I hope you can help me with the following problem:
The Situation
I need to find files in various folders and copy them to another folder. The files and folders can contain white spaces and umlauts.
The filenames contain an ID and a string like:
"2022-01-11-02 super important file"
The filenames I need to find are collected in a textfile named ids.txt. This file only contains the IDs but not the whole filename as a string.
What I want to achieve:
I want to read out ids.txt line by line.
For every line in ids.txt I want to do a find search and copy cp the result to destination.
So far I tried:
for n in $(cat ids.txt); do find /home/alex/testzone/ -name "$n" -exec cp {} /home/alex/testzone/output \; ;
while read -r ids; do find /home/alex/testzone -name "$ids" -exec cp {} /home/alex/testzone/output \; ; done < ids.txt
The output folder remains empty. Not using -exec also gives no (search)results.
I was thinking that -name "$ids" is the root cause here. My files contain the ID + a String so I should search for names containing the ID plus a variable string (star)
As argument for -name I also tried "$ids *" "$ids"" *" and so on with no luck.
Is there an argument that I can use in conjunction with find instead of using the star in the -name argument?
Do you have any solution for me to automate this process in a bash script to read out ids.txt file, search the filenames and copy them over to specified folder?
In the end I would like to create a bash script that takes ids.txt and the search-folder and the output-folder as arguments like:
my-id-search.sh /home/alex/testzone/ids.txt /home/alex/testzone/ /home/alex/testzone/output
EDIT:
This is some example content of the ids.txt file where only ids are listed (not the whole filename):
2022-01-11-01
2022-01-11-02
2020-12-01-62
EDIT II:
Going on with the solution from tripleee:
#!/bin/bash
grep . $1 | while read -r id; do
echo "Der Suchbegriff lautet:"$id; echo;
find /home/alex/testzone -name "$id*" -exec cp {} /home/alex/testzone/ausgabe \;
done
In case my ids.txt file contains empty lines the -name "$id*" will be -name * which in turn finds all files and copies all files.
Trying to prevent empty line to be read does not seem to work. They should be filtered by the expression grep . $1 |. What am I doing wrong?
If your destination folder is always the same, the quickest and absolutely most elegant solution is to run a single find command to look for all of the files.
sed 's/.*/-o\n—name\n&*/' ids.txt |
xargs -I {} find -false {} -exec cp {} /home/alex/testzone/output +
The -false predicate is a bit of a hack to allow the list of actual predicates to start with -o (as in "or").
This could fail if ids.txt is too large to fit into a single xargs invocation, or if your sed does not understand \n to mean a literal newline.
(Here's a fix for the latter case:
xargs printf '-o\n-name\n%s*\n' <ids.txt |
...
Still the inherent problem with using xargs find like this is that xargs could split the list between -o and -name or between -name and the actual file name pattern if it needs to run more than one find command to process all the arguments.
A slightly hackish solution to that is to ensure that each pair is a single string, and then separately split them back out again:
xargs printf '-o_-name_%s*\n' <ids.txt |
xargs bash -c 'arr=("$#"); find -false ${arr[#]/-o_-name_/-o -name } -exec cp {} "$0"' /home/alex/testzone/ausgabe
where we temporarily hold the arguments in an array where each file name and its flags is a single item, and then replace the flags into separate tokens. This still won't work correctly if the file names you operate on contain literal shell metacharacters like * etc.)
A more mundane solution fixes your while read attempt by adding the missing wildcard in the -name argument. (I also took the liberty to rename the variable, since read will only read one argument at a time, so the variable name should be singular.)
while read -r id; do
find /home/alex/testzone -name "$id*" -exec cp {} /home/alex/testzone/output \;
done < ids.txt
Please try the following bash script copier.sh
#!/bin/bash
IFS=$'\n' # make newlines the only separator
set -f # disable globbing
file="files.txt" # name of file containing filenames
finish="finish" # destination directory
while read -r n ; do (
du -a | awk '{for(i=2;i<=NF;++i)printf $i" " ; print " "}' | grep $n | sed 's/ *$//g' | xargs -I '{}' cp '{}' $finish
);
done < $file
which copies recursively all the files named in files.txt from . and it's subfiles to ./finish
This new version works even if there are spaces in the directory names or file names.
