I have use ICSharpCode.SharpZipLib.Zip library for creating Zip file it's working fine, but when i have attach this Zip file to mail attachment, mail is not sent due to attached zip file,
here is code for attaching zip file to mail
System.Net.Mail.Attachment attachment = null;
try
{
MemoryStream memoryStream = new MemoryStream();
attachment = new System.Net.Mail.Attachment(memoryStream, "test.zip");
}
catch (Exception e)
{
return false;
}
please know me how i can send zip file through mail?.
Your code
MemoryStream memoryStream = new MemoryStream();
attachment = new System.Net.Mail.Attachment(memoryStream, "test.zip");
passes a Stream, but that Stream is empty (there's nothing in memoryStream).
If you want to use a MemoryStream, you must read the contents of the ZIP file into memory. You can also use a FileStream if the ZIP is already on disk.
If using a MemoryStream, be sure and set its position 0.
memoryStream.Position = 0;
Depending on how you are using SharpZipLib, you may have access to a ZipOutputStream. If you do, I think you could use that.
There could be lot of reasons for the issue,
Check file date & time
Size of the file ( Based on server settings)
Try different formats (rar,7z, etc)
Related
I'm having trouble using the System.IO.Compression library to create a zip file and then return the byte[] for that zip file. So I want the method to take an input string--the path for the file to be zipped--and return the corresponding byte array for the file once it's been zipped. Currently, I can save the zip file to some known path and then use System.IO.File.ReadAllBytes(path) to get the byte[].
But how can I do this if I don't know beforehand where that zip file should end up (if anywhere at all)? My attempt at the solution is below. It doesn't seem to be working because when I try to reconstruct the zip file from the byte array it says that the archive is in an unknown format or damaged. Note: I've been able to do this using the Ionic DotNetZip library; however, I'm trying to do the same with only the use of System.IO libraries.
Any ideas on how to go about this?
private static byte[] CreateZipAndFindBytes(string importfile)
{
byte[] retVal = null;
using (System.IO.MemoryStream memStream = new System.IO.MemoryStream())
{
//ZipArchive(Stream, ZipArchiveMode, Boolean) //true means leave stream open
using (System.IO.Compression.ZipArchive archive = new System.IO.Compression.ZipArchive(memStream, System.IO.Compression.ZipArchiveMode.Create, true))
{
archive.CreateEntryFromFile(importfile, "test.zip"); //adds the file to the zip
}
retVal = memStream.ToArray();
}
return retVal;
}
entryName, the second parameter of CreateEntryFromFile is supposed to be "The name of the entry to create in the zip archive." It's not the name of your zip file.
you're zipping the import file, but also renaming it with a .zip extension, and I'm assuming that the import file is not actually a zip.
do this instead
archive.CreateEntryFromFile(importfile, importfile); //adds the file to the zip
I am trying to first download a pdf as a string and then attach it to a MailMessage. Here is what I have done so far
string htmlAttachment = webClient.DownloadString((HttpUtility.HtmlDecode(dictIterator.Value)));
MemoryStream stream = new MemoryStream();
StreamWriter writer = new StreamWriter(stream);
writer.Write(htmlAttachment);
writer.Flush();
stream.Position = 0;
msg.Attachments.Add(new Attachment((Stream)stream, dictIterator.Key));
But whenever I open the pdf attachment, it says, Insufficient data for an image'. Is it something related to encoding and should I directly download it as a stream rather than first getting it asstring`??
Yes, you should download the file in Binary (Not as string.. No encoding involved.. Just pure binary).
PS - Try also saving the file locally to the disk, before attaching it. And then open it from the disk.
It will eliminate the unlikely possibility of MailMessage being the problem.
I'm needing to create a zipped document containing files that exist on the server. I am using the Ionic.Zip to do so, and to create a new Package (which is the zip file) I have to have either a path to a physical file or a stream. I am trying to not create an actual file that would be the zip file, instead just create a stream that would exist in memory or something. how can i do this?
Create the package using a MemoryStream then.
You can try the save method in the ZipFile Class. It can save to a stream
try this.
using (MemoryStream ms = new MemoryStream())
{
using (Ionic.Zip.ZipFile zipFile = new Ionic.Zip.ZipFile())
{
zipFile.AddFiles(filesToBeZipped, false, "NewFolder");//filesTobeZipped is a List<string>
zipFile.Save(ms);
}
}
You'll want to use the .AddEntry method on the ZipFile you've created specifying a name and the byte[] containing the actual file data.
ex.
ZipFile zipFile = new ZipFile();
zipFile.AddEntry(file.FileName, file.FileData);
where file.FileName will be the entry name (in the zip file) and file.FileData is the byte array.
I have been developing an web application with Asp.Net and I'm using SharpZipLib to work with odt files (from Open Office) and in the future docx files (for ms office). I need to open an odt file (like a zip file) change a xml file inside it, zip again and give it to the browser send to my client.
I can do this in file system but it will get a space in my disk temporarily and we don't want it. I would like to do this in memory (with a MemoryStream class), but I don't know how to unzip folders/files in a memory stream with SharpZipLib, change and use it to zip again. Is there any sample about how to do this?
Thank you
You can use something like
Stream inputStream = //... File.OpenRead(...);
//for read file
ZipInputStream zipInputStream = new ZipInputStream(inputStream));
//for output
MemoryStream memoryStream = new MemoryStream();
using ( ZipOutputStream zipStream = new ZipOutputStream(memoryStream))
{
ZipEntry entry = new ZipEntry("...");
//...
zipStream.PutNextEntry(entry);
zipStream.Write(data, 0, data.Length);
//...
zipStream.Finish();
zipStream.Close();
}
Edit ::
You need in general unZip your file, get ZipEntry , change , and write in ZipOutputStream with MemoryStream.
Use this article http://www.codeproject.com/KB/cs/Zip_UnZip.aspx
I am using the following C# code to compress a file:
// Open the stream we want to compress
FileStream fs = File.Create(#"C:\Projects\Samples\test\compressed.zip", 0);
// Creates the GZipStream
GZipStream gzip = new GZipStream(fs, CompressionMode.Compress);
// Reading the content to compress
byte[] bytes = File.ReadAllBytes(#"C:\Projects\Samples\samplefile.xml");
// Writing compressed content
gzip.Write(bytes, 0, bytes.Length);
gzip.Close(); // This also closes the FileStream (the underlying stream)
However, when I extract the file from windows explorer the file loses it's extension so instead of samplefile.xml it just becomes samplefile. Same thing happened with .txt file not just .xml file.
Can you help me see what I'm doing wrong?
ok found the problem:
Line 2 has to be as follows:
FileStream fs = File.Create(#"C:\Projects\Samples\test\compressed.xml.zip", 0);
GZipStream doesn't create zip archives. It creates a gzip file, which contains only one file, and doesn't necessarily store a filename at all. Normally you should use the .gz extension to identify a gzip file, and it's conventional to use the entire name of the original file with .gz appended on the end. See also here for more information about gzip format: http://en.wikipedia.org/wiki/Gzip#File_format
If you actually want to create zip archives, you might want to use a library like SharpZipLib: http://www.icsharpcode.net/opensource/sharpziplib/