I have a .zip file in the current directory, I want to get its file name using Groovy. e.g. if the file is myfile.zip, I want to get the "myfile" part. Can anyone give me a code snip? Thanks.
Something like this should work:
filename=new File("directory").listFiles().find{it.name.endsWith(".zip")}
If you don't want the .zip on the end, subtract it:
filename=new File("directory").listFiles().find{it.name.endsWith(".zip")}.name - ".zip"
(By the way, the first one ends up with a file object--you can do whatever you want with it. The second ends up with the string that is the name without the .zip)
Related
I have an open std::fs::File, and I want to get it's filename, e.g. as a PathBuf. How do I do that?
The simple solution would be to just save the path used in the call to File::open. Unfortunately, this does not work for me. I am trying to write a program that reads log files, and the program that writes the logs keep changing the filenames as part of it's log rotation. So the file may very well have been renamed since it was opened. This is on Linux, so renaming open files is possible.
How do I get around this issue, and get the current filename of an open file?
On a typical Unix filesystem, a file may have multiple filenames at once, or even none at all. The file metadata is stored in an inode, which has a unique inode number, and this inode number can be linked from any number of directory entries. However, there are no reverse links from the inode back to the directory entries.
Given an open File object in Rust, you can get the inode number using the ino() method. If you know the directory the log file is in, you can use std::fs::read_dir() to iterate over all entries in that directory, and each entry will also have an ino() method, so you can find the one(s) matching your open file object. Of course this approach is subject to race conditions – the directory entry may already be gone again once you try to do anything with it.
On linux, files handles held by the current process can be found under /proc/self/fd. These look and act like symlinks to the original files (though I think they may technically be something else - perhaps someone who knows more can chip in).
You can therefore recover the (possibly changed) file name by constructing the correct path in /proc/self/fd using your file descriptor, and then following the symlink back to the filesystem.
This snippet shows the steps:
use std::fs::read_link;
use std::os::unix::io::AsRawFd;
use std::path::PathBuf;
// if f is your std::fs::File
// first construct the path to the symlink under /proc
let path_in_proc = PathBuf::from(format!("/proc/self/fd/{}", f.as_raw_fd()));
// ...and follow it back to the original file
let new_file_name = read_link(path_in_proc).unwrap();
This question already has answers here:
SSIS - How to loop through files in folder and get path+file names and finally execute stored Procedure with parameter as Path + Filename
(2 answers)
Closed 3 years ago.
I have a xlsx file that will be dropped into a folder on a monthly basis. The filename will change every month (filename_8292019) based on the date, to which I cannot change.
I want to build a foreach loop to pick up the xlsx file and manipulate it (load into SQL server table, the move the file to an archive folder). I cannot figure out how to do this with a dynamic filename (where the date changes.
I was able to successfully run the package when converting the xlsx to CSV, and also when pointing directly to the xlsx filename.
[Flat File Destination [219]] Error: Cannot open the datafile "filename"
OR errors relating to file not found
The Files: entry on the Collection tab of the Foreach Loop container will accept wildcard characters.
The general pattern here is to create a variable, say, FileName. Set your Files: to something like:
Files:
BaseFileName*
or, if you want to be sure to only pick up spreadsheets, maybe:
Files:
BaseFileName*.xlsx
Select either Name and extension or Fully qualified, which will include the full file path. I usually just use Name and extension and put the file path into another variable so when Ops tells me they're moving my drop location, I can change a parameter instead of editing the package. This step tells the container to remember the name of the file it just found so you can use it later for a variable mapping.
On the Variable Mappings tab, select your variable name and assign it to Index 0.
Then, for each spreadsheet, the container will loop, pick up the name of the first file it finds that matches your pattern, and assign the full name, with the date extension (and path, if you go that way), to your variable. Pass the variable as in input parameter to the tasks inside the loop and use that to process the file, including moving it to the archive, or you'll get yourself into an infinite loop, processing the same file(s) over and over. <--Does that sound like the voice of experience? Yeah. Been there, done that.
Edit:
Here, the FullFilePath variable is just the folder name, without a file reference. (Red variable to red entry in the Folder box).
The FileBaseName variable drives what shows up in the Files box. (Blue to blue).
Another variable picks up the actual file name, with the date extension. Later, say in a File System Task, if I need the folder & file name together, I concatenate the variables.
As far as the Excel Connection Manager error you're getting, unfortunately I'm no help. I don't use it. We have SentryOne's Task Factory for SSIS which includes a much more resilient Excel connector.
I have multiple zip files in a folder with names like below:
"abc.zip-20181002084936558425"
How to rename all of them with one command to get result like below:
"abc-20181002084936558425.zip"
I want the timestamp before the extension for multiple filenames. Now every file has different timestamp so renaming should consider that. Can I rename multiple files like this using single command.
Providing your file are really all with the same name convention and you being in the right directory :
for i in *.zip-*; do newName=${i//.zip};mv $i $newName".zip";done
should do the trick.
I have a process that periodically gets files from a server and copy them with SFTP to a local directory. It should not overwrite the file if it already exists. I know with something like Winston I can automatically rotate the log file when it fills up, but in this case I need a similar functionality to rotate files if they already exist.
An example:
The routine copies a remote file called testfile.txt to a local directory. The next time it's run the same remote file is found and copied. But now I want to rename the first testfile.txt to testfile.txt.0 so it's not overwritten. And so on - after a while I'd have a directory of files with the name testfile.txt.N and the most recent testfile.txt.
What you can do is you can append date and time on the file name that gives every filename a unique name and also helps you archive it.
For example you text.txt can be either 20170202_181921_test.txt or test_20170202_181921.txt
You can use a JavaScript Date Object to get date and time.
P.S show your code of downloading files so that I can add more to that.
If I have file being copied using the following
SetOutPath "$FOO_DIR"
File "..\..\Bar.Dat"
...
SetOutPath "$OTHER_FOO_DIR"
File "..\..\Bar.Dat"
Note that the file Bar.Dat is supposed to be copied to other locations later on to during the installation. How do I rename it for this specific copy operation? If I do a Rename on it the later operations will not find it. I have been looking for an option to pass a destination file name to the File operation, but can't find one.
Actually tried the above but the nsis compiler complains.
a little bit of rearranging would make it work:
File /oname="DestinationNameOfFile.Dat" "..\..\Bar.Dat"
*the source filename is the 2nd argument for /oname
Of course I managed to find the answer straight after posting this question. *smack*
File does in fact take a flag of the destination name.
File "..\..\Bar.Dat" /oname="DestinationNameOfFile.Dat"