My problem is that I have to generate a zip file using the linux zip console command. My command is as follows:
zip -r /folder1/folder2/EXP_45.zip /folder1/folder2/EXP_45/
That returns a correct zip only that includes the root folders I want:
Returns
EXP_45.zip
-folder1
--folder2
---EXP_45
...
I want
EXP_45.zip
-EXP_45
...
EXP_45 is a folder that can contain files and folders and they must be present in the zip. I just want the tree structure to start with the EXP_45 folder.
Is there any solution?
The reason why I need it to be a single command is that it is an action of a job in a PL SQL function like:
BEGIN
DBMS_SCHEDULER.CREATE_JOB (
JOB_NAME=>'compress_files', --- job name
JOB_ACTION=>'/usr/bin/zip', --- executable file with path
JOB_TYPE=>'executable', ----- job type
NUMBER_OF_ARGUMENTS=>4, -- parameters in numbers
AUTO_DROP =>false,
CREDENTIAL_NAME=>'credentials' -- give credentials name which you have created before "credintial"
);
dbms_scheduler.set_job_argument_value('compress_files',1,'-r');
dbms_scheduler.set_job_argument_value('compress_files',2,'-m');
dbms_scheduler.set_job_argument_value('compress_files',3,'/folder1/folder2/EXP_45.zip');
dbms_scheduler.set_job_argument_value('compress_files',4,'/folder1/folder2/EXP_45/');
DBMS_SCHEDULER.RUN_JOB('compress_files');
END;
I haven't been able to find a solution to this problem using zip but I have found it using jar. The command would be:
jar cMf /folder1/folder2/EXP_45.zip -C /folder1/folder2/EXP_45 .
Also, the solution using a job in pl sql in case it works for someone would be:
BEGIN
DBMS_SCHEDULER.CREATE_JOB (
JOB_NAME=>'compress_files', --- job name
JOB_ACTION=>'/usr/bin/jar', --- executable file with path
JOB_TYPE=>'executable', ----- job type
NUMBER_OF_ARGUMENTS=>5, -- parameters in numbers
AUTO_DROP =>false,
CREDENTIAL_NAME=>'credentials' -- give credentials name which you have created before "credintial"
);
dbms_scheduler.set_job_argument_value('compress_files',1,'cMf');
dbms_scheduler.set_job_argument_value('compress_files',2,'/folder1/folder2/EXP_45.zip');
dbms_scheduler.set_job_argument_value('compress_files',3,'-C');
dbms_scheduler.set_job_argument_value('compress_files',4,'/folder1/folder2/EXP_45');
dbms_scheduler.set_job_argument_value('compress_files',5,'.');
DBMS_SCHEDULER.RUN_JOB('compress_files');
END;
You want to use the -j (or --junk-paths) option when you are creating the zip file. Below is from the zip man page.
-j
--junk-paths
Store just the name of a saved file (junk the path), and do not store directory names.
By default, zip will store the full path (relative to the current directory).
Update following Question Clarification
Why not put the equivalent to the code below in a shell script & get the SQL function to invoke that? You just need to pass the directory name to cd into and the name of the output zip.
cd folder1/folder2
zip -r /tmp/EXP_45.zip EXP_45
Related
root#VM-0-11-debian:~/linux/2023/01# ls
root#VM-0-11-debian:~/linux/2023/01# ls > list
root#VM-0-11-debian:~/linux/2023/01# ls
list
root#VM-0-11-debian:~/linux/2023/01# cat list
list
I know that > will redirect stdout to file. it will create the file if not present, otherwise replace it.
I would like to ask that is the shell command ls > list implementation process as I described below?
1)As the file named list not exists, so create a file named list first.
2)ls command will list the directory content(list). the content listed(list) will be in the standard output.
3)Add the content of the standard output(list) to the file named list in a replaced way.
My personal understanding of the implementation process as described above, I hope you can give me some guidance. Thank you.
The file redirection operator > is handled by your shell and any file to which you write will be created/truncated before the binary is started. That's why you can see the file name list in the content of the file: the file has already been created before the ls process was started.
