How can I replace ; with ;\n (semicolon followed by a newline) in sed?
I've tried building off of
sed s/;/\\n/g file
and
sed -e '/;/G' file
but I can't get either to work
You need to cheat a bit: in bash you can say
sed $'s/;/;\\\n/g'
or, portably (POSIX):
sed "s/;/;$(printf '\\\n')/g"
sed does not portably/reliably handle backslash-escapes anywhere but in the pattern, and even there it's limited (POSIX only requires that \n be handled, not \t or the others). Note that you also need a backslash before the \n so sed doesn't read it as the end of the command.
sed -ie 's/;/;\n/g' <file>
That's assuming you want to do it inline in the file, remove the "i" and just use "-e" if that's not the case.
Related
I stumble upon the command sed -e 's/ /\'$'\n/g'that supposedly takes an input and split all spaces into new lines. Still, I don't quite get how the '$' works in the command. I know that s stands for substitute, / / stands for the blank spac, \n stands for new line and /g is for global replacement, but not sure how \'$' fits in the picture. Anybody who can shed some light here will be much appreciated.
Basically it's meant for platform portability. With GNU sed it would be just
sed -e 's/ /\n/g'
because GNU sed is able to interpret \n as new line.
However, other versions of sed, like the BSD version (that comes with MacOS) do not interprete \n as newline.
That's why the command is build out of two parts
sed -e 's/ /\' part2: $'\n/g'
The $'\n/g' is an ANSI C string parsed by the shell before executing sed. Escape sequences like \n will get expanded in such strings. Doing so, the author of the command passed a literal new line (0xa) to the sed command rather than passing the escape sequence \n. (0x5c 0x6e).
One more thing, since the newline (0xa) is a command separator in sed, it needs to get escaped. That's why the \ at the end of the first part.
Alternatively you could just use a multiline version:
sed -e 's/ /\
/g'
Btw, I would have written the command like
sed -e 's/ /\'$'\n''/g'
meaning just putting the $'\n' into the ANSI C string. Imo that's better to understand.
how can I replace a tab or empty spaces with a single semicolon?
This is what I'm trying:
sed -i.bak 's/(\s)+|(\t)+/,/g' aws_lab.txt
+/ was unexpected at this time.
Thanks
Just use the appropriate POSIX character class:
sed 's/[[:space:]]\{1,\}/,/g' aws_lab.txt
Very few seds will recognize the \s shorthand for [[:space:]] and since a tab is one of the characters included in [:space:] you dont need to specify it separately. Also, + is an ERE metacharacter, not BRE as supported by sed by default so you'd need to add the -E arg to sed to use it (only supported by GNU and OSX sed variants).
This might work for you (GNU sed):
sed -i.bak 's/[ \t][ \t]*/:/' file
Or
sed -ri.bak 's/[ \t]+/:/' file
Where there is an explicit space infront of the tab:
sed -i.bak 's/[ \t][ \t]*/:/g' file
Will replace repeated patterns throughout a line.
I am trying to use sed to replace strings with special characters in a text file. The sed command is becoming too much complicated. If someone could please help me with the exact command.
Code -
sed -i 's;PS1='${HOSTNAME} [$ORACLE_SID] $PWD> ';PS1="${COL_YELLOW}'CUSTOMER TEST:${HOSTNAME}:[$ORACLE_SID]:$PWD> '${COL_END}";g'
I tried to escape the special characters as below but its not working.
sed -i 's;PS1=\'\${HOSTNAME} [\$ORACLE_SID] \$PWD> \';PS1="\${COL_YELLOW}\'CUSTOMER TEST:\${HOSTNAME}:[\$ORACLE_SID]:\$PWD> \'\${COL_END}";g' .bash_profile_backup
This might work for you (GNU sed):
sed -i 's|PS1='\''${HOSTNAME} \[$ORACLE_SID\] $PWD> '\''|PS1="${COL_YELLOW}'\''CUSTOMER TEST:${HOSTNAME}:[$ORACLE_SID]:$PWD> '\''${COL_END}"|g' file
N.B. ' need to be quoted in both the pattern and replacement whereas [] needs to be escaped in the pattern only.
I would like to use (GNU) sed to do a simple search and replace. The issue is that I'm searching for a special character and it might be the reason it failed for me.
The input is:
^A9=139^A35=V^A34=9^A49=xxxx^A52=20140527-06:18:43.759^A5
and I want to replace the ^A with ;. I used:
sed -i '/s/^A/;/g' file.log
but I didn't get anything.
Your command should be,
sed -i 's/\^A/;/g' file
Command you tried,
sed -i '/s/^A/;/g' file.log
| |
| |______________You have to escape this special character. Because in general(regex) it means the starting point.
[No need to use `/` before s]
Example:
$ sed 's/\^A/;/g' file
;9=139;35=V;34=9;49=xxxx;52=20140527-06:18:43.759;5
^ has a special meaning with regular expressions. Use \^ (or potentially \\^, depending on how bash escapes things, I never quite remember it).
sed 's_((checksum|compressed)=\").*(\")_\1\2_' -i filename
I am using this command to replace the checksum and compressed filed with empty? But it didn't change anything?
for example, I want change this line " checksum="XXXXX" with checksum="", and also replace
compressed="XXXX" with compressed=""
What is wrong with my sed command?
It's because sed uses a funny regex dialect by default: you have to escape capturing brackets.
If you want to use "normal" regex that you're familiar with, use the -r flag (if you're on unix, GNU sed) or the -E flag (Mac OS X BSD sed):
sed -r 's_((checksum|compressed)=\").*(\")_\1\3_' -i filename
Additionally, note that you have three sets of capturing brackets in your sed, and I think you want to change the \1\2 to \1\3. (\1 contains checksum=", \2 contains checksum, and \3 contains ").
(For interest, here's how you would do it without the extended-regexp (-r/-E) flag, note that capturing brackets and the OR | are only considered in the regex sense if they are escaped:
sed 's_\(\(checksum\|compressed\)=\"\).*\(\"\)_\1\3_' -i filename
)
This might work for you:
echo 'checksum="XXXXX" compressed="YYYYYYY"' |
sed 's/\(checksum\|compressed\)="[^"]*"/\1=""/g'
checksum="" compressed=""
In sed (without the -r switch), ()|+?{}'s must have a \ prepended to give them the qualities of grouping. alternation, one or more, zero or one and intervals. .[]* work as metacharacters either way.
Try:
sed 's/\(\(checksum\|compressed\)\)="[^"]*"/\1=""/' -i filename