The following code gives me a System.IO.IOException with the message 'The process cannot access the file'.
private void UnPackLegacyStats()
{
DirectoryInfo oDirectory;
XmlDocument oStatsXml;
//Get the directory
oDirectory = new DirectoryInfo(msLegacyStatZipsPath);
//Check if the directory exists
if (oDirectory.Exists)
{
//Loop files
foreach (FileInfo oFile in oDirectory.GetFiles())
{
//Check if file is a zip file
if (C1ZipFile.IsZipFile(oFile.FullName))
{
//Open the zip file
using (C1ZipFile oZipFile = new C1ZipFile(oFile.FullName, false))
{
//Check if the zip contains the stats
if (oZipFile.Entries.Contains("Stats.xml"))
{
//Get the stats as a stream
using (Stream oStatsStream = oZipFile.Entries["Stats.xml"].OpenReader())
{
//Load the stats as xml
oStatsXml = new XmlDocument();
oStatsXml.Load(oStatsStream);
//Close the stream
oStatsStream.Close();
}
//Loop hit elements
foreach (XmlElement oHitElement in oStatsXml.SelectNodes("/*/hits"))
{
//Do stuff
}
}
//Close the file
oZipFile.Close();
}
}
//Delete the file
oFile.Delete();
}
}
}
I am struggling to see where the file could still be locked. All objects that could be holding onto a handle to the file are in using blocks and are explicitly closed.
Is it something to do with using FileInfo objects rather than the strings returned by the static GetFiles method?
Any ideas?
I do not see problems in your code, everything look ok. To check is the problem lies in C1ZipFile I suggest you initialize zip from stream, instead of initialization from file, so you close stream explicitly:
//Open the zip file
using (Stream ZipStream = oFile.OpenRead())
using (C1ZipFile oZipFile = new C1ZipFile(ZipStream, false))
{
// ...
Several other suggestions:
You do not need to call Close() method, with using (...), remove them.
Move xml processing (Loop hit elements) outsize zip processing, i.e. after zip file closeing, so you keep file opened as least as possible.
I assume you're getting the error on the oFile.Delete call. I was able to reproduce this error. Interestingly, the error only occurs when the file is not a zip file. Is this the behavior you are seeing?
It appears that the C1ZipFile.IsZipFile call is not releasing the file when it's not a zip file. I was able to avoid this problem by using a FileStream instead of passing the file path as a string (the IsZipFile function accepts either).
So the following modification to your code seems to work:
if (oDirectory.Exists)
{
//Loop files
foreach (FileInfo oFile in oDirectory.GetFiles())
{
using (FileStream oStream = new FileStream(oFile.FullName, FileMode.Open))
{
//Check if file is a zip file
if (C1ZipFile.IsZipFile(oStream))
{
// ...
}
}
//Delete the file
oFile.Delete();
}
}
In response to the original question in the subject: I don't know if it's possible to know if a file can be deleted without attempting to delete it. You could always write a function that attempts to delete the file and catches the error if it can't and then returns a boolean indicating whether the delete was successful.
I'm just guessing: are you sure that oZipFile.Close() is enough? Perhaps you have to call oZipFile.Dispose() or oZipFile.Finalize() to be sure it has actually released the resources.
More then Likely it's not being disposed, anytime you access something outside of managed code(streams, files, etc.) you MUST dispose of them. I learned the hard way with Asp.NET and Image files, it will fill up your memory, crash your server, etc.
In the interest of completeness I am posing my working code as the changes came from more than one source.
private void UnPackLegacyStats()
{
DirectoryInfo oDirectory;
XmlDocument oStatsXml;
//Get the directory
oDirectory = new DirectoryInfo(msLegacyStatZipsPath);
//Check if the directory exists
if (oDirectory.Exists)
{
//Loop files
foreach (FileInfo oFile in oDirectory.GetFiles())
{
//Set empty xml
oStatsXml = null;
//Load file into a stream
using (Stream oFileStream = oFile.OpenRead())
{
//Check if file is a zip file
if (C1ZipFile.IsZipFile(oFileStream))
{
//Open the zip file
using (C1ZipFile oZipFile = new C1ZipFile(oFileStream, false))
{
//Check if the zip contains the stats
if (oZipFile.Entries.Contains("Stats.xml"))
{
//Get the stats as a stream
using (Stream oStatsStream = oZipFile.Entries["Stats.xml"].OpenReader())
{
//Load the stats as xml
oStatsXml = new XmlDocument();
oStatsXml.Load(oStatsStream);
}
}
}
}
}
//Check if we have stats
if (oStatsXml != null)
{
//Process XML here
}
//Delete the file
oFile.Delete();
}
}
}
The main lesson I learned from this is to manage file access in one place in the calling code rather than letting other components manage their own file access. This is most apropriate when you want to use the file again after the other component has finished it's task.
