I have DotNetZip installed and running fine on a Windows 2008 server.
Using a classic ASP page, I want to bundle a bunch of comma-delimited files to a user and send it over in a zip file.
The following code works fine but it stores all the path information so the files inside the zip file are located in some ridiculous directory like C:\Inetpub\wwwroot\appname\_temp\
I'm using the following code:
Set objZip = CreateObject("Ionic.Zip.ZipFile")
sFileArray = Split(sFileArray, "|")
For iCount = 0 To UBound(sFileArray)
If sFileArray(iCount) <> "" Then
objZip.AddFile sFileArray(iCount)
End If
Next
objZip.Name = sFilePath & "test.zip"
objZip.Save()
objZip.Dispose()
Set objZip = Nothing
I see that the AddFile method allows you to specify where you want the added file to reside in the zip file if you add a second parameter. According to the documentation objZip.AddFile sFileArray(iCount), "" should put the file in the root of the zip file.
However, when I add that parameter, I get the following error:
Wrong number of arguments or invalid property assignment: 'objZip.AddFile'
Anyone have any idea what I'm doing wrong?
Thanks.
I think you are misinterperting the documentation. If the second parameter is null then the directory path of the file being added is used. If the second parameter is an empty string "" then the file is added to the root level in the zip. A quick look into the Ioniz.zip.dll shows that the single parameter override of AddFile method simply calls the the double parameter override with the second parameter set to null.
Hence your add file should look like:
objZip.AddFile sFileArray(iCount), ""
to get the result you are after.
Related
I tried PGP encrypting a file in ChoPGP library. At the end of the process it shows embedded file name along with the whole original file name path.
But I thought it will show only the filename without the whole path. Which I intend to work on and figure out a way to do so?
Doing the following:
using (ChoPGPEncryptDecrypt pgp = new ChoPGPEncryptDecrypt())
{
pgp.EncryptFile(#"\\appdevtest\c$\appstest\Transporter\test\Test.txt",
#"\\appdevtest\c$\appstest\Transporter\test\OSCTestFile_ChoPGP.gpg",
#"\\appdevtest\c$\appstest\Transporter\pgpKey\PublicKey\atorres\atorres_publicKey.asc",
true,
true);
}
which will result in:
But I would like to only extract the Test.txt in the end something
like this:
Looking at this line from the ChoPGP sources: https://github.com/Cinchoo/ChoPGP/blob/7152c7385022823013324408e84cb7e25d33c3e7/ChoPGP/ChoPGPEncryptDecrypt.cs#L221
You may find out that it uses internal function GetFileName, which ends up with this for the FileStream: return ((FileStream)stream).Name;.
And this one, according to documentation, Gets the absolute path of the file opened in the FileStream..
So you should either make fork of ChoPGP and modify this line to extract just filename, or submit a pull request to the ChoPGP. Btw, it's just a wrapper around the BouncyCastle so you may use that instead.
I am working on an ASP classic website, which the client reported suddenly exhibited an issue with a previously working function which listed only "image" file types. Upon reading the code, I found that the loop that lists the files in a folder, uses the InStr() function to identify files by their type, which should be "image." However, I found that something must have changed in the OS, as the type is not longer "image", but "JPG", or "PNG", etc. This dramatically changes the way the code works. Following is a snipped of the code:
Set oFSO = Server.CreateObject("Scripting.FileSystemObject")
Set oFolder = oFSO.GetFolder(Server.MapPath(sCurrentDirectoryPath))
Set oSubFolder = oFolder.Files
iFileCount = 0
For Each oFileName in oSubFolder
If InStr(1, LCase(oFileName.Type),"image") > 0 Then
iFileCount = iFileCount + 1
End If
Next
Because the InStr() function is trying to find a file type of "image", no files are counted up, and the function returns zero files found. Whilst debugging, I found that the value being returned by oFileName.Type, was as follows:
This is the file type:JPG File
This is the file type:JPG File
This is the file type:Text Document
This is the file type:Data Base File
Files in the folder were two "whatever.jpg" files, a "whatever.txt" file, and a "thumbs.db" file. So, it appears that the OS (Windows Server 2019) may have changed to be less generic with reporting an "image" file, and is now reporting "JPG file" or "PNG file", etc. This of course, breaks this code! Are there any suggestions from you'all on how I could go about modifying this code to work on reporting exactly how many image files are present?
