.net core create on memory zipfile
I am working on an MVC project where I create dynamically pdf files(wkhtmltopdf) which I want to return in a zip file. The pdf files are generated on fly - I don't need to store them, so my code to return a single file is:
File(pdfBytes, "application/pdf", "file_name")
Taking a look on Microsoft docs their example goes through stored files:
string startPath = @"c:examplestart";
string zipPath = @"c:exampleresult.zip";
string extractPath = @"c:exampleextract";
ZipFile.CreateFromDirectory(startPath, zipPath);
ZipFile.ExtractToDirectory(zipPath, extractPath);
In my case I want to create N pdf files and return it to the view as a zip..
Something like:
ZipFile zip = new ZipFile();
foreach(var html in foundRawHTML)
{
//create pdf
//append pdf to zip
}
return zip;
Although this is not doable because:
- ZipFile and File are statics and cannot be instatiated
- There is no way to add a file into a zip on fly(in memory)
Any help is welcome
.net asp.net-core zip
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
add a comment |
I am working on an MVC project where I create dynamically pdf files(wkhtmltopdf) which I want to return in a zip file. The pdf files are generated on fly - I don't need to store them, so my code to return a single file is:
File(pdfBytes, "application/pdf", "file_name")
Taking a look on Microsoft docs their example goes through stored files:
string startPath = @"c:examplestart";
string zipPath = @"c:exampleresult.zip";
string extractPath = @"c:exampleextract";
ZipFile.CreateFromDirectory(startPath, zipPath);
ZipFile.ExtractToDirectory(zipPath, extractPath);
In my case I want to create N pdf files and return it to the view as a zip..
Something like:
ZipFile zip = new ZipFile();
foreach(var html in foundRawHTML)
{
//create pdf
//append pdf to zip
}
return zip;
Although this is not doable because:
- ZipFile and File are statics and cannot be instatiated
- There is no way to add a file into a zip on fly(in memory)
Any help is welcome
.net asp.net-core zip
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
add a comment |
I am working on an MVC project where I create dynamically pdf files(wkhtmltopdf) which I want to return in a zip file. The pdf files are generated on fly - I don't need to store them, so my code to return a single file is:
File(pdfBytes, "application/pdf", "file_name")
Taking a look on Microsoft docs their example goes through stored files:
string startPath = @"c:examplestart";
string zipPath = @"c:exampleresult.zip";
string extractPath = @"c:exampleextract";
ZipFile.CreateFromDirectory(startPath, zipPath);
ZipFile.ExtractToDirectory(zipPath, extractPath);
In my case I want to create N pdf files and return it to the view as a zip..
Something like:
ZipFile zip = new ZipFile();
foreach(var html in foundRawHTML)
{
//create pdf
//append pdf to zip
}
return zip;
Although this is not doable because:
- ZipFile and File are statics and cannot be instatiated
- There is no way to add a file into a zip on fly(in memory)
Any help is welcome
.net asp.net-core zip
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
I am working on an MVC project where I create dynamically pdf files(wkhtmltopdf) which I want to return in a zip file. The pdf files are generated on fly - I don't need to store them, so my code to return a single file is:
File(pdfBytes, "application/pdf", "file_name")
Taking a look on Microsoft docs their example goes through stored files:
string startPath = @"c:examplestart";
string zipPath = @"c:exampleresult.zip";
string extractPath = @"c:exampleextract";
ZipFile.CreateFromDirectory(startPath, zipPath);
ZipFile.ExtractToDirectory(zipPath, extractPath);
In my case I want to create N pdf files and return it to the view as a zip..
