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.
Related
I am trying to rename a file from multiple directories
I have a file called slave.log in multiple directories like slave1,slave2......slave17. so daily log rotation happens and creates a new file with current dateformat in it whereas the file data contains a previous day data . I want to rename those files with previous date format .
I have written shell script which works fine but the problem here is , I need to pass the path as parameter but likewise I have 17 directories so i cant schedule 17 cron enteries to run ...I have only basic knowledge about scripting . Please help me with best solution for this scenario
I want to use fetch to gather a line of info from multiple nodes, and store them in the same txt file.
right now I have:
fetch:
src: /path/to/file.txt
dest: /ansible/path/to/file.txt
flat: yes
Instead of adding info to the existing txt file, it overrides the file and deletes the old info.
According to the official documentation of fetch module
Files that already exist at dest will be overwritten if they are different than the src.
https://docs.ansible.com/ansible/latest/collections/ansible/builtin/fetch_module.html
You maybe could use lineinfile or blockinfile module.
Fetch all the files that you want to append while renaming them with some combination of ansible_hostname in the name string. All files need to be added in the same folder, the destination name used will make the difference since you get the same file name from all sources it might be ending up overwriting the name. Use a changing variable like ansible_hostname or some sort of node identifier like IP address. Use this variable in building the file name for your fetched file
get a list of all the fetched files in a variable
Iterate thru that variable and then try lookup for each file
block={{lookup('file', 'sourceFile')}}
You can also iterate over all files in a folder, while appending to the end of the destination file. In your case, I believe blockinfile will be appropriate for this operation.
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)
I have been using Node.JS file system module for performing various file related operations. I have a need to verify the file name if exists in a directory and if exists i would need to keep a suffix at the end of the file. Typically how windows does with duplicate file names..
if TestFile.txt already exists and another file with same names comes in during processing the new file should be renamed as TestFile (1).txt and next file with same name should be renamed as TestFile (2).txt.
What could be the best way to achieve this. Do i have to use a temporary array to keep all file names and traverse through for each? This is a multi threaded environment and there could be 50,000+ documents coming for processing.
Thanks a ton.
I want to implement a file system using FUSE. When the contents of the directory is requested, only the types of files in that directory will be reported as subdirectories. For example, if there are ugur.PDF, guler.JPG and devatate.PNG files in the directory, the files themselves will not be reported, but types (PDF, JPG and PNG) will be reported instead. I tried to implement this file system. My problem is i can not imagine that how can i report without changing ls-l command? how does ls-l command work? (I have a readdir function and it loads file_type and file_name to buffer. I tried to change but i couldn't achive)
How does ls -l work? That seems to be the crux of the issue. It calls fstat(2). That system call fills the stat structure. When you implement a fuse provider, you are providing the data that ends up in the stat structure. If you want to change the directory structure, you return the necessary fabricated information for the various items.
I can think of two approaches to this:
1. Use a database like SQLite
you can store the files, path and file types in the database. Then when the user gets into some directory, you can say do some query like select file_types where path="" and populate as directories using filler()
2. Recursively traverse original path
you can create a list of all file types in the current directory, then use filler() to post them as directories. Then, when user enters any one directory, you can again do a check or something to see if cur_path is in last_path(orig directory), and you can select those file types and display them