I found the following very useful command on this site:
find -type f -name '*.bz2' -execdir bzgrep "pattern" {} \;
But I don't understand what the {} \; means, can anyone explain?
Thanks
Placeholders in find.
{} denotes 'whatever you found'.
; means end of statement. The \ lets find see it, without the shell interpolating it.
It's often considered sensible to use '{}' (e.g. with single quotes) because the shell will interpolate {}.
{} is the filename find found ant to substituted in the exec(dir) command.
\; is end of command given after execdir. You need the backslash, since it is not used to show the end of the complete unix command (find).
-execdir bzgrep "pattern" {} \;
^ ^ ^ ^ ^^
| | | | ||
| | | | |end of the execdir flag
| | | | |
| | | | shell escape
| | | |
| | | 2. argument to bzgrep
| | | {} is substituted with the current filename
| | 1. argument to bzgrep
| |
| the command to run
execute a command
i.e. for each file that find finds, it runs bzgrep where {} is substituted with the file name.
The ; is needed to end the -execdir, so you can e.g. have other flags to the find command after it. the \ is used to escape the ;, since a ; on the shell would otherwise be interpreted as a command separator (as in e.g. the oneline cd /etc ; ls -l). Single quoting the ; would also work, ';' instead of \; - at least in bash.
Or as the manpage sayes:
-exec command ;
Execute command; 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
Related
I have the following command to fuzzy find files in the command line and to open the selected file in VSCode.:
fzf --print0 -e | xargs -0 -r code
Now I want to be able to search also file contents for a string. I am able to find the searched string in the command line:
rg . | fzf --print0 -e
but now it does not work anymore to open the file in VSCode using this command:
rg . | fzf --print0 -e | xargs -0 -r code
because to VSCode is passed a file name which contains the file name itself and the search string which is of course an empty file.
How can I combine to two above commands to pass the file name to VSCode which contains the searched string?
The --vimgrep option to rg returns just what the doctor ordered. That's what I used in the vscode extension issue request you put in:
https://github.com/rlivings39/vscode-fzf-quick-open/commit/101a6d8e44b707d11e661ca10aaf37102373c644
It returns data like:
$ rg --vimgrep
extension.ts:5:5:let fzfTerminal: vscode.Terminal | undefined = undefined;
extension.ts:6:5:let fzfTerminalPwd: vscode.Terminal | undefined = undefined;
Then you can cut out the first 3 fields and pass them to code -g:
rg --vimgrep --color ansi | fzf --ansi --print0 | cut -z -d : -f 1-3 | xargs -0 -r code -g
In a part of my script I am trying to generate a list of the year and month that a file was submitted. Since the file contains the timestamp, I should be able to cut the filenames to the month position, and then do a sort+uniq filtering. However sed is generating an outlier for one of the files.
I am using this command sequence
ls -1 service*json | sed -e "s|\(.*201...\).*json$|\1|g" | sort |uniq
And this works for most of time except in some cases it outputs the whole timestamp:
$ ls
service-parent-20181119092630.json service-parent-20181123134132.json service-parent-20181202124532.json service-parent-20190121091830.json service-parent-20190125124209.json
service-parent-20181119101003.json service-parent-20181126104300.json service-parent-20181211095939.json service-parent-20190121092453.json service-parent-20190128163539.json
service-parent-20181120095850.json service-parent-20181127083441.json service-parent-20190107035508.json service-parent-20190122093608.json
service-parent-20181120104838.json service-parent-20181129155835.json service-parent-20190107042234.json service-parent-20190122115053.json
$ ls -1 service*json | sed -e "s|\(.*201...\).*json$|\1|g" | sort |uniq
service-parent-201811
service-parent-201811201048
service-parent-201812
service-parent-201901
I have also tried this variation but the second output line is still returned:
ls -1 service*json | sed -e "s|\(.*201.\{3\}\).*json$|\1|g" | sort |uniq
Can somebody explain why service-parent-201811201048 is returned past the requested 3 characters?
Thanks.
service-parent-201811201048 happens to have 201048 to match 201....
Might try ls -1 service*json | sed -e "s|\(.*-201...\).*json$|\1|g" | sort |uniq to ask for a dash - before 201....
It is not recommended to parse the output of ls. Please try instead:
for i in service*json; do
sed -e "s|^\(service-.*-201[0-9]\{3\}\).*json$|\1|g" <<< "$i"
done | sort | uniq
Your problem is explained at https://stackoverflow.com/a/54565973/1745001 (i.e. .* is greedy) but try this:
$ ls | sed -E 's/(-[0-9]{6}).*/\1/' | sort -u
service-parent-201811
service-parent-201812
service-parent-201901
The above requires a sed that supports EREs via -E, e.g. GNU sed and OSX/BSD sed.
I want to rename all the files in my home directory (example abc), in the format (abc_bkp) without using any loops and it should be a single line command in unix (bash script).