On Windows 10, the Type values for .jpg and .png files are JPEG image and PNG image respectively. What OS are you using?
Also, Type doesn't actually analyze the file, you could have a virus.exe file in the folder that has been renamed to virus.jpg, and the Type value will still show it as JPEG. So if the function is indented to check user uploaded content to ensure images are actually images, the Type value will be of no use. If you have root access you could install a COM DLL that uses a program such as ExifTool to properly analyze files (https://github.com/as08/ClassicASP.ExifTool), however that would be a complete rewrite.
But assuming you're not looking to check if an image file is actually an image file, you could just split the filename extensions and use Select Case to count the image files if your OS is returning just XXX file and no longer XXX image in the Type value (alternatively you could split the Type value, but you'd still need to check for valid image file extensions):
Dim oFSO, oFolder, oSubFolder, oFileName, iFileCount, oFileNameSplit
Set oFSO = Server.CreateObject("Scripting.FileSystemObject")
Set oFolder = oFSO.GetFolder(Server.MapPath(sCurrentDirectoryPath))
Set oSubFolder = oFolder.Files
iFileCount = 0
For Each oFileName In oSubFolder
oFileNameSplit = Split(oFileName.Name,".")
If uBound(oFileNameSplit) > 0 Then
Select Case Trim(lCase(oFileNameSplit(uBound(oFileNameSplit))))
Case "jpg", "jpeg", "png", "gif" ' Maybe add some more extensions...
iFileCount = iFileCount + 1
End Select
End If
Next
Set oFSO = Nothing
Set oFolder = Nothing
Set oSubFolder = Nothing
Response.Write(iFileCount)
I am trying to save an image to server.Server directory structure is as follows
httpdocs-
Folder-
Page.aspx
and
httpdocs-
Images-
Subimages
The Folder and Images are under the httpdocs. I need to save the image to subimages folder from pages.aspx. Saving image code is on pages.aspx.
I tried
string CroppedImagePath = Server.MapPath("~/Images/Subimages"+file)
But not get the exact result
Unless file contains a leading / character, that's going to end up saving as something like:
~/Images/Subimagesfilename.ext
Use Path.Combine() to build a path name. Something like this:
var CroppedImagePath = Path.Combine(Server.MapPath("~/Images/Subimages"), file);
try this
string CroppedImagePath = Server.MapPath("~/Images/Subimages/"+file)
/ after Subimages
I have two EXE files , one for encryption and other for decryption.
Now i want to call these exe's in my ASP pages built in VB script to encrypt these values and save then in a file.
These exe's take two or three values as parameters and return two values in XML form.
for exapmle :
Calling exe :: c:\IKish\Shared\Encrypt>Encrypt.exe /iv:"xx xx xx" /text:user
Output ::
<Evalue>
<key>784cjidbdk77fkdl161c</key>
<Result>9kflo53jvo1ce4801kskbf399b7</Result>
</Evalue>
Could someone pls help me with the code to execute the exe and catch its output in the ASP file?
Set oShell = Server.CreateObject("WScript.Shell")
Set oExec = oShell.Exec("Encrypt.exe /iv:""xx xx xx"" /text:user")
If Not (oExec Is Nothing) Then
strOutput = oExec.StdOut.ReadAll
End If
I have 3 jars: jar1, jar2 and jar3, in the same path who can change in other pc (ex: c:\prova)
When I run jar1, it moves jar2 in the Windows Sturtup folder.
I want that jar2 simply activate jar3 at every windows startup, but of course it doesn't find jar3 who is remained in the first path.
So I want that jar1 pass a reference (in this case the path c:\prova) to the jar2, when moving it, or at least on the first call to it.
I find it difficoult because:
I can't write the path in a text file in jar2: text files in jars aren't writable.
I can't write the text file in the windows Startup folder: it will be opened at every win startup..
I can't pass the path as a parameter, it will be good for the first call but I can't store this value for the succesive calls.
Sorry for my bad english, thanks for any help!
To add the file Path.txt (with jar3's path) in jar2:
Runtime.getRuntime().exec("jar uf jar2.jar Path.txt");
To read the file in jar2 (Startup is my class name):
String s = "/Path.txt";
is = Startup.class.getResourceAsStream(s);
br = new BufferedReader(new InputStreamReader(is));
while (null != (line = br.readLine())) {
list.add(line);
}
Thank me!