I am making a script to check and list the outputs of certain files in our job folder. I am wanting to check files that have been there today (or in the past 24 hours I guess).
Currently I am doing the following:
find /folder/jobfolder/rrz* -type f -mtime 0
so I get something like this:
/folder/jobfolder/rrzabc_1234.lis
/folder/jobfolder/rrzdef_4567.lis
/folder/jobfolder/rrzgre_8901.log
ideally I would like to use the printf command to add the date to it so I can see if it was today and separate it line by line. If I do
find /folder/jobfolder/rrz* -type f -mtime 0 -printf %p\t%AD\n It does not take use the escape characters.
/folder/jobfolder/rrzabc_1234.lis/folder/jobfolder/rrzdef_4567.lis/folder/jobfolder/rrzgre_8901.log
Is there a better way to approach this? also being able to use -iname to match capital or lowercase letters might be helpful
you are missing quotes; try
-printf '%p\t%AD\n'
so that bash does not interpret the \
Related
I have a directory which contains data for several years and several months.
Filenames have the format yy/mm/dd, f.e.
20150415,
20170831,
20121205
How can I find all data with month = 3?
F.e.
20150302,
20160331,
20190315
Thanks for your help!
ls -ltra ????03??
A question mark is a wildcard which stands for one character, so as your format seems to be YYYYmmDD, the regular expression ????03?? should stand for all files having 03 as mm.
Edit
Apparently the files have format YYYYmmDDxxx, where xxx is the rest of the filename, having an unknown length. This would correspond with regular expression *, so instead of ????03?? you might use ????03??*.
As far as the find is concerned: the same regular expression holds here, but as you seem to be working inside a directory (no subdirectories, at first sight), you might consider the -maxdepth switch):
find . -name "????03??*" | wc -l // including subdirectories
find . -maxdepth 1 -name "????03??*" | wc -l // only current directory
I would highly advise you to check without wc -l first for checking the results. (Oh, I just see the switch -type f, that one might still be useful too :-) )
I see a lot of similar questions, but answers of those questions don't quite meet my needs. The most common answer is to use command like find . -mmin -60 -maxdepth 1. But this command only outputs matched files, the thing I want is to output all files like what ls -al does, and highlight (like different color) all files which are modified in last N minutes. I don't know if it is possible to combine find and grep to achieve this goal, thank you for your help in advance!!
Assuming your version of find supports the printf action and you would like to highlight with, e.g., a leading ***, the following should work:
find . -maxdepth 1 -mmin -60 -printf '*** %p\n' -o -print
If you prefer colors, and if your terminal supports it, you can use ANSI escape codes. Example for green:
find . -maxdepth 1 -mmin -60 -printf '\033[32m%p\033[0m\n' -o -print
Explanations:
If the file or directory was modified less than 60 minutes ago it is the printf action that will we executed, else it is the print action.
By default find tests and actions are combined with logical AND. -o is the logical OR. It has lower precedence than AND. Here find stops as soon as one of two ORed terms evaluates as true.
I have a set of files that are named similarly:
TEXT_TEXT_YYYYMMDD
Example file name:
My_House_20170426
I'm trying to delete all files that don't match this format. Every file should have a string of text followed by an underscore, followed by another string of text and another underscore, then a date stamp of YYYYMMDD.
Can someone provide some advice on how to build a find or a remove statement that will delete files that don't match this format?
Using find, add -delete to the end once you're sure it works.
# gnu find
find . -regextype posix-egrep -type f -not -iregex '.*/[a-z]+_[a-z]+_[0-9]{8}'
# OSX find
find -E . -type f -not -iregex '.*/[a-z]+_[a-z]+_[0-9]{8}'
Intentionally only matching alphabetical characters for TEXT. Add 0-9 to each TEXT area like this [a-z0-9] if you need numbers.
grep -v '(pattern)'
will filter out lines that match a pattern, leaving those that don't match. You might try piping in the output of ls. And if you're particularly brave, you could pipe the output to something like xargs rm. But deleting is kinda scary, so maybe save the output to a file first, look at it, then delete the files listed.
I am figuring out a command to copy files that are modified on a Saturday.
find -type f -printf '%Ta\t%p\n'
This way the line starts with the weekday.
When I combine this with a 'egrep' command using a regular expression (starts with "za") it shows only the files which start with "za".
find -type f -printf '%Ta\t%p\n' | egrep "^(za)"
("za" is a Dutch abbreviation for "zaterdag", which means Saturday,
This works just fine.
Now I want to copy the files with this command:
find -type f -printf '%Ta\t%p\n' -exec cp 'egrep "^(za)" *' /home/richard/test/ \;
Unfortunately it doesn't work.
Any suggestions?
