RE: for Solution
RE: for Solution
Hi mohan.r1980
I think I don't understand your question but I'm going to try to help you:
When you talk about the current file name, are you talking about one DataInput file or are you talking about one file read in Scripting?
If the answer is one DataInput you can access to the name of the file with WorkFlow.Parameter["DataInput1","dif"] where DataInput1 is the name of the Data Input module you want to know. To separate the file name from this value you can do the following script:
For the other the method getName() can help you. The sintax is:
String baseName = fileList.Name;
or
String baseName = fileList.getName();
I hope i help you.
You can ask to GMC directly from his comunity
Regards.
RE: for Solution
Hi mohan.r1980
I think I don't understand your question but I'm going to try to help you:
When you talk about the current file name, are you talking about one DataInput file or are you talking about one file read in Scripting?
If the answer is one DataInput you can access to the name of the file with WorkFlow.Parameter["DataInput1","dif"] where DataInput1 is the name of the Data Input module you want to know. To separate the file name from this value you can do the following script:
Code:
String path = WorkFlow.Parameter["DataInput1","dif"];
Int endPath = path.findBackwards('\\'); //find the last directory
Int fileNameExtLength = path.Length - endPath; //calculate the file_name.ext lenght
String fileNameExt = path.mid(endPath,fileNameExtLength); //extract the file_name.ext
Int startExt = fileNameExt.findBackwards('.'); //find the extension
Int fileNameLength = fileNameExt.Length - startExt; //calculate the extension lenght
String fileName = fileNameExt.mid(startExt,fileNameLength); extract the file_name
String baseName = fileList.Name;
or
String baseName = fileList.getName();
I hope i help you.
You can ask to GMC directly from his comunity
Regards.