While perusing some AWS docs I noticed the following command:
find /var/www -type d -exec sudo chmod 2775 {} +
I'm familiar with the \; ending to exec in a find string but have never seen the '+'. Can anyone shed some light on this?
Here's the original page: http://docs.aws.amazon.com/AWSEC2/latest/UserGuide/install-LAMP.html
Thanks!
If you use a plus (+) instead of the escaped semicolon, the arguments will be grouped together before being passed to the command. For example:
$ find . -type f -exec echo {} +
. ./bar.txt ./foo.txt
In this case, only one child process (echo . ./bar.txt ./foo.txt)is created which is much more efficient, because it avoids a fork/exec for each single argument.
Using the escaped semi-colon, you will get a child process created for each argument.
$ find . -type f -exec echo {} \;
.
./bar.txt
./foo.txt
Related
AIM: to find all JS|TS excluding *.spec.js files in a directory but replace the base path with ./
I have this command
find src/app/directives -name '*.[j|t]s' ! -name '*.spec.js' -exec printf "import \"%s\";\n" {} \;
which in said directory prints the marked JS files. However I want to replace the src/app with ./
I've tried playing with [[]] and this command but they don't work.
find src/app/components -name '*.[j|t]s' ! -name '*.spec.js' -exec printf "import \"%s\";\n" ${{}/src
/hi} \;
zsh: bad substitution
Given your "AIM", all you really need is:
find src/app/directives -type f -name "*.[jt]s" ! -name "*.spec.js" -printf "./%f\n"
The reason being is the '|' in your character-class isn't matching anything, but isn't hurting anything for that matter. Your second ! -name "*.spec.js" is fine. You don't need -exec and can simply use -printf "./%f\n" (where "%f" provides the filename only for the current file). You simply prepend the "./" as part of the -printf format-string.
Let me know if I misunderstood your AIM or if you have further questions.
Removing src/app/directives While Preserving Remaining Path
If you want to preserve the remainder of the path after src/app/directives (essentially just replacing it with '.'), you can use a short helper-script with the POSIX parameter expansion to trim src/app/directives from the front of the string replacing it with '.' using printf in the helper script. For example the helper could be:
#!/bin/zsh
printf ".%s" "${1#./src/app/directives}"
(note: the leading "./" being removed along with src/app/directives is prepended by find, the '.' added by the printf format-string will result in the returned filename being ./rest/of/path/to/filename)
Call the script whatever you like, helper.sh below. Make it executable chmod +x helper.sh.
The find call would then be:
find src/app/directives -type f -name "*.[jt]s" ! -name "*.spec.js" -exec path/to/helper.sh '{}' \;
Give that a go and let me know if it does what you are needing.
I want to copy some files with string "ABC" in their name and without string "DEF" in their name using the find -exec command. This post mentions how to do it for the without case using find command. I want to know how to do it for both with and without cases.
You can use -and or -a for multiple true condition as follows. ! for negation can be applied after and
find . \( -regex ".*ABC.*" -and ! -regex ".*DEF.*" \) -exec cp {} /destination/folder/ \;
I'm looking to find a way to match an exact string
for example:
I have these cmd that I run on unix server
1.)find ./ -name "*.jsp" -type f -exec grep -m1 -l '50.000' {} + >> 50dotcol.txt
2.)find ./ -name ".jsp" -type f -exec grep -m1 -l '\<50.000>' {} + >> 50dotcol.txt
Edited after Georges response:
find ./ -name "*.jsp" -type f -exec grep -m1 -l '(50.000)' {} + >> 50dotcol.txt
Still didn't pull in any results
The first one will find any string containing "50" the second will omit double digit strings but will pull in $50,000 $50.000. But I'm just looking to pull in "50.000" and that's it, no other variations of this
Am I missing something in my find cmd?
Use
grep -m1 -l '(50\.000)'
instead. That backslash before the < interprets it literally, which you don't want to do. And you need to use parenthesis to have it considered as an exact group of characters.
I've got a set of files in a web root that all contain special characters that I'd like to remove (Â,€,â,etc).
My command
find . -type f -name '*.*' -exec grep -il "Â" {} \;
finds & lists out the files just fine, but my command
find . -type f -name '*.*' -exec tr -d 'Â' '' \;
doesn't produce the results I'm looking for.
Any thoughts?
to replace all non-ascii characters in all files inside the current directory you could use:
find . -type f | xargs perl -pi.bak -e 's,[^[:ascii:]],,g'
afterwards you will have to find and remove all the '.bak' files:
find . -type f -a -name \*.bak | xargs rm
I would recommend looking into sed. It can be used to replace the contents of the file.