Although this takes a little more code you can clearly see where the stream is disposed (at the end of the using), compared to having to trust that a component has correctly disposed of the stream.
Related
When I open an excel file, a hidden temporary file is generated in the same folder. I can open it with the TotalCommander Viewer, but I always get an IO exception when trying to open with powershell or c#.
new FileStream(#"D:\~$test.xlsx", FileMode.Open, FileAccess.Read, FileShare.ReadWrite);
System.IO.IOException: 'The process cannot access the file 'D:~$test.xlsx' because it is being used by another process.'
So how can I get the content?
Unfortunately for some reason you can not open the direct file, so I suggest another method when you copy a the file to a temp file, then read it and finally you delete the temp file, this way you can read it, I suppose TotalCommander uses the same method for opening files in Viewer.
static void Main(string[] args)
{
CopyReadAndDelete(#"c:\Documents\~$test.xlsx");
}
static void CopyReadAndDelete(string filePath)
{
var tempFileFullPath = Path.Combine(Path.GetDirectoryName(filePath), Guid.NewGuid().ToString());
File.Copy(filePath, tempFileFullPath);
try
{
using (var sr = new StreamReader(tempFileFullPath))
{
Console.WriteLine(sr.ReadToEnd()); //or do anything with the content
}
}
finally
{
File.Delete(tempFileFullPath);
}
}
I am trying to export the strings in a nested list to a txt or csv file of the users choice and everything seems to be working but when I actually go to check the file after I have exported it the file is absolutely blank. I went and did it on a separate test program to mock my problem and it worked on that program but when I moved the code over it would still not export anything.
This is just my initialized nested list in case its needed.
List<List<string>> aQuestion = new List<List<string>>();
This is the problem area for the code.
static void writeCSV(List<List<string>> aQuestion, List<char> aAnswer)
{
StreamWriter fOut = null;
string fileName = "";
//export questions
//determine if the file can be found
try
{
Console.Write("Enter the file path for where you would like to export the exam to: ");
fileName = Console.ReadLine();
if (!File.Exists(fileName))
{
throw new FileNotFoundException();
}
}
catch (FileNotFoundException)
{
Console.WriteLine("File {0} cannot be found", fileName);
}
//writes to the file
try
{
fOut = new StreamWriter(fileName, false);
//accesses the nested lists
foreach (var line in aQuestion)
{
foreach (var value in line)
{
fOut.WriteLine(string.Join("\n", value));
}
}
Console.WriteLine("File {0} successfully written", fileName);
}
catch (IOException ioe)
{
Console.WriteLine("File {0} cannot be written {1}", fileName, ioe.Message);
}
So if any of you guys can help me with this problem that would be great because it seems like such a small problem but I can't figure it out for the life of me.
It may happen that the buffer was not flushed to the disk. You should dispose the stream writer and it will push everything out to disk:
using (StreamWriter writer = new StreamWriter(fileName, false)) // <-- this is the change
{
//accesses the nested lists
foreach (var line in aQuestion)
{
foreach (var value in line)
{
writer.WriteLine(string.Join("\n", value));
}
}
}
On a more elaborate level, streams that may cause performance loss are normally buffered. File streams are definitely buffered, because it would be very inefficient to push each separate piece of data to the IO immediately.
When you're working with file streams, you can flush their content explicitly using the StreamWriter.Flush() method - that is useful if you want to debug code and wish to see how far it has gone writing the data.
However, you normally do not flush the stream yourself but just let its internal mechanisms choose the best moment to do that. Instead, you make sure to dispose the stream object, and that will force buffer to be flushed before closing the stream.
Use this simple method instead, it is much easier and it will take care of creating and disposing StreamWriter.
