This code is giving me an ArgumentExeption when the correct values are put in both ComboBoxes, executing the code. The code basically just deletes a file and replaces it with a modified version taken from another folder.
Here is the exact text of the error message:
An unhandled exception of type 'System.ArgumentException' occurred in Microsoft.VisualBasic.dll
Additional information: The given file path ends with a directory separator character.
Here's the code:
If ComboBox1.Text = "Nokia" And ComboBox2.Text = "HTC" And My.Computer.FileSystem.FileExists("C:\Users\" + user + "\Documents\Fiddler2\Scripts\CustomRules.js") Then
My.Computer.FileSystem.DeleteFile("C:\Users\" + user + "\Documents\Fiddler2\Scripts\CustomRules.js")
My.Computer.FileSystem.CopyFile("Config\OEM\NokiaHTC.js", destinationFileName:="C:\Users\" + user + "\Documents\Fiddler2\Scripts\")
Else
My.Computer.FileSystem.CopyFile("Config\OEM\NokiaHTC.js", destinationFileName:="C:\Users\" + user + "\Documents\Fiddler2\Scripts\")
End If
The problem is that the destination file path ends with a "\" value. This isn't legal for the CopyFile API. Switch it to include the file name and this should fix the problem
My.Computer.FileSystem.CopyFile( _
"Config\OEM\NokiaHTC.js", _
destinationFileName:="C:\Users\" + user + "\Documents\Fiddler2\Scripts\NokiaHTC.js")
Related
I have two file names in following format:
Gontagrator_1.xml
Gontagrator_2.xml
As of now i am picking just Gontagrater_1.xml and rename it to processing and failed once done.
+ "&fileName=" + sftpFileName
+ "&preMove="+sftpFileName+".$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}.processing"
+ "&move="+sftpFileName+".$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}"
+ "&moveFailed="+sftpFileName+".$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}.failed"
Now Gontagrator_2 comes into picture. It is suggested to create separate routes for both.
Can we download both in one route and rename accordingly? If yes, what values i need to pass?
Update1: There are multiple files with different names but i need to use both above file names only
update 2: Whole from component is:
"{{m.protocol}}://{{m.hostname}}{{t.directory}}"
+ "?username={{m.username}}"
+ "&password={{m.password}}"
+ "&download=true"
+ "&useList=false"
+ "&stepwise=false"
+ "&disconnect=true"
+ "&passiveMode=true"
+ "&reconnectDelay=10000"
+ "&bridgeErrorHandler=true"
+ "&delay=30000"
//+ "&fileName=" + sftpFileName
+ "&include="+ sftpFileName
+ "&preMove=$simple{file:onlyname}.$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}.processing"
+ "&move=$simple{file:onlyname}.$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}"
+ "&moveFailed=$simple{file:onlyname}.$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}.failed"
+ "&readLock=idempotent-changed"
+ "&idempotentRepository=#infinispan"
+ "&readLockRemoveOnCommit=true")
.onException(GenericFileOperationFailedException.class)
.onWhen(exchange -> {
Throwable cause = exchange.getException(GenericFileOperationFailedException.class).getCause();
return (cause != null && cause.toString().equalsIgnoreCase("2: No such file"));
}).handled(true)
.logExhausted(true)
.logExhaustedMessageHistory(true)
.log("Could not find file")
.end()
.log("Downloading xml file")
//.to(archiveReceivedFile(sftpFileName))
.split(body().tokenizeXML("ERequest", "ERequests")).streaming()
.inOnly(E_QUEUE)
Yes, you can download all files in one route.
