I need to hide the IP addresses in the log files for security reasons. The IP addresses are of version 4 and 6. How do I hide the addresses in a way that, IPv4 example 123.4.32.16 is replaced by x.x.x.x and IPv6 example 232e:23o5:te43:5423:5433:0000:ef09:23ff is replaced by x:x:x:x:x:x:x:x? Is it possible to do this using a single sed command?
You might want to use find and sed for this.
Let's assume your logs have the extension ".log":
find /path/to/logs -type f -name '*.log' -exec \
sed -i -e 's,[0-9]\+\(\.[0-9]\+\)\{3\},x.x.x.x,g' \
-e 's,[0-9a-f]\+\(:[0-9a-f]\+\)\{7\},x:x:x:x:x:x:x:x,gi' {} \;
How does this work?
First, we ask find to recursively locate files with the .log extension starting from /path/to/logs. -type f tells find we wan't to find regular files.
For each file, it will execute sed. The -i argument tells sed you want to edit the file in place. (Check out http://www.grymoire.com/Unix/Sed.html)
One solution using find and perl:
find /the/directory -type f -exec perl -pi -e '
s/\b\d{1,3}(\.\d{1,3}){3}\b/x.x.x.x/g;
s/\b[a-f\d]{1,4}(:[a-f\d]{1,4}){7}\b/x:x:x:x:x:x:x:x/gi' {} \;
(type on one line)
Well, first you should probably just fix whatever is doing the logging to log the way you want to.
Now if you need to go back and modify historical files, you might consider using sed
sed -e 's/\b(\d{1,3}\.){3}\d{1,3}\b/x.x.x.x/' /path/to/file
sed -e 's/\b([:xdigit:]{4}:){7}[:xdigit:]{4}\b/x.x.x.x.x.x.x.x/' /path/to_file
I use this:
find . -name "*.log" -exec grep -izl PATTERN {} \; | xargs perl -i.orig -e -n 's/PATTERN/REPLACEMENT/g'
You'd want to insert your PATTERN(s) and replace *.log with something else depending on the name of your log files.
The -i.orig backs up the files being replaced with an extension of .orig.
I found that this was relatively faster than other things I tried. find/grep combo to indentify candidates, then perl to do the work.
Related
I have a specific case with sed.
As part of the build process, we are required to find and replace the version numbers of required artifacts with the newly created tag name across multiple modules (pom.xml) files.
The command we use is this:
find . -name "*.xml" -print0 | xargs -0 sed -i 's/6\.0\.0\.0/6.0.0.0.001/g'
The edge case we have is this:
There are some modules that have similar version numbers but are already frozen versions coming in from our repository. These need not be changed.
Entries such as <modulename-version>modulename-6.0.0.0.016</modulename-version> are present in the pom.xml's but do not need to be changed.
Is there a way to ignore a pattern of 6\.0\.0\.0\.\d{3} with sed?
The entire setup is intended to run un-attended via python-fabric on our remote build server and we really dont want to wake up in the night to try and solve a problem where a module modulename-6.0.0.0.001.016.jar was not found!
Any help in this space would be most appreciated!
Change your sed command to:
'/6\.0\.0\.0\.\d{3}/!s/6\.0\.0\.0/6.0.0.0.001/g'
Or
'/6\.0\.0\.0\.\d{3}/b; s/6\.0\.0\.0/6.0.0.0.001/g'
Sed may also not accept \d, so you can just use [0-9]:
'/6\.0\.0\.0\.[0-9]{3}/!s/6\.0\.0\.0/6.0.0.0.001/g'
{3} also may need -r
`sed -r ...`
Complete commands:
find . -name "*.xml" -print0 | xargs -0 sed -ri '/6\.0\.0\.0\.[0-9]{3}/!s/6\.0\.0\.0/6.0.0.0.001/g'
find . -name "*.xml" -print0 | xargs -0 sed -ri '/6\.0\.0\.0\.[0-9]{3}/b; s/6\.0\.0\.0/6.0.0.0.001/g'
I would like to swap out every instance of "/this/name/" with "/that/name/" (within the files of a directory) - I'm just not sure how.
Is there a good way of combining the two commands below (or something equivalent) to search/regex an entire directory of files recursively?
• perl -pi -e "s/2f\x74\x68\x69\x73\x2f\x6e\x61\x6d\x65\x2f/\2f\x74\x68\x61\x74\x2f\x6e\x61\x6d\x65\x2f/g" /some/directory
• find . -name "*"
The first Perl example works fine for an individual file, just not a bunch of them. The Find example is of course just an example. I've used these two previous questions as some reference:
[Find] Unix find: multiple file types
[Perl] RegEx within perl -pi -e
Sure, use the -exec flag of find
find /path -type f -exec perl -pi -e "..." {} \;
I added -type f because I think you want to execute this for files only.
I'm refactoring some code, and I decided to replace one name by another, let's say foo by bar. They appear in multiple .cc and .h files, so I would like to change from:
Foo key();
to
Bar key();
that's it, replace all the occurrences of Foo by Bar in Unix. And the files are in the same directory. I thought about
sed -e {'s/Foo/Bar/g'}
but I'm unsure if that's going to work.
