I am trying to find the whole source code for occurrences of, say, "MY_NAME" and want to replace it with, say, "YOUR_NAME". I already know the files and the line numbers where they occur and i want to make a patch for the same so that anyone running the patch can do the same. Can anyone please help?
You can do it by console. Just use find to locate destination files, and then you can declare what you want to replace with what sentence. In example:
find -name '*' | xargs perl -pi -e 's/MY_NAME/YOUR_NAME/g'
It might be easier to do a sed command, and then generate a patch.
sed -e '12s/MY_NAME/YOUR_NAME/g;32s/MY_NAME/YOUR_NAME/g' file > file2
This will replace MY_NAME with YOUR_NAME on lines 12 and 32, and save the output into file2.
You can also generate a sed script if there are many changes:
#!/bin/sed -f
12s/MY_NAME/YOUR_NAME/g
32s/MY_NAME/YOUR_NAME/g
Then, for applying to many files, you should use find:
find -type f '(' -iname "*.c" -or -iname "*.h" ')' -exec "./script.sed" '{}' \;
Hope this helps =)
Use the command diff to create a patch-file that can then be distributed and applied with the patch-command.
man diff Will give you a lot of information on the process.
Related
Please help me with advice
I need to go through the files in this $tmp directory and make substitutions in the lines according to the rule described in sed
I trained on one file and this sed worked it out normally.
But when I perform a replacement in for , I get some nonsense on the output
for file in $tmp/*.txt; do
find $tmp/ -type f -name "*.txt" -exec sed -i -e 's/, /\n''/g; s/"//g; s/$/\x0A/; s/:[ ]/;/g; s/\n''/;\n''/g; s/^[ t]*//' {} \;
done
How can I fix it?
Thanks…
UP
I did it.
find - It was really unnecessary
#Rob Sweet Thanks!
You don't need the for loop AND the find command. Both essentially do the same thing, which is to iterate through all the files in /tmp that match *.txt.
When you don't need to edit files in a subdir of /tmp, than give sed all files in once.
sed -i your_command *.txt
Off topic:
I don't understand your sed command. Why two single quotes? When you want to replace a newline, consider option -z.
I want to insert a line into the start of multiple specified type files, which the files are located in current directory or the sub dir.
I know that using
find . -name "*.csv"
can help me to list the files I want to use for inserting.
and using
sed -i '1icolumn1,column2,column3' test.csv
can use to insert one line at the start of file,
but now I do NOT know how to pipe the filenames from "find" command to "sed" command.
Could anybody give me any suggestion?
Or is there any better solution to do this?
BTW, is it work to do this in one line command?
Try using xargs to pass output of find and command line arguments to next command, here sed
find . -type f -name '*.csv' -print0 | xargs -0 sed -i '1icolumn1,column2,column3'
Another option would be to use -exec option of find.
find . -type f -name '*.csv' -exec sed -i '1icolumn1,column2,column3' {} \;
Note : It has been observed that xargs is more efficient way and can handle multiple processes using -P option.
This way :
find . -type f -name "*.csv" -exec sed -i '1icolumn1,column2,column3' {} +
-exec do all the magic here. The relevant part of man find :
-exec command ;
Execute command; true if 0 status is returned. 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 everywhere it occurs in the arguments to the command, not just
in arguments where it is alone, as in some versions of find. Both of
these constructions might need to be escaped (with a `\') or quoted to protect
them from expansion by the shell. See the EXAMPLES section for examples of
the use of the -exec option. The specified command is run once for each
matched file. The command is executed in the starting directory. There
are unavoidable security problems surrounding use of the -exec action;
you should use the -execdir option instead
I'm trying to rename a load of files (I count over 200) that either have the company name in the filename, or in the text contents. I basically need to change any references to "company" to "newcompany", maintaining capitalisation where applicable (ie "Company becomes Newcompany", "company" becomes "newcompany"). I need to do this recursively.
Because the name could occur pretty much anywhere I've not been able to find example code anywhere that meets my requirements. It could be any of these examples, or more:
company.jpg
company.php
company.Class.php
company.Company.php
companysomething.jpg
Hopefully you get the idea. I not only need to do this with filenames, but also the contents of text files, such as HTML and PHP scripts. I'm presuming this would be a second command, but I'm not entirely sure what.