The immediate problem is that -printf and -exec are independent of each other. You want to process the result of -printf to decide whether or not to actually run the -exec part. Also, of course, passing an expression in single quotes simply passes a static string, and does not evaluate the expression in any way.
The immediate fix to the evaluation problem is to use a command substitution instead of single quotes, but the problem that the -printf function's result is not available to the command substitution still remains (and anyway, the command substitution would happen before find runs, not while it runs).
A common workaround would be to pass a shell script snippet to -exec, but that still doesn't expose the -printf function to the -exec part.
find whatever -printf whatever -exec sh -c '
case $something in za*) cp "$1" "$0"; esac' "$DEST_DIR" {} \;
so we have to figure out a different way to pass the $something here.
(The above uses a cheap trick to pass the value of $DEST_DIR into the subshell so we don't have to export it. The first argument to sh -c ... ends up in $0.)
Here is a somewhat roundabout way to accomplish this. We create a format string which can be passed to sh for evaluation. In order to avoid pesky file names, we print the inode numbers of matching files, then pass those to a second instance of find for performing the actual copying.
find \( -false $(find -type f \
-printf 'case %Ta in za*) printf "%%s\\n" "-o -inum %i";; esac\n' |
sh) \) -exec cp -t "$DEST_DIR" \+
Using the inode number means any file name can be processed correctly (including one containing newlines, single or double quotes, etc) but may increase running time significantly, because we need two runs of find. If you have a large directory tree, you will probably want to refactor this for your particular scenario (maybe run only in the current directory, and create a wrapper to run it in every directory you want to examine ... thinking out loud here; not sure it helps actually).
This uses features of GNU find which are not available e.g. in *BSD (including OSX). If you are not on Linux, maybe consider installing the GNU tools.
What you can do is a shell expansion. Something like
cp $(find -type f -printf '%Ta\t%p\n' | egrep "^(za)") $DEST_DIR
Assuming that the result of your find and grep is just the filenames (and full paths, at that), this will copy all the files that match your criteria to whatever you set $DEST_DIR to.
EDIT As mentioned in the comments, this won't work if your filenames contain spaces. If that's the case, you can do something like this:
find -type f -printf '%Ta\t%p\n' | egrep "^(za)" | while read file; do cp "$file" $DEST_DIR; done
I just stumbled upon Perl today while playing around with Bash scripting. When I tried to remove blank spaces in multiple file names, I found this post, which helped me a lot.
After a lot of struggling, I finally understand the rename and substitution commands and their syntax. I wanted to try to replace all "_(x)" at the end of file names with "x", due to duplicate files. But when I try to do it myself, it just does not seem to work. I have three questions with the following code:
Why is nothing executed when I run it?
I used redirection to show me the success note as an error, so I know what happened. What did I do wrong about that?
After a lot of research, I still do not entirely understand file descriptors and redirection in Bash as well as the syntax for the substitute function in Perl. Can somebody give give me a link for a good tutorial?
find -name "*_(*)." -type f | \
rename 's/)././g' && \
find -name "*_(*." -type f | \
rename 's/_(//g' 2>&1
You either need to use xargs or you need to use find's ability to execute commands:
find -name "*_(*)." -type f | xargs rename 's/)././g'
find -name "*_(*." -type f | xargs rename 's/_(//g'
Or:
find -name "*_(*)." -type f -exec rename 's/)././g' {} +
find -name "*_(*." -type f -exec rename 's/_(//g' {} +
In both cases, the file names are added to the command line of rename. As it was, rename would have to read its standard input to discover the file names — and it doesn't.
Does the first find find the files you want? Is the dot at the end of the pattern needed? Do the regexes do what you expect? OK, let's debug some of those too.
You could do it all in one command with a more complex regex:
find . -name "*_(*)" -type f -exec rename 's/_\((\d+)\)$/$1/' {} +
The find pattern is corrected to lose the requirement of a trailing .. If the _(x) is inserted before the extension, then you'd need "*_(*).*" as the pattern for find (and you'll need to revise the Perl regexes).
The Perl substitute needs dissection:
The \( matches an open parenthesis.
The ( starts a capture group.
The \d+ looks for 'one or more digits'.
The ) stops the capture group. It is the first and only, so it is given the number 1.
The \) matches a close parenthesis.
The $ matches the end of the file name.
The $1 in the replacement puts the value of capture group 1 into the replacement text.
In your code, the 2>&1 sent the error messages from the second rename command to standard output instead of standard error. That really doesn't help much here.
You need two separate tutorials; you are not going to find one tutorial that covers I/O redirection in Bash and regular expressions in Perl.
The 'official' Perl regular expression tutorial is:
perlretut, also available as perldoc perlretut on your machine.
The Bash manual covers I/O redirection, but it is somewhat terse:
I/O Redirections.