So yes, your understanding is correct.
This is why it is not possible to do something like sort txt > txt – the file txt will be truncated before sort reads it. You will end up with an empty file.
Upon looping a directory to delete txt files ONLY - a message is returned indicating The System cannot find the file specified: 'File.txt'.
I've made sure the txt files that I'm attempting to delete exist in the directory I'm looping. I've also checked my code and to make sure it can see my files by printing them in a list with the print command.
import os
fileLoc = 'c:\\temp\\files'
for files in os.listdir(fileLoc):
if files.endswith('.txt'):
os.unlink(files)
Upon initial execution, I expected to see all txt files deleted except for other non-txt files. The actual result was an error message "FileNotFoundError: [WinError 2] The system cannot find the file specified: 'File.txt'.
Not sure what I'm doing wrong, any help would be appreciated.
It isn't found because the the path you intended to unlink is relative to fileLoc. In fact with your code, the effect is to unlink the file relative to the current working directory. If there were *.txt files
in the cwd then the code would have unfortunate side-effects.
Another way to look at it:
Essentially, by analogy, in the shell what you're trying to do is equivalent to this:
# first the setup
$ mkdir foo
$ touch foo/a.txt
# now your code is equvalent to:
$ rm *.txt
# won't work as intended because it removes the *.txt files in the
# current directory. In fact the bug is also that your code would unlink
# any *.txt files in the current working directory unintentionally.
# what you intended was:
$ rm foo/*.txt
The missing piece was the path to the file in question.
I'll add some editorial: The Old Bard taught us to "when in doubt, print variables". In other words, debug it. I don't see from the OP an attempt to do that. Just a thing to keep in mind.
Anyway the new code:
Revised:
import os
fileLoc = 'c:\\temp\\files'
for file in os.listdir(fileLoc):
if file.endswith('.txt'):
os.unlink(os.path.join(fileLoc,file))
The fix: os.path.join() builds a path for you from parts. One part is the directory (path) where the file exists, aka: fileLoc. The other part is the filename, aka file.
os.path.join() makes a whole valid path from them using whatever OS directory separator is appropriate for your platform.
Also, might want to glance through:
https://docs.python.org/2/library/os.path.html
I have the following preexisting folder in my machine
D:\scripts\myfolder
I want my script to create a folder named logs and create a file log.txt in it. So the path would look like
D:\scripts\myfolder\logs\somelog.txt
So I used
p = pathlib.Path("D:\scripts\myfolder\logs\somelog.txt")
p.mkdir(parents=True, exisit_ok=True)
Now
print(p.parents[0]) ==> D:\scripts\myfolder\logs
print(p.parents[1]) ==> D:\scripts\myfolder
print(p.parents[2]) ==> D:\scripts
So, as per Path.mkdir documentation
p.mkdir(parents=True, exisit_ok=True) should create the folders logs, myfolder or scripts and so on if they don't exist.
But it creates a folder by the name some.txt inside logs folder, although it is none of the parents. Why is that so?
I understand that the workaround is to use pathlib.Path("D:\scripts\myfolder\logs")
The entire point of mkdir is to create the directory pointed to by its argument. Passing in parents=True creates the parent folders in addition.
Create a new directory at this given path. [...] If parents is true, any missing parents of this path are created as needed; [1]
If you want to ensure the containing directory exists, create the parent of your path:
p = pathlib.Path("D:\scripts\myfolder\logs\somelog.txt")
p.parent.mkdir(parents=True, exist_ok=True)
That's the way Pathlib.mkdir works. It can't tell if the final component should be a file or a directory. parents=True means to create also parents, not only parents. If the final path component is always a file, you could avoid it like
p.parents[0].mkdir(parents=True)
I am trying to zip a file using shell script command. I am using following command:
zip ./test/step1.zip $FILES
where $FILES contain all the input files. But I am getting a warning as follows
zip warning: name not matched: myfile.dat
and one more thing I observed that the file which is at last in the list of files in a folder has the above warning and that file is not getting zipped.
