I'm making a bash script which would call like
script.sh 172.16.1.1
trying to replace . and search files to delete them but it won't happen
echo $1 | find -name '*.`sed 's/\.*//g'`' -printf "%f\n" -delete
files look like
eth0-2:120.1721611 eth1-2:120.1721611
Try this command inside that script.
I think this may help you for your requirement.
$ find -name "*echo "$1" | sed 's/\.*//g'" -printf "%f\n" -delete
I am passing the name only for the particular field, If you passed for whole command it produce the different result.
The given command is searched from current directory to end.
If you need to search from root or home use / or ~ in find command like
$ find ~ -name "*echo "$1" | sed 's/\.*//g'" -printf "%f\n" -delete
Related
I am having a hard time getting find to look for matches in the current directory as well as its subdirectories.
When I run find *test.c it only gives me the matches in the current directory. (does not look in subdirectories)
If I try find . -name *test.c I would expect the same results, but instead it gives me only matches that are in a subdirectory. When there are files that should match in the working directory, it gives me: find: paths must precede expression: mytest.c
What does this error mean, and how can I get the matches from both the current directory and its subdirectories?
Try putting it in quotes -- you're running into the shell's wildcard expansion, so what you're acually passing to find will look like:
find . -name bobtest.c cattest.c snowtest.c
...causing the syntax error. So try this instead:
find . -name '*test.c'
Note the single quotes around your file expression -- these will stop the shell (bash) expanding your wildcards.
What's happening is that the shell is expanding "*test.c" into a list of files. Try escaping the asterisk as:
find . -name \*test.c
From find manual:
NON-BUGS
Operator precedence surprises
The command find . -name afile -o -name bfile -print will never print
afile because this is actually equivalent to find . -name afile -o \(
-name bfile -a -print \). Remember that the precedence of -a is
higher than that of -o and when there is no operator specified
between tests, -a is assumed.
“paths must precede expression” error message
$ find . -name *.c -print
find: paths must precede expression
Usage: find [-H] [-L] [-P] [-Olevel] [-D ... [path...] [expression]
This happens because *.c has been expanded by the shell resulting in
find actually receiving a command line like this:
find . -name frcode.c locate.c word_io.c -print
That command is of course not going to work. Instead of doing things
this way, you should enclose the pattern in quotes or escape the
wildcard:
$ find . -name '*.c' -print
$ find . -name \*.c -print
Try putting it in quotes:
find . -name '*test.c'
I see this question is already answered. I just want to share what worked for me. I was missing a space between ( and -name. So the correct way of chosen a files with excluding some of them would be like below;
find . -name 'my-file-*' -type f -not \( -name 'my-file-1.2.0.jar' -or -name 'my-file.jar' \)
I came across this question when I was trying to find multiple filenames that I could not combine into a regular expression as described in #Chris J's answer, here is what worked for me
find . -name one.pdf -o -name two.txt -o -name anotherone.jpg
-o or -or is logical OR. See Finding Files on Gnu.org for more information.
I was running this on CygWin.
You can try this:
cat $(file $( find . -readable) | grep ASCII | tr ":" " " | awk '{print $1}')
with that, you can find all readable files with ascii and read them with cat
if you want to specify his weight and no-executable:
cat $(file $( find . -readable ! -executable -size 1033c) | grep ASCII | tr ":" " " | awk '{print $1}')
In my case i was missing trailing / in path.
find /var/opt/gitlab/backups/ -name *.tar
I'm Using find command in shell script to search a file from a directory.
FileList=`find /users/tst/CS/FSP/data/out/ -name "file*.gz"`
and I'm getting result as
echo $FileList
/users/tst/CS/FSP/data/out/filename.gz
I want result as "FileList=filetname.gz". I Would like to know how can we get it.
If you are using bash, you can use ## with parameter expansion and so:
FileList=$(while read fil;do echo ${fil##*/};done <<< $(find /users/tst/CS/FSP/data/out/ -name "file*.gz")"
ALternatively, if you have "basename" available, you could use it as follows:
FileList=$(while read fil;do echo "$(basename $fil)";done <<< $(find /users/tst/CS/FSP/data/out/ -name "file*.gz")"
Other possibilities (apart from the solutions already shown in Raman Sailopal's answer)
If you use GNU find as it should be the case on Linux, you can use find's -printf action.
FileList=$(find /users/tst/CS/FSP/data/out/ -name "file*.gz" -printf "%f")
If you only have a POSIX compatible find, you can filter the output.