I have json files in the current directory, and subdirectories. All the files have a single line of content.
I want to a list of all files that contain the word XYZ, and the number of times it occurs in that file.
I want to print the list according to the following format:
file_name pattern_occurence_times
It should look something like:
.\x1\x2\file1.json 3
.\x1\file3.json 2
The problem is that grep counts the NUMBER of lines containing XYZ, not the number of occurrences.
Since the whole content of the files is always contained in a single line, the count is always 1 (if the pattern occurs in the file).
I used this command for that:
find . -type f -name "*.json" -exec grep --files-with-match -i 'xyz' {} \; -exec grep -wci 'xyz' {} \;
I wrote a python code, and it works, but I would like to know if there is any way of doing that using find and grep or any other command line tools.
Thanks
The classical approach to this problem is the pipeline grep -o regex file | wc -l. However, to execute a pipeline in find's -exec you have to run a shell (e.g. sh -c ... ). But all these things together will only print the number of matches, not the file names. Also, files with no matches have to be filtered out.
Because of all of this I think a single awk command would be preferable:
find ... -type f -exec awk '{$0=tolower($0); c+=gsub(/xyz/,"")}
END {if(c>0) print FILENAME " " c}' {} \;
Here the tolower($0) emulates grep's -i option. Make sure to write your search pattern xyz only in lowercase.
If you want to combine this with subsequent filters in find you can add else exit 1 at the end of the last awk block to continue (inside find) only with the printed files.
Use the -o option of grep, e.g. in conjunction with wc, e.g.
find . -name "*.json" | while read -r f ; do
echo $f : $(grep -ow XYZ "$f" | wc -l)
done
How do I find and replace every occurrence of:
subdomainA.example.com
with
subdomainB.example.com
in every text file under the /home/www/ directory tree recursively?
find /home/www \( -type d -name .git -prune \) -o -type f -print0 | xargs -0 sed -i 's/subdomainA\.example\.com/subdomainB.example.com/g'
-print0 tells find to print each of the results separated by a null character, rather than a new line. In the unlikely event that your directory has files with newlines in the names, this still lets xargs work on the correct filenames.
\( -type d -name .git -prune \) is an expression which completely skips over all directories named .git. You could easily expand it, if you use SVN or have other folders you want to preserve -- just match against more names. It's roughly equivalent to -not -path .git, but more efficient, because rather than checking every file in the directory, it skips it entirely. The -o after it is required because of how -prune actually works.
For more information, see man find.
The simplest way for me is
grep -rl oldtext . | xargs sed -i 's/oldtext/newtext/g'
Note: Do not run this command on a folder including a git repo - changes to .git could corrupt your git index.
find /home/www/ -type f -exec \
sed -i 's/subdomainA\.example\.com/subdomainB.example.com/g' {} +
Compared to other answers here, this is simpler than most and uses sed instead of perl, which is what the original question asked for.
All the tricks are almost the same, but I like this one:
find <mydir> -type f -exec sed -i 's/<string1>/<string2>/g' {} +
find <mydir>: look up in the directory.
-type f:
File is of type: regular file
-exec command {} +:
This variant of the -exec action runs the specified command on the selected files, but the command line is built by appending
each selected file name at the end; the total number of invocations of the command will be much less than the number of
matched files. The command line is built in much the same way that xargs builds its command lines. Only one instance of
`{}' is allowed within the command. The command is executed in the starting directory.
For me the easiest solution to remember is https://stackoverflow.com/a/2113224/565525, i.e.:
sed -i '' -e 's/subdomainA/subdomainB/g' $(find /home/www/ -type f)
NOTE: -i '' solves OSX problem sed: 1: "...": invalid command code .
