I have the following .sed script:
# replace female-male with F-M
s/female/F/
s/male/M/
# capitalize the name when the sport is volleyball or taekwondo
s/^([^,]*,)([^,]+)((,[^,]*){5},(volleyball|taekwondo),)/\1\U\2\L\3/
And the following csv file (first 10 lines)
id,name,nationality,sex,date_of_birth,height,weight,sport,gold,silver,bronze,info
736041664,A Jesus Garcia,ESP,male,1969-10-17,1.72,64,athletics,0,0,0,
532037425,A Lam Shin,KOR,female,1986-09-23,1.68,56,handball,0,0,0,
435962603,Aaron Brown,CAN,male,1992-05-27,1.98,79,athletics,0,0,1,
521041435,Aaron Cook,MDA,male,1991-01-02,1.83,80,taekwondo,0,0,0,
33922579,Aaron Gate,NZL,male,1990-11-26,1.81,,cycling,0,0,0,
173071782,Aaron Royle,AUS,male,1990-01-26,1.80,67,triathlon,0,0,0,
266237702,Aaron Russell,USA,male,1993-06-04,,98,volleyball,0,0,1,
382571888,Aaron Younger,AUS,male,1991-09-25,1.93,100,football,0,0,0,
87689776,Aauri Lorena Bokesa,ESP,female,1988-12-14,1.80,62,athletics,0,0,0,
The output must be done by the following command
sed -f script.sed ./file.csv
The problem I have is that despite making sure the regex is matching all the pertinent lines, I can only get it to replace the female-male values with F-M, the rest of the file is still the exact same. The names are not being capitalized.
If I run each regex directly (i.e 'sed -E 's/^([^,],)([^,]+)((,[^,]){5},(volleyball|taekwondo),)/\1\U\2\L\3/' file.csv') it works. But I need to do it via script, and with -f.
What am I missing? Thank you.
You still need to indicate that you're using extended regular expresssions:
sed -Ef script.sed file.csv
Otherwise, sed uses basic regular expressions, where escaping rules are different, specifically for () for capture groups, and {} for counts.
Have you tried using sed -Ef <script> <csv file>? You need -E to use extended regex expressions.
sed -i '/#if UTS_UBUNTU_RELEASE_ABI > 255/c\/*#if UTS_UBUNTU_RELEASE_ABI > 255' /usr/src/ixgbevf-2.16.4/src/kcompat.h
I am trying to understand the above command but couldn't figure out what c\ is doing here?
This sed command says to test lines for the regular expression #if UTS_UBUNTU_RELEASE_ABI > 255 (because it starts with forward slash), if so, use the change command to replace the whole line with whatever follows. (the -i means in place.)
In this case, it will change the matching line to be the beginning of a block comment (inserts /*) per my local testing.
I would request some help with a basic shell script that should do the following job.
File a particular word from a given file (file path is always constant)
Backup the file
Delete the specific word or replace the word with ;
Save the file changes
Example
File Name - abc.cfg
Contains the following lines
network;private;Temp;Windows;System32
I've used the following SED command for the operation
sed -i -e "/Temp;/d" abc.cfg
The output is not as expected. The complete line is removed instead of just the word Temp;
Any help would be appreciated. Thank you
sed matches against lines, and /d is the delete directive, which is why you get a deleted line. Instead, use substitution to replace the offending word with nothing:
sed 's/Temp;//g' abc.cfg
The /g modifier means "globlal", in case the offending word appears more than once. I would hold off on the -i (inline) flag until you are sure of your command, in general, or use -i .backup.
Thank you. I used your suggestion but couldn't get through. I appreciate the input though.
I was able to achieve this using the following SED syntax
sed -e "s/Temp//g" -i.backup abc.cfg
I wanted to take the backup before the change & hence -i was helpful.
I am trying to use a sed command to replace specials characters in my file.
The characters are %> to replace by ].
I'am using sed -r s/\%>\/\]\/g but i have this error bash: /]/g: No such file or directory, looks like sed doesn't like it.