Something like:
ZipFile zip = new ZipFile();
foreach(var html in foundRawHTML)
{
//create pdf
//append pdf to zip
}
return zip;
Although this is not doable because:
- ZipFile and File are statics and cannot be instatiated
- There is no way to add a file into a zip on fly(in memory)
Any help is welcome
.net asp.net-core zip
.net asp.net-core zip
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
asked Nov 19 '18 at 16:03
George GeorgeGeorge George
797
797
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
You can use in memory byte arrays and ZipArchive from System.IO.Compression, there is no need to map the local drive:
public static byte GetZipArchive(List<InMemoryFile> files)
{
byte archiveFile;
using (var archiveStream = new MemoryStream())
{
using (var archive = new ZipArchive(archiveStream, ZipArchiveMode.Create, true))
{
foreach (var file in files)
{
var zipArchiveEntry = archive.CreateEntry(file.FileName, CompressionLevel.Fastest);
using (var zipStream = zipArchiveEntry.Open())
zipStream.Write(file.Content, 0, file.Content.Length);
}
}
archiveFile = archiveStream.ToArray();
}
return archiveFile;
}
public class InMemoryFile
{
public string FileName { get; set; }
public byte Content { get; set; }
}
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can use in memory byte arrays and ZipArchive from System.IO.Compression, there is no need to map the local drive:
public static byte GetZipArchive(List<InMemoryFile> files)
{
byte archiveFile;
using (var archiveStream = new MemoryStream())
{
using (var archive = new ZipArchive(archiveStream, ZipArchiveMode.Create, true))
{
foreach (var file in files)
{
var zipArchiveEntry = archive.CreateEntry(file.FileName, CompressionLevel.Fastest);
using (var zipStream = zipArchiveEntry.Open())
zipStream.Write(file.Content, 0, file.Content.Length);
}
}
archiveFile = archiveStream.ToArray();
}
return archiveFile;
}
public class InMemoryFile
{
public string FileName { get; set; }
public byte Content { get; set; }
}
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
add a comment |
You can use in memory byte arrays and ZipArchive from System.IO.Compression, there is no need to map the local drive:
public static byte GetZipArchive(List<InMemoryFile> files)
{
byte archiveFile;
using (var archiveStream = new MemoryStream())
{
using (var archive = new ZipArchive(archiveStream, ZipArchiveMode.Create, true))
{
foreach (var file in files)
{
var zipArchiveEntry = archive.CreateEntry(file.FileName, CompressionLevel.Fastest);
using (var zipStream = zipArchiveEntry.Open())
zipStream.Write(file.Content, 0, file.Content.Length);
}
}
archiveFile = archiveStream.ToArray();
}
return archiveFile;
}
public class InMemoryFile
{
public string FileName { get; set; }
public byte Content { get; set; }
}
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
add a comment |
You can use in memory byte arrays and ZipArchive from System.IO.Compression, there is no need to map the local drive:
public static byte GetZipArchive(List<InMemoryFile> files)
{
byte archiveFile;
using (var archiveStream = new MemoryStream())
{
using (var archive = new ZipArchive(archiveStream, ZipArchiveMode.Create, true))
{
foreach (var file in files)
{
var zipArchiveEntry = archive.CreateEntry(file.FileName, CompressionLevel.Fastest);
using (var zipStream = zipArchiveEntry.Open())
zipStream.Write(file.Content, 0, file.Content.Length);
}
}
archiveFile = archiveStream.ToArray();
}
return archiveFile;
}
public class InMemoryFile
{
public string FileName { get; set; }
public byte Content { get; set; }
}
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
You can use in memory byte arrays and ZipArchive from System.IO.Compression, there is no need to map the local drive:
public static byte GetZipArchive(List<InMemoryFile> files)
{
byte archiveFile;
using (var archiveStream = new MemoryStream())
{
using (var archive = new ZipArchive(archiveStream, ZipArchiveMode.Create, true))
{
foreach (var file in files)
{
var zipArchiveEntry = archive.CreateEntry(file.FileName, CompressionLevel.Fastest);
using (var zipStream = zipArchiveEntry.Open())
zipStream.Write(file.Content, 0, file.Content.Length);
}
}
archiveFile = archiveStream.ToArray();
}
return archiveFile;
}
public class InMemoryFile
{
public string FileName { get; set; }
public byte Content { get; set; }
}
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
answered Nov 19 '18 at 16:45
Andrei FilimonAndrei Filimon
558210
558210
add a comment |
add a comment |
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