I'm trying to count the total lines in the files within a directory. To do this I am trying to use a combination of find and wc. However, when I run find . -exec wc -l {}\;, I recieve the error find: missing argument to -exec. I can't see any apparent issues, any ideas?
You simply need a space between {} and \;
find . -exec wc -l {} \;
Note that if there are any sub-directories from the current location, wc will generate an error message for each of them that looks something like that:
wc: ./subdir: Is a directory
To avoid that problem, you may want to tell find to restrict the search to files :
find . -type f -exec wc -l {} \;
Another note: good idea using the -exec option . Too many times people pipe commands together thinking to get the same result, for instance here it would be :
find . -type f | xargs wc -l
The problem with piping commands in such a manner is that it breaks if any files has spaces in it. For instance here if a file name was "a b" , wc would receive "a" and then "b" separately and you would obviously get 2 error messages: a: no such file and b: no such file.
Unless you know for a fact that your file names never have any spaces in them (or non-printable characters), if you do need to pipe commands together, you need to tell all the tools you are piping together to use the NULL character (\0) as a separator instead of a space. So the previous command would become:
find . -type f -print0 | xargs -0 wc -l
With version 4.0 or later of bash, you don't need your find command at all:
shopt -s globstar
wc -l **/*
There's no simple way to skip directories, which as pointed out by Gui Rava you might want to do, unless you can differentiate files and directories by name alone. For example, maybe directories never have . in their name, while all the files have at least one extension:
wc -l **/*.*
Related
I would like to go through all the files in the current directory (or sub-directories) and echoes me back the name of files only if they contain certain words.
More detail:
find -type f -name "*hello *" will give me all file names that have "hello" in their names. But instead of that, I want to search through the files and if that file's content contains "hello" then prints out the name of the file.
Is there a way to approach this?
You can use GNU find and GNU grep as
find /path -type f -exec grep -Hi 'hello' {} +
This is efficient in a way that it doesn't invoke as many grep instances to as many files returned from find. This works in an underlying assumption that find returns a set of files for grep to search on. If you are unsure if the files may not be available, as a fool-proof way, you can use xargs with -r flag, in which case the commands following xargs are executed only if the piped commands return any results
find /path -type f | xargs -r0 grep -Hi 'hello'
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 am running -> "find . -name '*.txt'" command and getting list of files.
I am getting below mention output:
./bsd/contrib/amd/ldap-id.txt
./bsd/contrib/expat/tests/benchmark/README.txt
./bsd/contrib/expat/tests/README.txt
./bsd/lib/libc/softfloat/README.txt
and so on,
Out of these files how can i run grep command and read contents and filter only those files which have certain keyword? for e.g. "version" in it.
xargs is a great way to accomplish this, and its already been covered.
The -exec option of find is also useful for this. It will perform a command over all files returned from find.
To invoke grep as few times as possible, passing multiple filenames to each call:
find . -name '*.txt' -exec grep -H 'foo' {} +
Alternately, to invoke grep exactly once for each file found:
find . -name '*.txt' -exec grep -H 'foo' {} ';'
In either case, {} is like a placeholder for the values from find; if your shell is zsh, it may be necessary to escape it, as in '{}'.
There are several ways to accomplish this.
If there are non-.txt files which might usefully contain the keyword:
grep -r KEYWORD *
This uses the recursive directory search option of grep.
To search only .txt files:
find . -name '*.txt' -exec grep KEYWORD {} \;
or
find . -name '*.txt' -exec grep KEYWORD {} +
or
find . -execdir grep KEYWORD {}
The first runs grep for each matching file. The second runs grep much fewer times, accumulating many matched files before invoking grep. The third form runsgrep` once in every directory.
There is usually a function built into find for that, but to be portable across platforms, I typically use xargs. Say you want to find all the xml files in or below the current directly and get a list of each occurrence of 'foo', you can do this:
find ./ -type f -name '*.xml' -print0 | xargs -0 -n 1 grep -H foo
It should be self-explanatory except for the -print0, which separates filenames with NULs rather than newlines, and the -0, which tells xargs to use those NULs rather than interpreting spaces and quotes as syntax (which can confuse it if filenames contain either).
I want to insert a line into the start of multiple specified type files, which the files are located in current directory or the sub dir.
I know that using
find . -name "*.csv"
can help me to list the files I want to use for inserting.
and using
sed -i '1icolumn1,column2,column3' test.csv
can use to insert one line at the start of file,
but now I do NOT know how to pipe the filenames from "find" command to "sed" command.
Could anybody give me any suggestion?
Or is there any better solution to do this?
BTW, is it work to do this in one line command?
Try using xargs to pass output of find and command line arguments to next command, here sed
find . -type f -name '*.csv' -print0 | xargs -0 sed -i '1icolumn1,column2,column3'
Another option would be to use -exec option of find.
find . -type f -name '*.csv' -exec sed -i '1icolumn1,column2,column3' {} \;
Note : It has been observed that xargs is more efficient way and can handle multiple processes using -P option.
This way :
find . -type f -name "*.csv" -exec sed -i '1icolumn1,column2,column3' {} +
-exec do all the magic here. The relevant part of man find :
-exec command ;
Execute command; true if 0 status is returned. All following arguments
to find are taken to be arguments to the command until an argument consisting
of `;' is encountered. The string `{}' is replaced by the current file name
being processed everywhere it occurs in the arguments to the command, not just
in arguments where it is alone, as in some versions of find. Both of
these constructions might need to be escaped (with a `\') or quoted to protect
them from expansion by the shell. See the EXAMPLES section for examples of
the use of the -exec option. The specified command is run once for each
matched file. The command is executed in the starting directory. There
are unavoidable security problems surrounding use of the -exec action;
you should use the -execdir option instead
I have a list of file names as output of certain command.
I need to find each of these files in a given directory.
I tried following command:
ls -R /home/ABC/testDir/ | grep "\.java" | xargs find /home/ABC/someAnotherDir -iname
But it is giving me following error:
find: paths must precede expression: XYZ.java
What would be the right way to do it?
ls -R /home/ABC/testDir/ | grep -F .java |
while read f; do find . -iname "$(basename $f)"; done
You can also use ${f##*/} instead of basename. Or;
find /home/ABC/testDir -iname '*.java*' |
while read f; do find . -iname "${f##*/}"; done
Note that, undoubtedly, many people will object to parsing the output of ls or find without using a null byte as filename separater, claiming that whitespace in filenames will cause problems. Those people usually ignore newlines in filenames, and their objections can be safely ignored. (As long as you don't allow whitespace in your filenames, that is!)
A better option is:
find /home/ABC/testDir -iname '*.java' -exec find . -iname {}
The reason xargs doesn't work is that is that you cannot pass 2 arguments to -iname within find.
find /home/ABC/testDir -name "\.java"