NOTE: If there are too many files to process you'll get Argument list too long. The workaround - use find -exec or xargs solution described above.
cd /home/www && find . -type f -print0 |
xargs -0 perl -i.bak -pe 's/subdomainA\.example\.com/subdomainB.example.com/g'
For anyone using silver searcher (ag)
ag SearchString -l0 | xargs -0 sed -i 's/SearchString/Replacement/g'
Since ag ignores git/hg/svn file/folders by default, this is safe to run inside a repository.
This one is compatible with git repositories, and a bit simpler:
Linux:
git grep -l 'original_text' | xargs sed -i 's/original_text/new_text/g'
Mac:
git grep -l 'original_text' | xargs sed -i '' -e 's/original_text/new_text/g'
(Thanks to http://blog.jasonmeridth.com/posts/use-git-grep-to-replace-strings-in-files-in-your-git-repository/)
To cut down on files to recursively sed through, you could grep for your string instance:
grep -rl <oldstring> /path/to/folder | xargs sed -i s^<oldstring>^<newstring>^g
If you run man grep you'll notice you can also define an --exlude-dir="*.git" flag if you want to omit searching through .git directories, avoiding git index issues as others have politely pointed out.
Leading you to:
grep -rl --exclude-dir="*.git" <oldstring> /path/to/folder | xargs sed -i s^<oldstring>^<newstring>^g
A straight forward method if you need to exclude directories (--exclude-dir=..folder) and also might have file names with spaces (solved by using 0Byte for both grep -Z and xargs -0)
grep -rlZ oldtext . --exclude-dir=.folder | xargs -0 sed -i 's/oldtext/newtext/g'
An one nice oneliner as an extra. Using git grep.
git grep -lz 'subdomainA.example.com' | xargs -0 perl -i'' -pE "s/subdomainA.example.com/subdomainB.example.com/g"
Simplest way to replace (all files, directory, recursive)
find . -type f -not -path '*/\.*' -exec sed -i 's/foo/bar/g' {} +
Note: Sometimes you might need to ignore some hidden files i.e. .git, you can use above command.
If you want to include hidden files use,
find . -type f -exec sed -i 's/foo/bar/g' {} +
In both case the string foo will be replaced with new string bar
find /home/www/ -type f -exec perl -i.bak -pe 's/subdomainA\.example\.com/subdomainB.example.com/g' {} +
find /home/www/ -type f will list all files in /home/www/ (and its subdirectories).
The "-exec" flag tells find to run the following command on each file found.
perl -i.bak -pe 's/subdomainA\.example\.com/subdomainB.example.com/g' {} +
is the command run on the files (many at a time). The {} gets replaced by file names.
The + at the end of the command tells find to build one command for many filenames.
Per the find man page:
"The command line is built in much the same way that
xargs builds its command lines."
Thus it's possible to achieve your goal (and handle filenames containing spaces) without using xargs -0, or -print0.
I just needed this and was not happy with the speed of the available examples. So I came up with my own:
cd /var/www && ack-grep -l --print0 subdomainA.example.com | xargs -0 perl -i.bak -pe 's/subdomainA\.example\.com/subdomainB.example.com/g'
Ack-grep is very efficient on finding relevant files. This command replaced ~145 000 files with a breeze whereas others took so long I couldn't wait until they finish.
or use the blazing fast GNU Parallel:
grep -rl oldtext . | parallel sed -i 's/oldtext/newtext/g' {}
grep -lr 'subdomainA.example.com' | while read file; do sed -i "s/subdomainA.example.com/subdomainB.example.com/g" "$file"; done
I guess most people don't know that they can pipe something into a "while read file" and it avoids those nasty -print0 args, while presevering spaces in filenames.
Further adding an echo before the sed allows you to see what files will change before actually doing it.