Put your sed code inside quotes and also add the file-path you want to work with and finally don't escape the sed delimiters.
$ echo '%>' | sed 's/%>/]/g'
]
ie,
sed 's/%>/]/g' file
To complement Avinash Raj's correct and helpful answer:
Since you were using an overall unquoted string (neither single- nor double-quoted), you were on the right track by \-escaping individual characters in your sed command.
However, you neglected to \-quote >, which is what caused your problem:
> is one of the shell's so-called metacharacters
Metacharacters have special meaning and separate words
Thus, s/\%>\/\]\/g is mistakenly split into 2 arguments by >:
s/\% is passed to sed - as s/%, because the shell removes the \ instances (a process called quote removal).
As you can see, this is not a valid sed command, but that doesn't even come into play - see below.
>\/\]\/g is interpreted by the shell (bash), because it starts with output-redirection operator >; after quote removal, the shell sees >/]/g, tries to open file /]/g for writing, and fails, because your system doesn't have a subdirectory named ] in its root directory.
bash tries to open an output file specified by a redirection before running the command and, if it fails to open the file, does not run the command - which is what happened here:
bash complained about the nonexistent target directory and aborted processing of the command - sed was never even invoked.
Upshot:
In a string that is neither enclosed in single nor in double-quotes, you must \-quote:
all metacharacters: | & ; ( ) < > space tab
additionally, to prevent accidental pathname expansion (globbing): * ? [
Also note that if you need to quote (escape) characters for sed,you need to add an extra layer of quoting; for instance to instruct sed to use a literal . in the regex, you must pass \\. - two backslashes - so that sed sees the properly escaped \..
Given the above, it is much simpler to (habitually) use single quotes around your sed command, because it ensures that the string is passed as is to sed.
Let's compare a working version of your command to the one from Avinash Raj's answer (leaving out the -r for brevity):
sed s/\%\>\/\]\/g # ok - all metachars. \-quoted, others are, but needn't be quoted
sed s/%\>/]/g # ok - minimum \-quoting
sed 's/%>/]/g' # simplest: single-quoted command
I'm not sure whether I got the question correctly. If you want to replace either % or > by ] then sed is not required here. Use tr in this case:
tr '%>' ']' < input.txt
If you want to replace the sequence %> by ] then the sed command as shown by #AvinashRaj is the way to go.
I couldn't find an answer for this exact problem, so I'll ask it.
I'm working in Cygwin and want to reference previous commands using !n notation, e.g., if command 5 was which ls, then !5 runs the same command.
The problem is when trying to do substitution, so running:
!5:s/which \([a-z]\)/\1/
should just run ls, or whatever the argument was for which for command number 5.
I've tried several ways of doing this kind of substitution and get the same error:
bash: :s/which \([a-z]*\)/\1/: substitution failed
As far as I can tell the s/old/new/ history substitution syntax only does simple string substitution; it does not support full regexes. Here's what man bash has to say:
s/old/new/
Substitute new for the first occurrence of old in the event line. Any delimiter can be used in place of /. The final delimiter is optional if it is the last character of the event line. The delimiter may be quoted in old and new with a single backslash. If & appears in new, it is replaced by old. A single backslash will quote the &. If old is null, it is set to the last old substituted, or, if no previous history substitutions took place, the last string in a !?string[?] search.
Never fear, though. There are in fact easier ways to accomplish what you are trying to do:
!$ evaluates to the last argument of the previous command:
# ls /etc/passwd
/etc/passwd
# vim !$
vim /etc/passwd
!5:$ evaluates to the last argument of command #5:
# history
...
5: which ls
...
# !5:$
ls
You can also use Alt+. to perform an immediate substitution equivalent to !$. Alt+. is one of the best bash tricks I know.
This worked for me using Bash in Cygwin (note that my which ls command was number 501 in my history list; not 5 like yours):
$(!501 | sed 's/which \([a-z]\)/\1/')
You could also do it this way (which is shorter/cleaner):
$(!501 | sed 's/which //')