So you could use the command:
find . -type f -name '*.*' -exec sed -i "s/Â//" {} \;
I have tested this with a simple example and it seems to work. The -exec should handle files with whitespace in their name, but there may be other vulnerabilities I'm not aware of.
Use
tr -d 'Â'
What does the ' ' stands for? On my system using your command produces this error:
tr: extra operand `'
Only one string may be given when deleting without squeezing repeats.
Try `tr --help' for more information.
sed 's/ø//' file.txt
That should do the trick for replacing a special char with an empty string.
find . -name "*.*" -exec sed 's/ø//' {} \
It would be helpful to know what "doesn't produce the results I'm looking for" means. However, in your command tr is not provided with the filenames to process. You could change it to this:
find . -type f -name '*.*' -exec tr -d 'Â' {} \;
Which is going to output everything to stdout. You probably want to modify the files instead. You can use Grundlefleck's answer, but one of the issues alluded to in that answer is if there are large numbers of files. You can do this:
find . -type f -name '*.*' -print0 | xargs -0 -I{} sed -i "s/Â//" \{\}
which should handle files with spaces in their names as well as large numbers of files.
with bash shell
for file in *.*
do
case "$file" in
*[^[:ascii:]]* )
mv "$file" "${file//[^[:ascii:]]/}"
;;
esac
done
I would use something like this.
for file in `find . -type f`
do
# Search for char end remove it. Save file as file.new
sed -e 's/[ۉ]//g' $file > $file.new
# mv file.new to file DON'T RUN IF YOU WILL NOT OVERITE ORIGINAL FILE
mv $file.new $file
done
The above script will fail as levislevis85 has mentioned it with spaces in filenames. This would not be the case if you use the following code.
find . -type f | while read file
do
# Search for char end remove it. Save file as file.new
sed -e 's/[ۉ]//g' "$file" > "$file".new
# mv file.new to file DON'T RUN IF YOU WILL NOT OVERITE ORIGINAL FILE
mv "$file".new "$file"
done
I can do this:
$ find .
.
./b
./b/foo
./c
./c/foo
And this:
$ find . -type f -exec cat {} \;
This is in b.
This is in c.
But not this:
$ find . -type f -exec cat > out.txt {} \;
Why not?
find's -exec argument runs the command you specify once for each file it finds. Try:
$ find . -type f -exec cat {} \; > out.txt
or:
$ find . -type f | xargs cat > out.txt
xargs converts its standard input into command-line arguments for the command you specify. If you're worried about embedded spaces in filenames, try:
$ find . -type f -print0 | xargs -0 cat > out.txt
Hmm... find seems to be recursing as you output out.txt to the current directory
Try something like
find . -type f -exec cat {} \; > ../out.txt
You could do something like this :
$ cat `find . -type f` > out.txt
How about just redirecting the output of find into a file, since all you're wanting to do is cat all the files into one large file:
find . -type f -exec cat {} \; > /tmp/out.txt
Maybe you've inferred from the other responses that the > symbol is interpreted by the shell before find gets it as an argument. But to answer your "why not" lets look at your command, which is:
$ find . -type f -exec cat > out.txt {} \;
So you're giving find these arguments: "." "-type" "f" "-exec" "cat" you're giving the redirect these arguments: "out.txt" "{}" and ";". This confuses find by not terminating the -exec arguments with a semi-colon and by not using the file name as an argument ("{}"), it possibly confuses the redirection too.
Looking at the other suggestions you should really avoid creating the output in the same directory you're finding in. But they'd work with that in mind. And the -print0 | xargs -0 combination is greatly useful. What you wanted to type was probably more like:
$ find . -type f -exec cat \{} \; > /tmp/out.txt
Now if you really only have one level of sub directories and only normal files, you can do something silly and simple like this:
cat `ls -p|sed 's/\/$/\/*/'` > /tmp/out.txt
Which gets ls to list all your files and directories appending '/' to the directories, while sed will append a '*' to the directories. The shell will then interpret this list and expand the globs. Assuming that doesn't result in too many files for the shell to handle, these will all be passed as arguments to cat, and the output will be written to out.txt.
Or just leave out the find which is useless if you use the really great Z shell (zsh), and you can do this:
setopt extendedglob
(this should be in your .zshrc)
Then:
cat **/*(.) > outfile
just works :-)
Try this:
(find . -type f -exec cat {} \;) > out.txt
In bash you could do
cat $(find . -type f) > out.txt
with $( ) you can get the output from a command and pass it to another