Try this:
sed -i 's/subdomainA/subdomainB/g' `grep -ril 'subdomainA' *`
According to this blog post:
find . -type f | xargs perl -pi -e 's/oldtext/newtext/g;'
#!/usr/local/bin/bash -x
find * /home/www -type f | while read files
do
sedtest=$(sed -n '/^/,/$/p' "${files}" | sed -n '/subdomainA/p')
if [ "${sedtest}" ]
then
sed s'/subdomainA/subdomainB/'g "${files}" > "${files}".tmp
mv "${files}".tmp "${files}"
fi
done
If you do not mind using vim together with grep or find tools, you could follow up the answer given by user Gert in this link --> How to do a text replacement in a big folder hierarchy?.
Here's the deal:
recursively grep for the string that you want to replace in a certain path, and take only the complete path of the matching file. (that would be the $(grep 'string' 'pathname' -Rl).
(optional) if you want to make a pre-backup of those files on centralized directory maybe you can use this also: cp -iv $(grep 'string' 'pathname' -Rl) 'centralized-directory-pathname'
after that you can edit/replace at will in vim following a scheme similar to the one provided on the link given:
:bufdo %s#string#replacement#gc | update
You can use awk to solve this as below,
for file in `find /home/www -type f`
do
awk '{gsub(/subdomainA.example.com/,"subdomainB.example.com"); print $0;}' $file > ./tempFile && mv ./tempFile $file;
done
hope this will help you !!!
For replace all occurrences in a git repository you can use:
git ls-files -z | xargs -0 sed -i 's/subdomainA\.example\.com/subdomainB.example.com/g'
See List files in local git repo? for other options to list all files in a repository. The -z options tells git to separate the file names with a zero byte, which assures that xargs (with the option -0) can separate filenames, even if they contain spaces or whatnot.
A bit old school but this worked on OS X.
There are few trickeries:
• Will only edit files with extension .sls under the current directory
• . must be escaped to ensure sed does not evaluate them as "any character"
• , is used as the sed delimiter instead of the usual /
Also note this is to edit a Jinja template to pass a variable in the path of an import (but this is off topic).
First, verify your sed command does what you want (this will only print the changes to stdout, it will not change the files):
for file in $(find . -name *.sls -type f); do echo -e "\n$file: "; sed 's,foo\.bar,foo/bar/\"+baz+\"/,g' $file; done
Edit the sed command as needed, once you are ready to make changes:
for file in $(find . -name *.sls -type f); do echo -e "\n$file: "; sed -i '' 's,foo\.bar,foo/bar/\"+baz+\"/,g' $file; done
Note the -i '' in the sed command, I did not want to create a backup of the original files (as explained in In-place edits with sed on OS X or in Robert Lujo's comment in this page).
Happy seding folks!
just to avoid to change also
NearlysubdomainA.example.com
subdomainA.example.comp.other
but still
subdomainA.example.com.IsIt.good
(maybe not good in the idea behind domain root)
find /home/www/ -type f -exec sed -i 's/\bsubdomainA\.example\.com\b/\1subdomainB.example.com\2/g' {} \;
Here's a version that should be more general than most; it doesn't require find (using du instead), for instance. It does require xargs, which are only found in some versions of Plan 9 (like 9front).
du -a | awk -F' ' '{ print $2 }' | xargs sed -i -e 's/subdomainA\.example\.com/subdomainB.example.com/g'
If you want to add filters like file extensions use grep:
du -a | grep "\.scala$" | awk -F' ' '{ print $2 }' | xargs sed -i -e 's/subdomainA\.example\.com/subdomainB.example.com/g'
For Qshell (qsh) on IBMi, not bash as tagged by OP.