I've searched the codebase and found nearly 2000 mentions of the company name in nearly 300 files, so I don't fancy doing it manually.
Please help! :)
bash has powerful looping and substitution capabilities:
for filename in `find /root/of/where/files/are -name *company*`; do
mv $filename ${filename/company/newcompany}
done
for filename in `find /root/of/where/files/are -name *Company*`; do
mv $filename ${filename/Company/Newcompany}
done
For the file and directory names, use for, find, mv and sed.
For each path (f) that has company in the name, rename it (mv) from f to the new name where company is replaced by newcompany.
for f in `find -name '*company*'` ; do mv "$f" "`echo $f | sed s/company/nemcompany/`" ; done
For the file contents, use find, xargs and sed.
For every file, change company by newcompany in its content, keeping original file with extension .backup.
find -type f -print0 | xargs -0 sed -i .bakup 's/company/newcompany/g'
I'd suggest you take a look at man rename an extremely powerful perl-utility for, well, renaming files.
Standard syntax is
rename 's/\.htm$/\.html/' *.htm
the clever part is that the tool accept any perl-regexp as a pattern for a filename to be changed.
you might want to run it with the -n switch which will make the tool to only report what it would have changed.
Can't figure out a nice way to keep the capitalization right now, but since you already can search through the filestructure, issue several rename with different capitalization until all files are changed.
To loop through all files below current folder and to search for a particular string, you can use
find . -type f -exec grep -n -i STRING_TO_SEARCH_FOR /dev/null {} \;
The output from that command can be directed to a file (after some filtering to just extract the file names of the files that need to be changed).
find . /type ... > files_to_operate_on
Then wrap that in a while read loop and do some perl-magic for inplace-replacement
while read file
do
perl -pi -e 's/stringtoreplace/replacementstring/g' $file
done < files_to_operate_on
There are few right ways to recursively process files. Here's one:
while IFS= read -d $'\0' -r file ; do
newfile="${file//Company/Newcompany}"
newfile="${newfile//company/newcompany}"
mv -f "$file" "$newfile"
done < <(find /basedir/ -iname '*company*' -print0)
This will work with all possible file names, not just ones without whitespace in them.
Presumes bash.
For changing the contents of files I would advise caution because a blind replacement within a file could break things if the file is not plain text. That said, sed was made for this sort of thing.
while IFS= read -d $'\0' -r file ; do
sed -i '' -e 's/Company/Newcompany/g;s/company/newcompany/g'"$file"
done < <(find /basedir/ -iname '*company*' -print0)
For this run I recommend adding some additional switches to find to limit the files it will process, perhaps
find /basedir/ \( -iname '*company*' -and \( -iname '*.txt' -or -ianem '*.html' \) \) -print0
I'm trying to return a particular line from files found from this search:
find . -name "database.php"
Each of these files contains a database name, next to a php variable like $dname=
I've been trying to use -exec to execute a grep search on this file with no success
-exec "grep {\}\ dbname"
Can anyone provide me with some understanding of how to accomplish this task?
I'm running CentOS 5, and there are about 100 database.php files stored in subdirectories on my server.
Thanks
Jason
You have the arguments to grep inverted, and you need them as separate arguments:
find . -name "database.php" -exec grep '$dbname' /dev/null {} +
The presence of /dev/null ensures that the file name(s) that match are listed as well as the lines that match.
I think this will do it. Not sure if you need to make any adjustments for CentOS.
find . -name "database.php" -exec grep dbname {} \;
I worked it out using xargs
find . -name "database.php" -print | xargs grep \'database\'\=\> > list_of_databases
Feel free to post a better way if you find one (or what some rep for a good answer)
I tend to habitually avoid find because I've never learned how to use it properly, so the way I'd accomplish your task would be:
grep dbname **/database.php
Edit: This command won't be viable in all cases because it can potentially generate a very long argument list, whereas find executes its command on found files one by one like xargs. And, as I noted in my comment, it's possibly not very portable. But it's damn short ;)
I need to replace some string into another in files. I know how to do that with single file: sed -i 's/a/b/'. But what about recursive function? I think I have to use find . -name * with xargs somehow.
I need your help :)
You are correct, find and xargs are what you want to use. Here's an example which will find all files with the ".ext" file extension in the current folder and all subfolders ,and replace the letter a with the letter b in the files.
find . -name "*.ext" | xargs sed -i 's/a/b/g'