This should do the trick:
sed -i'.bak' 's/\bFoo\b/Bar/g' *.files
I would use sed:
sed -i.bak -e '/Foo/ s//Bar/g' /path/to/dir/*.cc
Repeat for the *.h files
I don't use sed alot, but iF you have access to Perl on the command line (which many unix's do) you can do:
perl -pi -e 's/Foo key/Bar key/g' `find ./ -name '*.h' -o -name '*.cc'`
This will find (recursively) all files in the current directory ending with .h or .cc and then use Perl to replace 'Foo key' with 'Bar key' in each file.
I like Jaypal's sed command. It useds \b to ensure that you only replace full words (Foo not Foobar) and it makes backup files in case something went wrong.
However, if all of your files are not in one directory, then you will need to use a more sophisticated method to list them all. Use the find command to send them all to sed:
find . -print0 -regex '.*\.\(cc\|h\)' | xargs -0 sed -i'.bak' 's/\bFoo\b/Bar/g'
You probably have perl installed (if its UNIX), so here's something that should work for you:
perl -e "s/Foo/Bar/g;" -pi.save $(find path/to/DIRECTORY -type f)
Note, this provides a backup of the original file, if you need that as a bit of insurance.
Otherwise, you can do what #Kevin mentioned and just use an IDE refactoring feature.
Note: I just saw you're using Vim, here's a quick tutorial on how to do it
I have changed up my director structure and I want to do the following:
Do a recursive grep to find all instances of a match
Change to the updated location string
One example (out of hundreds) would be:
from common.utils import debug --> from etc.common.utils import debug
To get all the instances of what I'm looking for I'm doing:
$ grep -r 'common.' ./
However, I also need to make sure common is preceded by a space. How would I do this find and replace?
It's hard to tell exactly what you want because your refactoring example changes the import as well as the package, but the following will change common. -> etc.common. for all files in a directory:
sed -i 's/\bcommon\./etc.&/' $(egrep -lr '\bcommon\.' .)
This assumes you have gnu sed available, which most linux systems do. Also, just to let you know, this will fail if there are too many files for sed to handle at one time. In that case, you can do this:
egrep -lr '\bcommon\.' . | xargs sed -i 's/\bcommon\./etc.&/'
Note that it might be a good idea to run the sed command as sed -i'.OLD' 's/\bcommon\./etc.&/' so that you get a backup of the original file.
If your grep implementation supports Perl syntax (-P flag, on e.g. Linux it's usually available), you can benefit from the additional features like word boundaries:
$ grep -Pr '\bcommon\.'
By the way:
grep -r tends to be much slower than a previously piped find command as in Rob's example. Furthermore, when you're sure that the file-names found do not contain any whitespace, using xargs is much faster than -exec:
$ find . -type f -name '*.java' | xargs grep -P '\bcommon\.'
Or, applied to Tim's example:
$ find . -type f -name '*.java' | xargs sed -i.bak 's/\<common\./etc.common./'
Note that, in the latter example, the replacement is done after creating a *.bak backup for each file changed. This way you can review the command's results and then delete the backups:
$ find . -type f -name '*.bak' | xargs rm
If you've made an oopsie, the following command will restore the previous versions:
$ find . -type f -name '*.bak' | while read LINE; do mv -f $LINE `basename $LINE`; done
Of course, if you aren't sure that there's no whitespace in the file names and paths, you should apply the commands via find's -exec parameter.
Cheers!
This is roughly how you would do it using find. This requires testing
find . -name \*.java -exec sed "s/FIND_STR/REPLACE_STR/g" {}
This translates as "Starting from the current directory find all files that end in .java and execute sed on the file (where {} is a place holder for the currently found file) "s/FIND_STR/REPLACE_STR/g" replaces FIND_STR with REPLACE_STR in each line in the current file.
I need to recursively search directories and replace a string (say http://development:port/URI) with another (say http://production:port/URI) in all the files where ever it's found. Can anyone help?
It would be much better if that script can print out the files that it modified and takes the search/replace patterns as input parameters.
Regards.
find . -type f | xargs sed -i s/pattern/replacement/g
Try this:
find . -type f | xargs grep -l development | xargs perl -i.bak -p -e 's(http://development)(http://production)g'
Another approach with slightly more feedback:
find . -type f | while read file
do
grep development $file && echo "modifying $file" && perl -i.bak -p -e 's(http://development)(http://prodution)g' $file
done
Hope this helps.
It sounds like you would benefit from a layer of indirection. (But then, who wouldn't?)
I'm thinking that you could have the special string in just one location. Either reference the configuration settings at runtime, or generate these files with the correct string at build time.
Don't try the above within a working SVN / CVS directory, since it will also patch the .svn/.cvs, which is definitely not what you want. To avoid .svn modifications, for example, use:
find . -type f | fgrep -v .svn | xargs sed -i 's/pattern/replacement/g'
Use zsh so with advanced globing you can use only one command.
E.g.:
sed -i 's:pattern:target:g' ./**
HTH