Limitations of qsh commands:
find does not have the -print0 option
xargs does not have -0 option
sed does not have -i option
Thus the solution in qsh:
PATH='your/path/here'
SEARCH=\'subdomainA.example.com\'
REPLACE=\'subdomainB.example.com\'
for file in $( find ${PATH} -P -type f ); do
TEMP_FILE=${file}.${RANDOM}.temp_file
if [ ! -e ${TEMP_FILE} ]; then
touch -C 819 ${TEMP_FILE}
sed -e 's/'$SEARCH'/'$REPLACE'/g' \
< ${file} > ${TEMP_FILE}
mv ${TEMP_FILE} ${file}
fi
done
Caveats:
Solution excludes error handling
Not Bash as tagged by OP
If you wanted to use this without completely destroying your SVN repository, you can tell 'find' to ignore all hidden files by doing:
find . \( ! -regex '.*/\..*' \) -type f -print0 | xargs -0 sed -i 's/subdomainA.example.com/subdomainB.example.com/g'
Using combination of grep and sed
for pp in $(grep -Rl looking_for_string)
do
sed -i 's/looking_for_string/something_other/g' "${pp}"
done
perl -p -i -e 's/oldthing/new_thingy/g' `grep -ril oldthing *`
to change multiple files (and saving a backup as *.bak):
perl -p -i -e "s/\|/x/g" *
will take all files in directory and replace | with x
called a “Perl pie” (easy as a pie)
I am trying to use the following way to replace my string
grep -rl matchstring . | xargs sed -i 's/string1/string2/g'
What I want to achieve is actually currently I got a hardcode domain name like
http://account.mysmallwebsite.com
https://account.mysmallwebsite.com
Because this is quite inflexible, I would prefer just
<?php echo $domainName;?>
the variable of domainName will be set in a php file and all the files will include this file so in future if I change domain, it will be easier to just change 1 place rather than multiple pages.
and replace all occurrence of the website
http://account.mysmallwebsite.com
https://account.mysmallwebsite.com
with the echo string, but how do I achieve it through command line at putty shell ssh
You can use the following:
find . -name "*.php" -exec sed -r -i.bak 's#http(s?)://account.mysmallwebsite.com#<?php echo $domainName;?>#g' {} \;
Explanation
find . -name "*.php" -exec ... {} \; this looks for files whose name ends with .php and performs the command indicated in ....
sed -r -i.bak 's#http(s?)://account.mysmallwebsite.com#<?php echo $domainName;?>#g' file:
this looks for http(s?)://account.mysmallwebsite.com, that is, http + maybe an s + ://account.mysmallwebsite.com and replaces it with <?php echo $domainName;?>.
Note the usage of # as separator instead of the typical /, so that we don't have to escape the slashes in the URL.
The -i.bak creates a backup file with .bak extension, while the original one gets in-place edited.
You can also "play" with -maxdepth value, to define how many levels of subdirectories you want to work on. For example, -maxdepth 1 will just check the current directory, while -maxdepth 2 will also include the sub folders of the given directory.
Graphically:
s#http(s?)://account.mysmallwebsite.com#<?php echo $domainName;?>#g
^ ^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^
| ^^^^^^|^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ substitution ^
| | look for this text replace all matches
| may be an "s"
^
search
See the sed command in action:
$ cat a
hello
http://account.mysmallwebsite.com
https://account.mysmallwebsite.com
httpss://account.mysmallwebsite.com
bye
$ sed -r 's#http(s?)://account.mysmallwebsite.com#<?php echo $domainName;?>#g' a
hello
<?php echo $domainName;?>
<?php echo $domainName;?>
httpss://account.mysmallwebsite.com
bye
In my hierarchy of directories I have many text files called STATUS.txt. These text files each contain one keyword such as COMPLETE, WAITING, FUTURE or OPEN. I wish to execute a shell command of the following form:
./mycommand OPEN
which will list all the directories that contain a file called STATUS.txt, where this file contains the text "OPEN"
In future I will want to extend this script so that the directories returned are sorted. Sorting will determined by a numeric value stored the file PRIORITY.txt, which lives in the same directories as STATUS.txt. However, this can wait until my competence level improves. For the time being I am happy to list the directories in any order.
