Optimal string replacing in files for AIX - string

I need to remove about 40 emails from several files in a distribution list.
One Address might appear in different files and need to be removed from all of them.
I am working in a directory with several .sh files which also have several lines.
I have done something like this in a couple of test files:
find . -type f -exec grep -li ADDRESS_TO_FIND {} 2>/dev/null \; | xargs sed -i 's/ADDRESS_TO_REMOVE/ /g' *
It works fine but once I try it in the real files, it takes a long time and just sits there. I need to run this in different servers, this is the main cause I want to optimize this.
I have tried to run something like this:
find . -type f -name '*sh' 2>/dev/null | xargs grep ADDRESS_TO_FIND
but that will return:
./FileContainingAddress.sh:ADDRESS_TO_FIND
How do I add something like this:
awk '{print substr($0,1,10)}'
But to return me everything before the ":"?
I can do the rest from there, but haven't found how to trim that part

You can use -exec as a predicate in find, as long as you don't use the multiple file + version, which means that you can provide several -exec clauses each of which will be dependent on the success of the previous one. This style will avoid the construction of lists of filenames, which makes it much more robust in the face of files with odd characters in their names.
For example:
find . -type f -name '*sh' \
-exec grep -qi ADDRESS_TO_FIND {} \; \
-exec sed -i 's/ADDRESS_TO_FIND/ /g' {} \;
You probably want to provide the address as a parameter rather than having to type it twice, unless you really meant for the two instance to be different (ADDRESS_TO_FIND vs. ADDRESS_TO_REMOVE):
clean() {
find . -type f -name '*sh' \
-exec grep -qi "$1" {} \; \
-exec sed -i "s/$1/ /g" {} \;
}
(Watch out for / in the argument to clean. I'll leave making the sed more robust as an exercise.)

After looking back at your question, I noticed something that's potentially quite important:
find -type f -exec grep -li ADDRESS {} \; | xargs sed -i 's/ADDRESS/ /g' *
# here! -----------------------------------------------------------------^
The asterisk is being expanded, so the sed line is operating on every file in the directory.
Assuming that this wasn't a typo in your question, I believe that this is the source of your poor performance. You should remove it!

Related

How to grep through many files of same file type

I wish to grep through many (20,000) text files, each with about 1,000,000 lines each, so the faster the better.
I have tried the below code and it just doesn't seem to want to do anything, it doesn't find any matches even after an hour (it should have done by now).
for i in $(find . -name "*.txt"); do grep -Ff firstpart.txt $1; done
Ofir's answer is good. Another option:
find . -name "*.txt" -exec grep -fnFH firstpart.txt {} \;
I like to add the -n for line numbers and -H to get the filename. -H is particularly useful in this case as you could have a lot of matches.
Instead of iterating through the files in a loop, you can just give the file names to grep using xargs and let grep go over all the files.
find . -name "*.txt" | xargs grep $1
I'm not quite sure whether it will actually increase the performance, but it's probably worth a try.
ripgrep is the most amazing tool. You should get that and use it.
To search *.txt files in all directories recursively, do this:
rg -t txt -f patterns.txt
Ripgrep uses one of the fastest regular expression engines out there. It uses multiple threads. It searches directories and files, and filters them to the interesting ones in the fastest way.
It is simply great.
For anyone stuck using grep for whatever reason:
find -name '*.txt' -type f -print0 | xargs -0 -P 8 -n 8 grep -Ff patterns.txt
That tells xargs to -n 8 use 8 arguments per command and to -P 8 run 8 copies in parallel. It has the downside that the output might become interleaved and corrupted.
Instead of xargs you could use parallel which does a fancier job and keeps output in order:
$ find -name '*.txt' -type f -print0 | parallel -0 grep --with-filename grep -Ff patterns.txt

Moving files with a specific modification date; "find | xargs ls | grep | -exec" fails w/ "-exec: command not found"

Iam using centos 7
If I want to find files that have specific name and specific date then moving these files to another folder iam issuing the command
find -name 'fsimage*' | xargs ls -ali | grep 'Oct 20' | -exec mv {} /hdd/fordelete/ \;
with the following error
-bash: -exec: command not found xargs: ls: terminated by signal 13
As another answer already explains, -exec is an action for find, you can't use it as a shell command. On contrary, xargs and grep are commands, and you can't use them as find actions, just like you can't use pipe | inside find.
But more importantly, even though you could use ls and grep on find's result just to move files older than some amount of time, you shouldn't. Such pipeline is fragile and fails on many corner cases, like symlinks, files with newlines in name, etc.
Instead, use find. You'll find it quite powerful.
For example, to mv files modified more than 7 days ago, use the -mtime test:
find -name 'fsimage*' -mtime +7 -exec mv '{}' /some/dir/ \;
To mv files modified on a specific/reference date, e.g. 2017-10-20, you can use the -newerXY test:
find -name 'fsimage*' -newermt 2017-10-20 ! -newermt 2017-10-21 -exec mv '{}' /some/dir/ \;
Also, if your mv supports the -t option (to give target dir first, multiple files after), you can use {} + placeholder in find for multiple files, reducing the total number of mv command invocations (thanks #CharlesDuffy):
find -name 'fsimage*' -mtime +7 -exec mv -t /some/dir/ '{}' +
the -exec as you wrote it is quite meaningless, moreover it seems you are mixing find syntax with shell oe (-exec as you wrote it should be passed to find)
there are probably more concise ways of doing, but this should do what you expect:
find -name 'fsimage*' -type f | xargs ls -ali | grep 'Oct 20' | awk '{ print $NF }' | while read file; do mv "$file" /hdd/fordelete/ ; done
nevertheless, you should take care of not just copy/paste things you do not really understand from the web, you may wreck you system...