You need to
Remove fileName option (such that the route will pick all file)
Use file expression language (file:onlyname) to refer to filename currently handled by Camel SFTP component
+ "&preMove=$simple{file:onlyname}.$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}.processing"
+ "&move=$simple{file:onlyname}.$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}"
+ "&moveFailed=$simple{file:onlyname}.$simple{date:now:yyyy-MM-dd'T'hh-mm-ss}.failed"
Use include option to control file to be picked up (REGEX to match file name pattern)
+ "&include=Gontagrator_(1|2)\.xml"
Hi I am using apache camel 2.15.2. I have got a consumer template so that I can copy file with dynamic file names:
if (fileInfo != null) {
filename = fileInfo.getFileName();
String camelUri = "file://" + fileInfo.getCopyFilePath() + "/?fileName=RAW("
+ filename + ")&noop=false&idempotent=false&readLock=changed";
System.out.println("Camel uri: " + camelUri);
logger.info("Camel uri: " + camelUri);
Exchange ex = consumerTemplate.receive(camelUri);
....
As you can see, I have set noop, and idempotent explicitly to achieve copying same file more than once. But, it does not do that. It hangs on receive method for subsequent tries to copy a file with same name. It can copy that, only if we restart the application. Any suggestions would be much appreciated. It might be something similar to this issue, but I do not have access to that solution. Thanks in advance.
When I debugged through Camel code, it seems, it is calling EventDrivenPollingConsumer's receive method, and hangs when calls queue.take() (line 110, EventDrivenPollingConsumer). And, even inside that, 'count' variable is zero in ArrayBlockingQueue:
while (count == 0)
notEmpty.await();
Added this, just in case it helps anyone having a clue.
Ok, If I call 'consumerTemplate.doneUoW(ex)', it does copy multiple times. But, at the same time it was deleting (actually moving to .camel folder) the source file, which I did not want to! Then, had to set noop=true:
if (fileInfo != null) {
filename = fileInfo.getFileName();
String camelUri = "file://" + fileInfo.getCopyFilePath() + "/?fileName=RAW("
+ filename + ")&noop=true&idempotent=false&readLock=none";
System.out.println("Camel uri: " + camelUri);
logger.info("Camel uri: " + camelUri);
Exchange ex = consumerTemplate.receive(camelUri);
// consumerTemplate.r
logger.info("File received: " + fileInfo.getFileName());
exchange.getOut().setBody(ex.getIn().getBody());
consumerTemplate.doneUoW(ex);
}
Now, it works as expected.
I am new to Active Directory and still learning some of the concepts.
The code below shows connecting to AD on my local machine and this code works properly
DirectoryEntry entry = new DirectoryEntry("LDAP://CN=testing1,CN=Users,DC=mydomain,DC=com");
DirectoryEntryConfiguration entryConfiguration = entry.Options;
Console.WriteLine("Server: " + entryConfiguration.GetCurrentServerName());
Console.WriteLine("Page Size: " + entryConfiguration.PageSize.ToString());
Console.WriteLine("Password Encoding: " + entryConfiguration.PasswordEncoding.ToString());
Console.WriteLine("Password Port: " + entryConfiguration.PasswordPort.ToString());
Console.WriteLine("Referral: " + entryConfiguration.Referral.ToString());
Console.WriteLine("Security Masks: " + entryConfiguration.SecurityMasks.ToString());
Console.WriteLine("Is Mutually Authenticated: " + entryConfiguration.IsMutuallyAuthenticated().ToString());
Console.WriteLine();
Console.ReadLine();
Here is my problem: when I replace mydomain in the LDAP path of another machine it gives me an error
LDAP://CN=testing1,CN=Users,DC=XXXX,DC=com
gives me this error
System.DirectoryServices.DirectoryServicesCOMException was unhandled
Message=A referral was returned from the server.
This was basically a teething error
Instead of this:
LDAP://CN=testing1,CN=Users,DC=XXXX,DC=com
I should have written
LDAP://XXX.com/CN=testing1,CN=Users,DC=XXXX,DC=com
I am trying to construct a Control template from code behind. Things were working fine till recently I found that the code was throwing an exception because of escape characters in string. The error message is dynamically constructed by retrieving from resource file.
The exception is
A first chance exception of type 'System.Windows.Markup.XamlParseException' occurred in PresentationFramework.dll
Additional information: Name cannot begin with the '#' character, hexadecimal value 0x40. Line 1, position 537.