I have searched Stack Overflow for the following, but to no avail:
unix filter by file contents
linux filter by file contents
shell traverse directory file contents
bash traverse directory file contents
shell traverse directory find
bash traverse directory find
linux file contents directory
unix file contents directory
linux find name contents
unix find name contents
shell read file show directory
bash read file show directory
bash directory search
shell directory search
I have tried the following shell commands:
This helps me identify all the directories that contain STATUS.txt
$ find ./ -name STATUS.txt
This reads STATUS.txt for every directory that contains it
$ find ./ -name STATUS.txt | xargs -I{} cat {}
This doesn't return any text, I was hoping it would return the name of each directory
$ find . -type d | while read d; do if [ -f STATUS.txt ]; then echo "${d}"; fi; done
... or the other way around:
find . -name "STATUS.txt" -exec grep -lF "OPEN" \{} +
If you want to wrap that in a script, a good starting point might be:
#!/bin/sh
[ $# -ne 1 ] && echo "One argument required" >&2 && exit 2
find . -name "STATUS.txt" -exec grep -lF "$1" \{} +
As pointed out by #BroSlow, if you are looking for directories containing the matching STATUS.txt files, this might be more what you are looking for:
fgrep --include='STATUS.txt' -rl 'OPEN' | xargs -L 1 dirname
Or better
fgrep --include='STATUS.txt' -rl 'OPEN' |
sed -e 's|^[^/]*$|./&|' -e 's|/[^/]*$||'
# ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
# simulate `xargs -L 1 dirname` using `sed`
# (no trailing `\`; returns `.` for path without dir part)
Maybe you can try this:
grep -rl "OPEN" . --include='STATUS.txt'| sed 's/STATUS.txt//'
where grep -r means recursive , -l means only list the files matching, '.' is the directory location. You can pipe it to sed to remove the file name.
You can then wrap this in a bash script file where you can pass in keywords such as 'OPEN', 'FUTURE' as an argument.
#!/bin/bash
grep -rl "$1" . --include='STATUS.txt'| sed 's/STATUS.txt//'
Try something like this
find -type f -name "STATUS.txt" -exec grep -q "OPEN" {} \; -exec dirname {} \;
or in a script
#!/bin/bash
(($#==1)) || { echo "Usage: $0 <pattern>" && exit 1; }
find -type f -name "STATUS.txt" -exec grep -q "$1" {} \; -exec dirname {} \;
You could use grep and awk instead of find:
grep -r OPEN * | awk '{split($1, path, ":"); print path[1]}' | xargs -I{} dirname {}
The above grep will list all files containing "OPEN" recursively inside you dir structure. The result will be something like:
dir_1/subdir_1/STATUS.txt:OPEN
dir_2/subdir_2/STATUS.txt:OPEN
dir_2/subdir_3/STATUS.txt:OPEN
Then the awk script will split this output at the colon and print the first part of it (the dir path).
dir_1/subdir_1/STATUS.txt
dir_2/subdir_2/STATUS.txt
dir_2/subdir_3/STATUS.txt
The dirname will then return only the directory path, not the file name, which I suppose it what you want.
I'd consider using Perl or Python if you want to evolve this further, though, as it might get messier if you want to add priorities and sorting.
Taking up the accepted answer, it does not output a sorted and unique directory list. At the end of the "find" command, add:
| sort -u
or:
| sort | uniq
to get the unique list of the directories.
Credits go to Get unique list of all directories which contain a file whose name contains a string.
IMHO you should write a Python script which:
Examines your directory structure and finds all files named STATUS.txt.
For each found file:
reads the file and executes mycommand depending on what the file contains.
If you want to extend the script later with sorting, you can find all the interesting files first, save them to a list, sort the list and execute the commands on the sorted list.
Hint: http://pythonadventures.wordpress.com/2011/03/26/traversing-a-directory-recursively/