If the directory contains nothing but files, this should do it:
ls | xargs -I {} mv {} {}_bkp
If it contains subdirectories, links, and other things you don't want to rename, you must filter the output of ls. Here is a crude way to do it; maybe someone can suggest a more elegant approach:
ls -l | grep ^- | cut -d' ' -f 13 | xargs -I {} mv {} {}_bkp
If you don't want to use loops then I believe the BEST way could be find command, try following command as a DRY run first and once you are satisfy with results then you could remove echo from it to give a real shot.
find -type f -or -type d | xargs -I % echo mv % %_bkp
-I: From man xargs page:
-I replace-str
Replace occurrences of replace-str in the initial-arguments with names read from standard input. Also, unquoted blanks do not
terminate
input items; instead the separator is the newline character. Implies -x and -L 1.
I'm using executing this bash commands inside a search script I've built with php:
find myFolder -type f -exec grep -r KEYWORD {} +
find myFolder -type f -exec grep -r KEYWORD {} + | wc -l
find myFolder -type f | wc -l
The first line gives me back the filenames where KEYWORD was found.
The second line gives me the number of occurrences and the third line the total number of files.
Is there a way to do this more elegantly and faster?
You can get more efficiency if you avoid -exec, which makes one fork per file match. xargs is a better choice here. So I would do something like this:
find myFolder -type f -print0 | xargs -0 grep KEYWORD
find myFolder -type f -print0 | xargs -0 grep KEYWORD | wc -l
The last one should be OK, at least with GNU find.
The -print0 and -0 ensure that filenames with spaces in them are handled correctly.
Note that grep -r` implies recursive grepping, but as you're only supplying one filename in each invocation it is redundant.
I'm new to sed, and need to grab just the filename from the output of find. I need to have find output the whole path for another part of my script, but I want to just print the filename without the path. I also need to match starting from the beginning of the line, not from the end. In english, I want to match, the first group of characters ending with ".txt" not containing a "/". Here's my attempt that doesn't work:
ryan#fizz:~$ find /home/ryan/Desktop/test/ -type f -name \*.txt
/home/ryan/Desktop/test/two.txt
/home/ryan/Desktop/test/one.txt
ryan#fizz:~$ find /home/ryan/Desktop/test/ -type f -name \*.txt | sed s:^.*/[^*.txt]::g
esktop/test/two.txt
ne.txt
Here's the output I want:
two.txt
one.txt
Ok, so the solutions offered answered my original question, but I guess I asked it wrong. I don't want to kill the rest of the line past the file suffix i'm searching for.
So, to be more clear, if the following:
bash$ new_mp3s=\`find mp3Dir -type f -name \*.mp3\` && cp -rfv $new_mp3s dest
`/mp3Dir/one.mp3' -> `/dest/one.mp3'
`/mp3Dir/two.mp3' -> `/dest/two.mp3'
What I want is:
bash$ new_mp3s=\`find mp3Dir -type f -name \*.mp3\` && cp -rfv $new_mp3s dest | sed ???
`one.mp3' -> `/dest'
`two.mp3' -> `/dest'
Sorry for the confusion. My original question just covered the first part of what I'm trying to do.
2nd edit:
here's what I've come up with:
DEST=/tmp && cp -rfv `find /mp3Dir -type f -name \*.mp3` $DEST | sed -e 's:[^\`].*/::' -e "s:$: -> $DEST:"
This isn't quite what I want though. Instead of setting the destination directory as a shell variable, I would like to change the first sed operation so it only changes the cp output before the "->" on each line, so that I still have the 2nd part of the cp output to operate on with another '-e'.
3rd edit:
I haven't figured this out using only sed regex's yet, but the following does the job using Perl:
cp -rfv `find /mp3Dir -type f -name \*.mp3` /tmp | perl -pe "s:.*/(.*.mp3).*\`(.*/).*.mp3\'$:\$1 -> \$2:"
I'd like to do it in sed though.
Something like this should do the trick:
find yourdir -type f -name \*.txt | sed 's/.*\///'
or, slightly clearer,
find yourdir -type f -name \*.txt | sed 's:.*/::'
Why don't you use basename instead?
find /mydir | xargs -I{} basename {}
No need external tools if using GNU find
find /path -name "*.txt" -printf "%f\n"
I landed on the question based on the title: using sed to grab filename from fullpath.
So, using sed, the following is what worked for me...
FILENAME=$(echo $FULLPATH | sed -n 's/^\(.*\/\)*\(.*\)/\2/p')
The first group captures any directories from the path. This is discarded.
The second group capture is the text following the last slash (/). This is returned.
Examples:
echo "/test/file.txt" | sed -n 's/^\(.*\/\)*\(.*\)/\2/p'
file.txt
echo "/test/asd/asd/entrypoint.sh" | sed -n 's/^\(.*\/\)*\(.*\)/\2/p'
entrypoint.sh
echo "/test/asd/asd/default.json" | sed -n 's/^\(.*\/\)*\(.*\)/\2/p'
default.json
find /mydir | awk -F'/' '{print $NF}'
path="parentdir2/parentdir1/parentdir0/dir/FileName"
name=${path##/*}