//In this case when exception is thrown,
//string errorMessage = "Name cannot contain any of the following characters $ \" # ; ^ | "
public static ControlTemplate GetErrorTemplate(string errorMessage)
{
string xamlString = "<ControlTemplate xmlns=\"http://schemas.microsoft.com/winfx/2006/xaml/presentation\" " +
"xmlns:x=\"http://schemas.microsoft.com/winfx/2006/xaml\" " +
"xmlns:nicefx=\"clr-namespace:NiceFx.Interop.UIComponents;assembly=NiceFx\" " +
"xmlns:wpfkit=\"http://schemas.microsoft.com/wpf/2008/toolkit\" >" +
" <DockPanel LastChildFill=\"True\">" +
"<TextBlock Foreground=\"White\" Background=\"Red\" FontSize=\"12\" Padding=\"2\" FontFamily=\"Trebuchet MS\" Margin=\"5,5,0,0\" TextWrapping=\"Wrap\" DockPanel.Dock=\"Bottom\" Text=\"" + errorMessage + "\"></TextBlock>" +
"<AdornedElementPlaceholder />" +
" </DockPanel>" +
" </ControlTemplate>";
//EXCEPTION OCCURS IN THIS LINE
ControlTemplate ct = (ControlTemplate)XamlReader.Load(XmlReader.Create(
new StringReader(xamlString)));
return ct;
}
How do I escape this string? I tried all possible ways but I am unable to do so.
According to the comment in your code, errorMessage contains a ", which will be inserted (without escaping it) into the XAML you are constructing. This " will then act as the closing quote of the Text attribute. At this point, the next non-whitespace character the parser encounters will be #, which is not an allowed character for the name of a XAML attribute, so it stops and reports the error.
That covers the why. As for how to escape it, you can use the XML entity for double quote: " Note that you may need to apply this escaping to multiple characters in your parameter.
I'm running a script made in Groovy from Soap UI and the script needs to generate lots of files.
Those files have also in the name two numbers from a list (all the combinations in that list are different), and there are 1303 combinations
available and the script generates just 1235 files.
A part of the code is:
filename = groovyUtils.projectPath + "\\" + "$file"+"_OK.txt";
targetFile = new File(filename);
targetFile.createNewFile();
where $file is actually that part of the file name which include those 2 combinations from that list:
file = "abc" + "-$firstNumer"+"_$secondNumber"
For those file which are not created is a message returned:"The filename, directory name or volume label syntax is incorrect".
I've tried puting another path:
filename = "D:\\rez\\" + "\\" + "$file"+"_OK.txt";
targetFile = new File(filename);
targetFile.createNewFile();
and also:
File parentFolder = new File("D:\\rez\\");
File targetFile = new File(parentFolder, "$file"+"_OK.txt");
targetFile.createNewFile();
(which I've found here: What are possible reasons for java.io.IOException: "The filename, directory name, or volume label syntax is incorrect")
but nothing worked.
I have no ideea where the problem is. Is strange that 1235 files are created ok, and the rest of them, 68 aren't created at all.
Thanks,
My guess is that some of the files have illegal characters in their paths. Exactly which characters are illegal is platform specific, e.g. on Windows they are
\ / : * ? " < > |
Why don't you log the full path of the file before targetFile.createNewFile(); is called and also log whether this method succeeded or not, e.g.
filename = groovyUtils.projectPath + "\\" + "$file"+"_OK.txt";
targetFile = new File(filename);
println "attempting to create file: $targetFile"
if (targetFile.createNewFile()) {
println "Successfully created file $targetFile"
} else {
println "Failed to create file $targetFile"
}
When the process is finished, check the logs and I suspect you'll see a common pattern in the ""Failed to create file...." messages
File.createNewFile() returns false when a file or directory with that name already exists. In all other failure cases (security, I/O) it throws an exception.
Evaluate createNewFile()'s return value or, additionally, use the File.exists() method:
File file = new File("foo")
// works the first time
createNewFile(file)
// prints an error message
createNewFile(file)
void createNewFile(File file) {
if (!file.createNewFile()) {
assert file.exists()
println file.getPath() + " already exists."
}
}