specifying the file path at run time
Your section code snippet is failing because when you reach the end of the first using statement:
using (Stream stream = entry.Open())
{
metaDataStream = stream;
}
... the stream will be disposed. That's the point of a using statment. You should be fine with this sort of code, but load the XML file while the stream is open:
XDocument doc;
using (Stream stream = entry.Open())
{
doc = XDocument.Load(stream);
}
That's to load it as XML... if you really just want the text, you could use:
string text;
using (Stream stream = entry.Open())
{
using (StreamReader reader = new StreamReader(stream))
{
text = reader.ReadToEnd();
}
}
Again, note how this is reading before it hits the end of either using statement.
Here is a sample of how to read a zip file using .net 4.5
private void readZipFile(String filePath)
{
String fileContents = "";
try
{
if (System.IO.File.Exists(filePath))
{
System.IO.Compression.ZipArchive apcZipFile = System.IO.Compression.ZipFile.Open(filePath, System.IO.Compression.ZipArchiveMode.Read);
foreach (System.IO.Compression.ZipArchiveEntry entry in apcZipFile.Entries)
{
if (entry.Name.ToUpper().EndsWith(".XML"))
{
System.IO.Compression.ZipArchiveEntry zipEntry = apcZipFile.GetEntry(entry.Name);
using (System.IO.StreamReader sr = new System.IO.StreamReader(zipEntry.Open()))
{
//read the contents into a string
fileContents = sr.ReadToEnd();
}
}
}
}
}
catch (Exception)
{
throw;
}
}