File.WriteAllLines(PathToYourFile,aQuestion.SelectMany(x=>x));
More reference on File.WriteAllLines Here
Also, in your code your not disposing StreamWrite. Enclose it in a Using block. Like this..
using(var writer = new StreamWriter(PathToYourFile,false)
{
//Your code here
}
I have a ZipArchive and am looking to access a file inside. I am not sure how I would do this, but I have a list
List<ZipContents> importList = new List<ZipContents>();
Which has two parameters:
ZipArchive which is called ZipFile
String which is called FileName
Inside the ZipArchive which is importList.ZipFile I need to find an XML file that has the same name as the Zip file name.
Currently I have this:
foreach (var import in importList)
{
var fn = import.FileName; // This is the actual file name of the zip file
// that was added into the ZipArchive.
// I will need to access the specific XML file need in the Zip by associating with
// fn
// ToDo: Extract XML file needed
// ToDo: Begin to access its contents...
}
So for example the code is looking into the ZipArchive with the name test.zip. there will be a file called test.xml that I will then need to be able to access its contents.
Like I said above I need to be able to access the contents of that file. I am sorry I have no code to support how to do this, but I have not been able to find anything else...
I have looked through a lot of the ZIpArchive documentation (including: http://msdn.microsoft.com/en-us/library/system.io.compression.ziparchive%28v=vs.110%29.aspx) and other posts on SO on how to do this, but I have come up empty. Would anyone have an idea on how to do this? Any help would be much appreciated. Thanks!
You need to extract the archive to a directory (may as well use temp since I assume you don't want to keep these):
archive.ExtractToDirectory("path string");
//Get the directory info for the directory you just extracted to
DirectoryInfo di = new DirectoryInfo("path string");
//find the xml file you want
FileInfo fi = di.GetFiles(string.Format("{0}.xml", archiveName)).FirstOrDefault();
//if the file was found, do your thing
if(fi != null)
{
//Do your file stuff here.
}
//delete the extracted directory
di.Delete();
Edit: To do the same thing just unpacking the file you care about:
//find your file
ZipArchiveEntry entry = archive
.Entries
.FirstOrDefault(e =>
e.Name == string.Format("{0}.xml", archiveName));
if(entry != null)
{
//unpack your file
entry.ExtractToFile("path to extract to");
//do your file stuff here
}
//delete file if you want
The MSDN you linked does a rather good job explaining how to access the files. Here it is applied to your example.
// iterate over the list items
foreach (var import in importList)
{
var fn = import.FileName;
// iterate over the actual archives
foreach (ZipArchiveEntry entry in import.ZipFile.Entries)
{
// only grab files that end in .xml
if (entry.FullName.EndsWith(".xml", StringComparison.OrdinalIgnoreCase))
{
// this extracts the file
entry.ExtractToFile(Path.Combine(#"C:\extract", entry.FullName));
// this opens the file as a stream
using(var stream = new StreamReader(entry.Open())){
// this opens file as xml stream
using(var xml = XmlReader.Create(stream){
// do some xml access on an arbitrary node
xml.MoveToContent();
xml.ReadToDescendant("my-node");
var item = xml.ReadElementContentAsString();
}
}
}
}
}
The following will extract a single xml file called file.xml and read it to an XDocument object:
var xmlEntry = importList.SelectMany(y => y.Entries)
.FirstOrDefault(entry => entry.Name.Equals("file.xml",
StringComparison.OrdinalIgnoreCase));
if (xmlEntry == null)
{
return;
}
// Open a stream to the underlying ZipArchiveEntry
using (XDocument xml = XDocument.Load(xmlEntry.Open()))
{
// Do stuff with XML
}
My current project uses a direct file path of this excel document to read information off of the excel file. I need to get my project ready for release so I cannot have the project hard code a file path in the form of a string.
I want to embed the Excel File in my resource, which I have done, but know how can I get the file path from Resource, and send a file path to the class which reads the Excel file. The class must be feed a filepath so I was thinking of making a copy of this Excel file, and in the Temp folder then referenceing the file path for the class to read the Excel file.
