I have to read a file in my shell script. I was using PL/SQL's UTL_FILE to open the file.
But I have to do a new change which will append timestamp to the file.
e.g import.data file becomes import_20152005101200.data
Now timestamp is the time at which file arrive at the server.
Since the file name changed I can't use the old way of file accessing.
I came up with below solution:
UTL_FILE.FOPEN ('path','import_${file_date}.data','r');
To achieve this I have to get filename and trim it using SUBSTR to get timestamp and pass to file_date variable.
However I am not able to find how to access filename in a particular path. I can use basename. But My file name keeps changing because of timestamp.
Any help/ alternate ideas are welcome.
PL/SQL isn't a good tool to solve this problem; UTL_FILE doesn't have any tools to list all the files in a folder.
A better solution is to define a stored procedure which uses UTL_FILE and pass the file name to process as an argument to the procedure. That way, you use the shell (which has many powerful commands and tools to examine folders and files) or a script language like Python to determine which file to process.
Related
I need to store data in a file in this format
word, audio, jpeg
How would I store that all in one file? Is it even possible do would I need to store links to other data files in place of the audio and jpeg. Would I need a custom file format?
1. Your own filetype
As mentioned by #Ken White you would need to be creating your own custom file format for this sort of thing, which would then mean creating your own parser type. This could be achieved in almost any language you wanted but since you are planning on using word format, then maybe C# would be best for you. However, this technique could be quite complicated and take a relatively large amount of time to thoroughly test your file compresser / decompressor, but may be best depending on your needs.
2. Command line utilities
Another way to go about this would be to use a bash script to combine all of the files into one file, and then decompress it at the other end. For example the steps could involve:
Combine files using windows copy / linux cat command on command line
Create a metdata file of your own that says how many files are in this custom file, and how much memory each one takes up (could be a short XML or JSON file for example...)
Use the linux split command or install a Windows command line file splitter program (here's just one example) to split the file back into whatever components have made it up.
This way you only have to create a really small file type, and let the OS utilities handle the combining of them for you.
Example on Windows:
Copy all of the files in your current directory into one output file called 'file.custom'
copy /b * file.custom
Generate custom file format describing metadata (i.e. get the file size on disk in C# example here). This is just maybe what I would do in JSON. SO formatting was being annoying so here's a link (Copy paste it into an editor or online JSON viewer).
Use a decompress windows / linux command line tool to decompress each files to the exact length (and export it back to the exact name) specified in the JSON (metadata) file. (More info on splitting files on this post).
3. ZIP files
You could always store all of the files in a compressed zip file, and then just use a zip compressor, expander as and when you like to retreive any number of file formats stored within.
I found a couple of examples of :
Combining multiple files into one ZIP file in only C# .net,
Unzipping ZIP files in C#
Zipping & Unzipping with only windows built-in utilities
Zipping & Unzipping in Linux command line
Good Zipping/Unzipping library in Java
Zipping/Unzipping in Python
I am currently facing an issue which I don't know how to fix. I got the following Julia code:
while true
print(watch_file("test"))
end
So this should get me all the file changes in the directory named "test". At least on windows.
Now thats all well and good, and it kinda works, at least for creating a file or moving a file to that directory. This is an example of what I get:
("New Textfile.txt",Base.FileEvent(true,false,false))
But when I delete or rename that file, I don't get the filename of the file deleted or renamed.
("",Base.FileEvent(true,false,false))
Is there a different method/function I can get the filename with, even when the file is deleted or renamed? Or even better, a way that archives this and is cross-platform-compatible? Any help appreciated.
EDIT: If you could give me an alternative that supports recursive monitoring, that would be even better.
In Linux, Julia 0.4.5 and 0.4.3 watch_file returns file name always. It is a very platform-dependent feature (like in Node.js https://nodejs.org/api/fs.html#fs_caveats) and only manual polling can be truly platform-independent solution.
Is it possible to open a flatfile when only part of the file name is known?
I have files in a directory that have a timestamp appended to the filename, is it possible to open it by specifiying the known part of the filename (excluding timestamp)?
Is it possible with a PLSQL only approach?
There is a dbms_ package which allows you to get a directory listing for the directory (or you can implement your own in a java stored procedure - google!) This will allow you to find the file you are looking for - if necessary choose which is the relevant file and then process.
See http://notdennis.wordpress.com/2013/07/03/listing-directory-files-plsql/
Description:
I am searching a very large server for files that is on a different server. right now I open command prompt and type
DIR [FILE NAME] /S/4
This returns the server location of the file with some other stuff that is not really needed.
Question:
I have a lot of files to search and one by one input into the above command could take forever. Is there a way I could input all of the names of all the files and only search once and the search results would only need to show file name and location?
First, I hope you don't mean DOS, but rather Windows cmd or batch.
You can certainly write a script that will run your DIR command once per file being sought.
But what you most likely want instead is to search once and print the path of each file found. For this you can use PowerShell's FindChildItem or the improved one posted here: http://windows-powershell-scripts.blogspot.in/2009/08/unix-linux-find-equivalent-in.html
It will be something like:
Find-ChildItem -Name "firstfile.txt|secondfile.txt|..."
Another approach is to install msys or cygwin or another Linux tools environment for Windows and use the Linux find command.
I wonder if anyone knows how to write a batch script to edit some text in a .cs file.
What I want to do is change "AssemblyVersion("1.0.0.0")" "AssemblyVersion("1.0.0.x")" where x++ for every time the job in jenkins is being built.
Best Regards Jan
Do you want to use only a batch script for this? You could also use Execute Groovy Script option and write some simple groovy script to achieve this
file = new File("folder/path/myfile.cs")
fileText = file.text;
fileText = fileText.replaceAll(srcExp, replaceText);
file.write(fileText);
You can also use the availabe environment variables from your jenkins job to construct your replace text. These variables will be present at /env-vars.html
Stay away from "batch-file automation" - will only cause you grief.
(for a starter, different versions of Windows support a different set of batch-commands)
You should incorporate the build-number in the script as an Environment Variable -
use either the "built-in" %BUILD_NUMBER% parameter or set your own format with
the Formatted Version-Number Plugin .
If you do need to edit that 'CS' file, I suggest using either Perl or PowerShell.
Cheers