deleting 10 days old folder in linux

#!/bin/bash
now=$(date +"%d-%m-%Y")
year=$(date +"%Y")
echo $year
find /var/backup/backup-scripts -name '*$year*' -ctime +10 -exec rm {} \;
~
~
This script is running with no errors but not removing 10 days old folder
Please advise
Thanks in anticipation
you're single quoting a variable, it can't be expanded by the shell. Try
find /var/backup/backup-scripts -name "*$year*" -ctime +10 -exec rm {} \;
IHTH
#shellter's answer is a more direct mapping to your methodology, but here's how I would code that:
find /var/backup/backup-scripts -name "*$(date +%Y)*" -mtime +10 -type f \
-print0 |xargs -0 rm -v
Some differences:
It's a one-liner (though if you want a cron job, make a script)
-mtime ("modified") replaces -ctime ("changed") because sometimes access is a change
I restricted the results to files
I passed the output to xargs (in a manner safe for spaced filenames) to distribute the task
xargs is useful because it won't be overwhelmed by too many files (you can only send so many arguments to a command line program). It also runs slightly faster since it will run its multiple instances in parallel. (If it doesn't run multiple instances, there won't be any speedup, but in that case it probably won't take too long either way.)
One of the main reasons to use pipes here is that you can edit the output even further. Let's say you wanted to keep anything from January:
find /var/backup/backup-scripts -name "*$(date +%Y)*" -mtime +10 -type f \
|egrep -v "$(date +%Y)-?01" |xargs -d "\n" rm -v
egrep -v will only show lines that do not match the given regex, so they won't be removed.
This uses \n as a delimiter rather than \0 (null), which will trip over files with line breaks in their names (don't do that!), but otherwise it excludes hits like file20150101.txt and file2015-01-01.txt (note the question mark in my extended posix regex; further explanation at regex101)

Bash script to recursively find and replace in files [duplicate]

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)

Piping find results into grep for fast directory exclusion

I am successfully using find to create a list of all files in the current subdirectory, excluding those in the subdirectory "cache." Here's my first bit of code:
find . -wholename './cach*' -prune -o -print
I now wish to pipe this into a grep command. It seems like that should be simple:
find . -wholename './cach*' -prune -o -print | xargs grep -r -R -i "samson"
... but this is returning results that are mostly from the cache directory. I've tried removing the xargs reference, but that does what you'd expect, running the grep on text of the file names, rather than on the files themselves. My goal is to find "samson" in any files that aren't cached content.
I'll probably get around this issue by just using doubled greps in this instance, but I'm very curious about why this one-liner behaves this way. I'd love to hear thoughts on a way to modify it while still using these two commands (as there are speed advantages to doing it this way).
(This is in CentOS 5, btw.)
The wholename match may be the reason why it's still including "cache" files. If you're executing the find command in the directory that contains the "cache" folder, it should work. If not, try changing it to -name '*cache*' instead.
Also, you do not need the -r or -R for your grep, that tells it to recurse through directories - but you're testing individual files.
You can update your command using the piped version, or a single-command:
find . -name '*cache*' -prune -o -print0 | xargs -0 grep -il "samson"
or
find . -name '*cache*' -prune -o -exec grep -iq "samson" {} \; -print
Note, the -l in the first command tells grep to "list the file" and not the line(s) that match. The -q in the second does the same; it tells grep to respond quietly so find will then just print the filename.
You've told grep itself to recurse (twice! -r and -R are synonyms). Since one of the arguments you're passing is . (the top directory), grep is searching in every file (some of them twice, or even more if they're in subdirectories).
If you're going to use find and grep, do this:
find . -path './cach*' -prune -o -print0 | xargs -0 grep -i "samson"
Using -print0 and -0 makes your script work even with file names that contain spaces or punctuation characters.
However, you probably don't need to bother with find here, since GNU grep is capable of excluding directories:
grep -R --exclude-dir='cach*' -i "samson" .
(This also excludes ./deeply/nested/directory/cache. If you only want to exclude cache directories at the toplevel, use find as you did.)
Use the -exec option on find instead of piping them to another command. From there you can use grep "samson" {} \; to look for samson in each file listed.
For example:
find . -wholename './cach*' -prune -o -exec grep "samson" "{}" +

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