Can anyone explain me why this is happening? I am new to shell script world.
zip warning: name not matched: myfile.dat
This means the file myfile.dat does not exist.
You will get the same error if the file is a symlink pointing to a non-existent file.
As you say, whatever is the last file at the of $FILES, it will not be added to the zip along with the warning. So I think something's wrong with the way you create $FILES. Chances are there is a newline, carriage return, space, tab, or other invisible character at the end of the last filename, resulting in something that doesn't exist. Try this for example:
for f in $FILES; do echo :$f:; done
I bet the last line will be incorrect, for example:
:myfile.dat :
...or something like that instead of :myfile.dat: with no characters before the last :
UPDATE
If you say the script started working after running dos2unix on it, that confirms what everybody suspected already, that somehow there was a carriage-return at the end of your $FILES list.
od -c shows the \r carriage-return. Try echo $FILES | od -c
Another possible cause that can generate a zip warning: name not matched: error is having any of zip's environment variables set incorrectly.
From the man page:
ENVIRONMENT
The following environment variables are read and used by zip as described.
ZIPOPT
contains default options that will be used when running zip. The contents of this environment variable will get added to the command line just after the zip command.
ZIP
[Not on RISC OS and VMS] see ZIPOPT
Zip$Options
[RISC OS] see ZIPOPT
Zip$Exts
[RISC OS] contains extensions separated by a : that will cause native filenames with one of the specified extensions to be added to the zip file with basename and extension swapped.
ZIP_OPTS
[VMS] see ZIPOPT
In my case, I was using zip in a script and had the binary location in an environment variable ZIP so that we could change to a different zip binary easily without making tonnes of changes in the script.
Example:
ZIP=/usr/bin/zip
...
${ZIP} -r folder.zip folder
This is then processed as:
/usr/bin/zip /usr/bin/zip -r folder.zip folder
And generates the errors:
zip warning: name not matched: folder.zip
zip I/O error: Operation not permitted
zip error: Could not create output file (/usr/bin/zip.zip)
The first because it's now trying to add folder.zip to the archive instead of using it as the archive. The second and third because it's trying to use the file /usr/bin/zip.zip as the archive which is (fortunately) not writable by a normal user.
Note: This is a really old question, but I didn't find this answer anywhere, so I'm posting it to help future searchers (my future self included).
eebbesen hit the nail in his comment for my case (but i cannot vote for comment).
Another possible reason missed in the other comments is file exceeding the file size limit (4GB).
I converted my script for unix environment using dos2unix command and executed my script as ./myscript.sh instead bash myscript.sh.
I just discovered another potential cause for this. If the permissions of the directory/subdirectory don't allow the zip to find the file, it will report this error. Actually, if you run a chmod -R 444 on the directory, and then try to zip it, you will reproduce this error, and also have a "stored 0%" report, like this:
zip warning: name not matched: borrar/enviar
adding: borrar/ (stored 0%)
Hence, try changing the permissions of the file. If you are trying to send them through email, and those email filters (like Gmail's) invent silly filters of not sending executables, don't forget that making permissions very strict when making zip compression can be the cause of the error you are reporting, of "name not matched".
spaces are not allowed:
it would fail if there are more than one files(s) in $FILES unless you put them in loop
I also encountered this issue. In my case, the line separate is CRLF in my zip shell script which causes the problem. Using LF fixed it.
I'm trying to create a cron job to create mysql backups. I would like to be able to first check how many files are there in the directory and if there are 5 or more remove one (the oldest) and create a new mysqldump. I know how to create the mysqldump, but not sure about the condition. I'm planning to store the procedure in the .sh file and trigger that file once a day with the cronjob.
Could someone show the example of what the procedure should look like?
You could wrap the following in a shell script
for file_to_delete in `ls -1ta test* | tail --lines=+6` ; do echo "ENTER_CMD_HERE $file_to_delete" ; done
where
- test is name of file (replace it with path + name of the mysql bkp files)
- replace ENTER_CMD_HERE with say rm