I want to search and replace string located in a several files via bash console.
Here is the command I use to find a string in a file:
grep "string" * -r
so the above is for searching, now I need a command to replace the string.
Is that even possible?
http://www.grymoire.com/Unix/Sed.html
It's cranky and difficult, but it's one way to do it.
Here's an example:
sed -i 's/ugly/beautiful/g' /home/bruno/old-friends/sue.txt
This replaces ugly with beautiful in sue.txt.
Related
My file extension is CSV file looks below format in unix server.
"Product_Package_Map_10302017.csv","451","2017-10-30 05:02:26"
"Targeting_10302017.csv","13","2017-10-30 05:02:26",
"Targeting_Options_10302017.csv","42","2017-10-30 05:02:27"
I want to delete a particular line based on filename keyword.
You can use grep -v:
grep -v '^"Product_Package_Map_10302017.csv"' file > file.filtered
'^"Product_Package_Map_10302017.csv"' matches the string "Product_Package_Map_10302017.csv" exactly at the line beginning
or sed can do it in-place:
sed -i '/^"Product_Package_Map_10302017.csv"/d' file
See this related post for other alternatives:
Delete lines in a text file that contain a specific string
See this previous question. A grep-based answer would be my first choice but, as you can see, there are many ways to address this one!
(Would have just commented, but my 'rep' is not yet high enough)
My question is basically the same as this previous question...
How to change all occurrences of a word in all files in a directory
...except I'm trying to change the reference to a header file.
For example,I'm trying to change [lessthan]abc/filename.h[greaterthan] to "filename.h", is this even possible using the same syntax, or should I be looking to whip myself up a quick program to do it?
Thanks
You can do it easily with sed:
sed -i -e 's,<abc/filename.h>,"filename.h",' *
I have a file with name
ROCKET_25_08:00.csv
I want to trim the name of the file to
ROCKET_25_.csv
I tried mv but mv is not what I required because there will be cases where the files may be more than one.
I want the name till the second _.
How to get that in unix.
Please advise.
There are some utilities that provide more flexible renaming. But one solution that won't use anything other but included UNIX tools (like sed) would be:
ls -d * | sed -re 's/^([^_]*_[^_]*_)(.*)(\....)$/mv -v \1\2\3 \1\3/' | bash
This will only work in one directory, it won't process subdirectories.
It's not at all clear what you are actually trying to do, but if you just want to remove text between the last underscore and the period, you can do:
f=ROCKET_25_08:00.csv
echo ${f%_*}_.csv
I have a parameter name like
PAR="DBS_OUT"
and I have a text file(Repla.txt) with below values:
DB_TECH
DB_ADMIN
DB_TERA
DB_APS
These values in the files can defer but the parameter value will remain the same.
Now I have some Unix shell script where I need to find all such values mentioned in the file (Repla.txt)
and replace them with the parameter (PAR). Since the values in the Repla.txt is not fixed I am not able to use the sed command. for eg:
sed 's/old/new/g' input.txt > output.txt
Can anyone please help me.
Thanks
I'm not sure I completely understand what you are trying to do but if you are trying to use the values contained in Repla.txt as the strings that you want to replace in other files then the following bash line will do what you want:
PAR="DBS_OUT"; for FIND in `cat Repla.txt`; do $( find /path/to/files -name 'test?.txt' -exec sed -i "s/$FIND/$PAR/g" '{}' \;); done;
It will replace the strings contained in Repla.txt with the string DBS_OUT in all files that match test?.txt in the dir (and subdirs) /path/to/files. You will need to understand how find works.
Also note that I am not telling sed to backup, you probably want to test this out on some test files before you execute it for real. Hopefully you also have your scripts in source control so its not a big deal if you mess things up.
I hope your replacement on Capital letters only.
sed 's/DBS_[A-Z]*/DBS_OUT/g' repla.txt > destination file
or
sed 's/DBS_[A-Z]*/DBS_OUT/g' repla.txt
I use grep to sort log big file into small one but still there is long dir path in output log file which is common every time.I have to do find and replace every time.
Isnt there any way i can grep -r "format" log.log | execute findnreplce thing?
Sed will do what you want. Basic syntax to replace all the matches of foo with bar in-place in $file is:
sed -i 's/foo/bar/g' $file
If you're just wanting to delete rather than replace, simply leave out the 'bar' (so s/foo//g).
See this tutorial for a lot more detail, such as regex support.
sed -n '/match/s/pattern/repl/p'
Will print all the lines that match the regex match, with all instances of pattern replaced by repl. Since your lines may contain paths, you will probably want to use a different delimeter. / is customary, but you can also do:
sed -n '\#match#s##repl#p`
In the second case, omitting pattern will cause match to be used for the pattern to be replaced.