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
Related
I am learning bash since 6 days I think I got some of the basics.
Anyway, for the wallpapers downloaded from Variety I've written two scripts. One of them moves downloaded photos older than 12 days to a folder and renames them all as "Aday 1,2,3..." and the other lets me select these and moves them to another folder and removes photos I didn't select. 1st script works just as I intended, my question is about the other
I think I should write the script down to better explain my problem
Script:
#!/bin/bash
#Move victors of 'Seçme-Eleme' to 'Kazananlar'
cd /home/eurydice/Bulunur\ Bir\ Şeyler/Dosyamsılar/Seçme-Eleme
echo "Select victors"
read vct
for i in $vct; do
mv -i "Aday $i.png" /home/eurydice/"Bulunur Bir Şeyler"/Dosyamsılar/Kazananlar/"Bahar $RANDOM.png" ;
mv -i "Aday $i.jpg" /home/eurydice/"Bulunur Bir Şeyler"/Dosyamsılar/Kazananlar/"Bahar $RANDOM.jpg" ;
done
#Now let's remove the rest
rm /home/eurydice/Bulunur\ Bir\ Şeyler/Dosyamsılar/Seçme-Eleme/*
In this script I originally intended to define another variable (let's call this "n") and so did I with copying and changing the variable from the first script. It was something like that
for i in $vct; do
n=1
mv "Aday $i.png" /home/eurydice/"Bulunur Bir Şeyler"/Dosyamsılar/Kazananlar/"Bahar $n.png" ;
mv "Aday $i.jpg" /home/eurydice/"Bulunur Bir Şeyler"/Dosyamsılar/Kazananlar/"Bahar $n.jpg" ;
n=$((n+1))
done
When I do that for the first time the script worked just as I intended. However, in my 2nd test run this script overwrote the files that already existed. I mean, for example in 1st run i had 5 files whose names are "Bahar 1,2,3,4,5" and the 2nd time I chose 3 files to add. I wanted their names to be "Bahar 6,7,8" but instead, my script made them the new 1,2 and 3. I tried many solutions and when I couldn't fix that I just assigned random numbers to them.
Is there a way to make this script work as I intended?
This command finds the biggest file name number amongst files in current directory. If no file is found, biggest number is assigned to 0.
biggest_number=$(ls -1 | sed -n 's/^[^0-9]*\([0-9]\+\)\(\.[a-zA-Z]\+\)\?$/\1/p' | sort -r -g | head -n 1)
[[ ! -z "$biggest_number" ]] || biggest_number=0
The regex in sed command assumes that there is no digit in filenames before the trailing number intended for increment.
As soon as you have found the biggest number, you can use it to start your loop to prevent overwrites.
n=$((biggest_number+1))
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
Assume you have a file called “heading” as follows
echo "Permissions^V<TAB>^V<TAB>Size^V<TAB>^V<TAB>File Name" > heading
echo "-------------------------------------------------------" >> heading
Write a (single) set of commands that will create a report as follows:
make a list of the names, permissions and size of all the files in your current directory,
matching (roughly) the format of the heading you just created,
put the list of files directly following the heading, and
save it all into a file called “file.list”.
All this is to be done without destroying the heading file.
I need to be able to do this all in a pipleline without altering the file. I can't seem to do this without destroying the file. Can somebody please make a pipe for me?
You can use command group:
{ cat heading; ls -l | sed 's/:/^V<tab>^V<tab>/g'; } > file.list
I working with linux, bash.
I have one directory with 100 folders in it, each one named different.
In each of these 100 folders, there is a file called first.bars (so I have 100 files named first.bars). Although all named first.bars, the files are actually slightly different.
I want to get all these files moved to one new folder and rename/number these files so that I know which file comes from which folder. So the first first.bars file must be renamed to 001.bars, the second to 002.bars.. etc.
I have tried the following:
ls -d * >> /home/directorywiththe100folders/list.txt
cat list.txt | while read line;
do cd $line;
mv first.bars /home/newfolder
This does not work because I can't have 100 files, named the same, in one folder. So I only need to know how to rename them. The renaming must be connected to the cat list.txt, because the first line is the folder containing the first file wich is moved and renamed. That file will be called 001.bars.
Try doing this :
$ rename 's/^.*?\./sprintf("%03d.", $c++)/e' *.bar
If you want more information about this command, see this recent response I gave earlier : How do I rename multiple files beginning with a Unix timestamp - imapsync issue
If the rename command is not available,
for d in /home/directorywiththe100folders/*/; do
newfile=$(printf "/home/newfolder/%d.bars" $(( c++ )) )
mv "$d/first.bars" "$newfile"
done
Lets say i have different files in a folder that contains the same day data such as :
ThisFile_2012-10-01.txt
ThatFile_2012-10-01.txt
AnotherSilly_2012-10-01.txt
InnovativeFilesEH_2012-10-01.txt
How to i append them to each other in any preferred order? Would below be the exact way i need to type in my shellscript? The folder gets same files everyday but with different dates. Old dates disappear so every day there are these 4 files.
InnovativeFilesEH_*.txt >> ThatFile_*.txt
ThisFile_*.txt >> ThatFile_*.txt
AnotherSilly_*.txt >> ThatFile_*.txt
Finally, a use for "cat" as intended :-):
cat InnovativeFilesEH_*.txt ThisFile_*.txt AnotherSilly_*.txt >> ThatFile_*.txt
Assumption:
Want to preserve some specific ordering in which these files are appended.
Using the example you provided:
#!/bin/sh
# First find the actual files we want to operate on
# and save them into shell variables:
final_output_file="Desired_File_Name.txt"
that_file=$(find -name ThatFile_*.txt)
inno_file=$(find -name InnovativeFilesEH_*.txt)
this_file=$(find -name ThisFile_*.txt)
another_silly_file=$(find -name AnotherSilly_*.txt)
# Now append the 4 files to Desired_File_Name.txt in the specific order:
cat $that_file > $final_output_file
cat $inno_file >> $final_output_file
cat $this_file >> $final_output_file
cat $another_silly_file >> $final_output_file
Adjust the ordering in which you want the files to be appended by reordering / modifying the cat statements