I have a programme that is generating files like this "Incoming11781Arp", and there is always Incoming, and there is always 5 numbers, but there are 3 letters/upper-case/lower-case/numbers/special case _ in any way. Like Incoming11781_pi, or Incoming11781rKD.
How can I delete them using a script run from a cron job please? I've tried -
#!/bin/bash
file=~/Mail/Incoming******
rm "$file";
but it failed saying that there was no matching file or directory.
You mustn't double-quote the variable reference for pathname expansion to occur - if you do, the wildcard characters are treated as literals.
Thus:
rm $file
Caveat: ~/Mail/Incoming****** doesn't work the way you think it does and will potentially match more files than intended, as it is equivalent to ~/Mail/Incoming*, meaning that any file that starts with Incoming will match.
To only match files starting with Incoming that are followed by exactly 6 characters, use ~/Mail/Incoming??????, as #Jidder suggests in a comment.
Note that you could make your glob (pattern) even more specific:
file=~/Mail/Incoming[0-9][0-9][0-9][0-9][0-9][[:alpha:]_][[:alpha:]_][[:alpha:]_]
See the bash manual for a description of pathname expansion and pattern syntax: http://www.gnu.org/software/bash/manual/bashref.html#index-pathname-expansion.
You can achieve the same effect with the find command...
$ directory='~/Mail/'
$ file_pattern='Incoming*'
$ find "${directory}" -name "${file_pattern}" -delete
The first two lines define the directory and the file pattern separately, the find command will then proceed to delete any matching files inside that directory.
Related
I'm using a naming convention with number prefixes to track some files. But I am running out with 2-digit prefix. So, instead of 11.abc 12.def I want to move to 011.abc 012.def. I already have some 013.xxx 014.yyy.
Trying this in an empty directory:
touch 11.abc 12.def 013.xxx 014.yyy
ls -1 gives:
013.xxx
014.yyy
11.abc
12.def
Try #1:
This should match anything that starts with 2 digits, but not 3.
rename -n 's/^\d\d[^\d]/0$1/' *
Now I was kind of hoping that $1 would hold the match, like 11, with 0$1 giving me 011.
No such luck:
Use of uninitialized value $1 in concatenation (.) or string at (eval 2) line 1.
'11.abc' would be renamed to '0abc'
Use of uninitialized value $1 in concatenation (.) or string at (eval 2) line 1.
'12.def' would be renamed to '0def'
On the positive side, it's willing to leave 013 and 014 alone.
Try #2 rename -n 's/^\d\d[^\d]/0/' *
'11.abc' would be renamed to '0abc'
'12.def' would be renamed to '0def'
Since this is regex based, can I somehow save the match group 11 and 12?
If I can't use rename I'll probably write a quick Python script. Don't want to loop with mv on it.
And, actually, my naming covention is 2-3 digits followed by a dot, so this is a good match too.
rename -n 's/^\d\d\./<whatever needs to go here>/' *
For what it's worth, I am using the Homebrew version of rename, as I am on a mac.
try this:
rename 's/^(\d{2}\..*)/0$1/' *
rename is problematic because it's not part of POSIX (so it isn't normally available on many Unix-like systems), and there are two very different forms of it in widespread use. See Why is the rename utility on Debian/Ubuntu different than the one on other distributions, like CentOS? for more information.
This Bash code does the renaming with mv (which is part of POSIX):
#! /bin/bash -p
shopt -s nullglob # Patterns that match nothing expand to nothing.
for f in [0-9][0-9].* ; do
mv "$f" "0$f"
done
shopt -s nullglob is to prevent problems if the code is run in a directory that has no files that need to be renamed. If nullglob isn't enabled the code would try to rename a file called '[0-9][0-9].*', which would have unwanted consequences whether or not such a file existed.
I have a Bash script that takes in a directory as a parameter, and after some processing will do some output based on the files in that directory.
The command would be like the following, where dir is a directory with the following structure inside
dir/foo
dir/bob
dir/haha
dir/bar
dir/sub-dir
dir/sub-dir/joe
> myscript ~/files/stuff/dir
After some processing, I'd like the output to be something like this
foo
bar
sub-dir/joe
The code I have to remove the path passed in is the following:
shopt -s extglob
for file in $files ; do
filename=${file#${1}?(/)}
This gets me to the following, but for some reason the optional / is not being taken care of. Thus, my output looks like this:
/foo
/bar
/sub-dir/joe
The reason I'm making it optional is because if the user runs the command
> myscript ~/files/stuff/dir/
I want it to still work. And, as it stands, if I run that command with the trailing slash, it outputs as desired.
So, why does my ?(/) not work? Based on everything I've read, that should be the right syntax, and I've tried a few other variations as well, all to no avail.
Thanks.
that other guy's helpful answer solves your immediate problem, but there are two things worth nothing:
enumerating filenames with an unquoted string variable (for file in $files) is ill-advised, as sjsam's helpful answer points out: it will break with filenames with embedded spaces and filenames that look like globs; as stated, storing filenames in an array is the robust choice.
there is no strict need to change global shell option shopt -s extglob: parameter expansions can be nested, so the following would work without changing shell options:
# Sample values:
file='dir/sub-dir/joe'
set -- 'dir/' # set $1; value 'dir' would have the same effect.
filename=${file#${1%/}} # -> '/sub-dir/joe'
The inner parameter expansion, ${1%/}, removes a trailing (%) / from $1, if any.