FileList=$(find /users/tst/CS/FSP/data/out/ -name "file*.gz" | sed 's#.*/##')
The sed command will remove the longest part that ends with a /. (I use # instead of the ususal / to avoid quoting the / in the search pattern.)
FileList=find /users/tst/CS/FSP/data/out/ -name "file*.gz" -type f -name "file*.gz" | cut -d "/" -f 8
So far I have this:
find -name ".txt"
I'm not quite sure how to use wc to find out the exact number of files. When using the command above, all the .txt files show up, but I need the exact number of files with the .txt extension. Please don't suggest using other commands as I'd like to specifically use find and wc. Thanks
Try:
find . -name '*.txt' | wc -l
The -l option to wc tells it to return just the number of lines.
Improvement (requires GNU find)
The above will give the wrong number if any .txt file name contains a newline character. This will work correctly with any file names:
find . -iname '*.txt' -printf '1\n' | wc -l
-printf '1\n tells find to print just the line 1 for each file name found. This avoids problems with file names having difficult characters.
Example
Let's create two .txt files, one with a newline in its name:
$ touch dir1/dir2/a.txt $'dir1/dir2/b\nc.txt'
Now, let's find the find command:
$ find . -name '*.txt'
./dir1/dir2/b?c.txt
./dir1/dir2/a.txt
To count the files:
$ find . -name '*.txt' | wc -l
3
As you can see, the answer is off by one. The improved version, however, works correctly:
$ find . -iname '*.txt' -printf '1\n' | wc -l
2
find -type f -name "*.h" -mtime +10 -print | wc -l
This worked out.
I have several directories with a pattern:
$find -name "*.out"
./trnascanse.out
./darn.out
./blast_rnaz.out
./erpin.out
./rnaspace_cli.out
./yass.out
./atypicalgc.out
./blast.out
./combine.out
./infernal.out
./ecoli.out
./athaliana.out
./yass_carnac.out
./rnammer.out
I can get the list into a file find -name "*.out" > files because I want to create for each directory a file ending with .ref instead of .out : trnascanse.ref, darn.ref, blast_rnaz.refand so on.
I would say that this is possible with some grep and touch but I don't know how to do it. Any idea? Or just create each one manually is the only way (as I did with this directories). Thanks
Here's one way:
for d in *.out ; do echo touch "${d%.out}.ref" ; done
The ${d%.out} expands $d and removes the trailing .out. Read about it in the bash man page.
If the output of above one-liner looks ok, pipe it to sh , or remove the echo and re-run it.
Use this:
find -maxdepth 1 -type d -printf "%f" -exec bash -c "mkdir $(echo '{}' | sed 's/\.out$//').ref" \;
I have some files in my folder /home/sample/* * /*.pdf and *.doc and * .xls etc ('**' means some sub-sub directory.
I need the shell script or linux command to list the files in following manner.
pdf_docs/xx.pdf
documents/xx.doc
excel/xx.xls
pdf_docs, documents and excel are directories, which is located in various depth in /home/sample. like
/home/sample/12091/pdf_docs/xx.pdf
/home/sample/documents/xx.doc
/home/excel/V2hm/1001/excel/xx.xls
You can try this:
for i in {*.pdf,*.doc,*.xls}; do find /home/sample/ -name "$i"; done | awk -F/ '{print $(NF-1) "/" $NF}'
I ve added a line of awk which will print the last 2 fields (seperated by '/' ) of the result alone
Something like this?
for i in {*.pdf,*.doc,*.xls}; do
find /home/sample/ -name "$i";
done | perl -lnwe '/([^\/]+\/[^\/]+)$/&&print $1'
How about this?
find /home/sample -type f -regex '^.*\.\(pdf\|doc\|xls\)$'
Takes into account spaces in file names, potential case of extension
for a in {*.pdf,*.doc,*.xls}; do find /home/sample/ -type f -iname "$a" -exec basename {} \; ; done
EDIT
Edited to take into account only files
You don't need to call out to an external program to chop the pathname like you're looking for:
$ filename=/home/sample/12091/pdf_docs/xx.pdf
$ echo ${filename%/*/*}
/home/sample/12091
$ echo ${filename#${filename%/*/*}?}
pdf_docs/xx.pdf
So,
find /home/sample -name \*.doc -o -name \*.pdf -o -name \*.xls -print0 |
while read -r -d '' pathname; do
echo "${pathname#${pathname%/*/*}?}"
done