FileName = #"D:\SomeFolder\ExcelFile.xlsx"; //This is the old code, hard coded
//I need code that is going to make a copy of this file from the Resources and save it somewhere in a temp folder, but then give me
the File path in the form of a string.
string FileName;
// I need the file name to have the directory of this excel that is in the Resource folder
//Call Class to Create XML File and store Data from BIN File Locally on Program
ReadExcel_CreateXML = new ExcelRecorder(FileName);
Something else to think about is that you are probably reading the current files using a FileStream and either a BinaryReader or StreamReader. If that's the case, the consumer of the file could be written to accept an arbitrary Stream instead and then you can create a MemoryStream to pass to the consuming class:
// The resource will be a byte array, I'm just creating a
// byte array manually for example purposes.
var fileData = System.Text.Encoding.UTF8.GetBytes("Hello\nWorld!");
using (var memoryStream = new MemoryStream(fileData))
using (var streamReader = new StreamReader(memoryStream))
{
// Do whatever you need with the file's contents
Console.WriteLine(streamReader.ReadLine());
Console.WriteLine(streamReader.ReadLine());
}
This approach means you won't be cluttering up the client computer with temporary files that you'll need to clean up. It also means your consuming class will become more flexible if you ever need to process data over any other type of Stream.
I'm not sure if this the best solution, but it will work:
1st get the byte[] array of the file in your resources:
byte[] fileByteArray = global::YourProjectNameSpace.Properties.Resources.ExcelFileName
2nd Export the file to a temporary location using this function:
(I got from here: Write bytes to file)
public bool ByteArrayToFile(string _FileName, byte[] _ByteArray)
{
try
{
// Open file for reading
System.IO.FileStream _FileStream =
new System.IO.FileStream(_FileName, System.IO.FileMode.Create,
System.IO.FileAccess.Write);
// Writes a block of bytes to this stream using data from
// a byte array.
_FileStream.Write(_ByteArray, 0, _ByteArray.Length);
// close file stream
_FileStream.Close();
return true;
}
catch (Exception _Exception)
{
// Error
Console.WriteLine("Exception caught in process: {0}",
_Exception.ToString());
}
// error occured, return false
return false;
}
And last access that temporary file like you normally would
Use:
Just create a button in a form and put this code in the button's click event
private void button1_Click(object sender, EventArgs e)
{
byte[] fileByteArray = global::YourProjectNameSpace.Properties.Resources.ExcelFileName;
if (ByteArrayToFile(#"C:\Temp\file.xlsx", fileByteArray))
{
//File was saved properly
}
else
{
//There was an error saving the file
}
}
Hope it works
Trying to extract files to a given folder ignoring the path in the zipfile but there doesn't seem to be a way.
This seems a fairly basic requirement given all the other good stuff implemented in there.
What am i missing ?
code is -
using (Ionic.Zip.ZipFile zf = Ionic.Zip.ZipFile.Read(zipPath))
{
zf.ExtractAll(appPath);
}
While you can't specify it for a specific call to Extract() or ExtractAll(), the ZipFile class has a FlattenFoldersOnExtract field. When set to true, it flattens all the extracted files into one folder:
var flattenFoldersOnExtract = zip.FlattenFoldersOnExtract;
zip.FlattenFoldersOnExtract = true;
zip.ExtractAll();
zip.FlattenFoldersOnExtract = flattenFoldersOnExtract;
You'll need to remove the directory part of the filename just prior to unzipping...
using (var zf = Ionic.Zip.ZipFile.Read(zipPath))
{
zf.ToList().ForEach(entry =>
{
entry.FileName = System.IO.Path.GetFileName(entry.FileName);
entry.Extract(appPath);
});
}
You can use the overload that takes a stream as a parameter. In this way you have full control of path where the files will be extracted to.
Example:
using (ZipFile zip = new ZipFile(ZipPath))
{
foreach (ZipEntry e in zip)
{
string newPath = Path.Combine(FolderToExtractTo, e.FileName);
if (e.IsDirectory)
{
Directory.CreateDirectory(newPath);
}
else
{
using (FileStream stream = new FileStream(newPath, FileMode.Create))
e.Extract(stream);
}
}
}
That will fail if there are 2 files with equal filenames. For example
files\additionalfiles\file1.txt
temp\file1.txt
First file will be renamed to file1.txt in the zip file and when the second file is trying to be renamed an exception is thrown saying that an item with the same key already exists