I suggested you change files to an array which is a possible workaround for non-standard filenames that may contain spaces.
files=("dir/A/B" "dir/B" "dir/C")
for filename in "${files[#]}"
do
echo ${filename##dir/} #replace dir/ with your param.
done
Output
A/B
B
C
Here's the documentation from man bash under "Parameter Expansion":
${parameter#word}
${parameter##word}
Remove matching prefix pattern. The word is
expanded to produce a pattern just as in pathname
expansion. If the pattern matches the beginning of
the value of parameter, then the result of the
expansion is the expanded value of parameter with
the shortest matching pattern (the ``#'' case) or
the longest matching pattern (the ``##'' case)
deleted.
Since # tries to delete the shortest match, it will never include any trailing optional parts.
You can just use ## instead:
filename=${file##${1}?(/)}
Depending on what your script does and how it works, you can also just rewrite it to cd to the directory to always work with paths relative to .
I am completely new to bash script. I am trying to do something really basic before using it for my actual requirement. I have written a simple code, which should print test code as many times as the number of files in the folder.
My code:
for variable in `ls test_folder`; do
echo test code
done
"test_folder" is a folder which exist in the same directory where the bash.sh file lies.
PROBLEM: If the number of files are one then, it prints single time but if the number of files are more than 1 then, it prints a different count. For example, if there are 2 files in "test_folder" then, test code gets printed 3 times.
Just use a shell pattern (aka glob):
for variable in test_folder/*; do
# ...
done
You will have to adjust your code to compensate for the fact that variable will contain something like test_folder/foo.txt instead of just foo.txt. Luckily, that's fairly easy; one approach is to start the loop body with
variable=${variable#test_folder/}
to strip the leading directory introduced by the glob.
Never loop over the output of ls! Because of word splitting files having spaces in their names will be a problem. Sure, you could set IFS to $\n, but files in UNIX can also have newlines in their names.
Use find instead:
find test_folder -maxdepth 1 -mindepth 1 -exec echo test \;
This should work:
cd "test_folder"
for variable in *; do
#your code here
done
cd ..
variable will contain only the file names
I am trying to make heads or tails of a shell script. Could someone please explain this line?
$FILEDIR is a directory containing files. F is a marker in an array of files that is returned from this command:
files=$( find $FILEDIR -type f | grep -v .rpmsave\$ | grep -v .swp\$ )
The confusing line is within a for loop.
for f in $files; do
target=${f:${#FILEDIR}}
<<do some more stuff>>
done
I've never seen the colon, and the hash before in a shell script for loop. I haven't been able to find any documentation on them... could someone try and enlighten me? I'd appreciate it.
There are no arrays involved here. POSIX sh doesn't have arrays (assuming you're not using another shell based upon the tags).
The colon indicates a Bash/Ksh substring expansion. These are also not POSIX. The # prefix expands to the number of characters in the parameter. I imagine they intended to chop off the directory part and assign it to target.
To explain the rest of that: first find is run and hilariously piped into two greps which do what could have been done with find alone (except breaking on possible filenames containing newlines), and the output saved into files. This is also something that can't really be done correctly if restricted only to POSIX tools, but there are better ways.
Next, files is expanded unquoted and mutalated by the shell in more ridiculous ways for the for loop to iterate over the meaningless results. If the rest of the script is this bad, probably throw it out and start over. There's no way that will do what's expected.
The colon can be as a substring. So:
A=abcdefg
echo ${A:4}
will print the output:
efg
I'm not sure why they would use a file directory as the 2nd parameter though...
If you are having problems understanding the for loop section, try http://www.dreamsyssoft.com/unix-shell-scripting/loop-tutorial.php
I am trying to iterate through files in the same directory with only one 4 in them.
Here is what I have so far. The problem with my current script is that files with any number of 4's get selected, not files with only one 4.
for i in *4*.cpp;
do
...
Sort of like [!4] but for any number of non 4 characters.
*http://www.tuxfiles.org/linuxhelp/wildcards.html
I want to iterate through file names such as me4.cpp, 4.cpp, and hi4hi.cpp
I want to ignore file names such as lala.cpp, 44.cpp, 4hi4.cpp
Thank you!
Figured it out. I tried [!4]* on a whim.
Oops turned out I didn't. That is interpreted as ([!4]) then (*)
The grep style regex you need is:
^[^4]*4[^4]*$
A bunch of not-4's after the start of the line, a 4, and another bunch of not-4's to the end of the line.
In pure shell, consider using a case statement:
for file in *4*.cpp
do
case "$file" in
(*4*4*) : Ignore;;
(*) : Process;;
esac
done
That looks for names containing 4's, and then ignores those containing 2 or more 4's.
How about using find
find ./ -regex "<regular expression>"
Assuming bash:
shopt -s extglob
for file in *([^4])4*([^4]).cpp; ...
where *([^4]) means zero or more characters that are not "4"