I'd like to do something similar to sed in Perl, namely be able to delete lines matching a certain pattern.
Given this input:
abcd
edfd
abcd
derder
abcd
erre
I want to remove the lines containing bc. How can I do this?
I had to use double quotes on Windows:
perl -ne "print unless /bc/" file
This is a FAQ.
How do I change, delete, or insert a line in a file, or append to the beginning of a file?
If you're programming in Perl then it's well worth taking a couple of hours to familiarise yourself with the FAQ.
Related
I am struggling to work out how to get a , out from inbetween various patterns such as:
500,000
xyz ,CA
I have tried something like:
sed -E "s/\([a-zA-Z]*\),([a-zA-Z]*\)/\([a-zA-Z]*\) ([a-zA-Z]*\)/g" $file -i
It picks up the first pattern, but then over writes it with the second pattern, I feel like I am missing something very simple and I can't work it out, any help really appreciated.
You're missing the notion of capture groups, I think. To refer to a parenthesized portion of the search within the replacement string, use \1 for the first group, \2 for the second group, etc.
The modified line would be:
sed -E "s/([a-zA-Z]),([a-zA-Z])/\1 \2/g" $file -i
Rather than replacing the part that matches the first ([a-zA-Z]) with the literal text "([a-zA-Z])", this modified line just copies the matched portion into the output (and likewise for the second group).
I have this text file where I need to first find a string "BEGINNING" and then find a string "HERE" after the first "BEGINNING" but only once. And there can be any amount of strings in between. This must be done with SED commands so no awk. I know I can simply do /BEGINNING/ to find the first one but I don't know how to put the two together in one SED command.
something like this?
$ sed -n '/BEGINNING/,${/HERE/{p;q}}' file
may be supported only by GNU sed, not sure.
I want to insert a range of lines from a file, say something like 210,221r before the first occurrence of a pattern in a bunch of other files.
As I am clearly not a GNU sed expert, I cannot figure how to do this.
I tried
sed '0,/pattern/{210,221r file
}' bunch_of_files
But apparently file is read from line 210 to EOF.
Try this:
sed -r 's/(FIND_ME)/PUT_BEFORE\1/' test.text
-r enables extendend regular expressions
the string you are looking for ("FIND_ME") is inside parentheses, which creates a capture group
\1 puts the captured text into the replacement.
About your second question: You can read the replacement from a file like this*:
sed -r 's/(FIND_ME)/`cat REPLACEMENT.TXT`\1/' test.text
If replace special characters inside REPLACEMENT.TXT beforehand with sed you are golden.
*= this depends on your terminal emulator. It works in bash.
In https://stackoverflow.com/a/11246712/4328188 CodeGnome gave some "sed black magic" :
In order to insert text before a pattern, you need to swap the pattern space into the hold space before reading in the file. For example:
sed '/pattern/ {
h
r file
g
N
}' in
However, to read specific lines from file, one may have to use a two-calls solution similar to dummy's answer. I'd enjoy knowing of a one-call solution if it is possible though.
I have a list of usernames and i would like add possible combinations to it.
Example. Lets say this is the list I have
johna
maryb
charlesc
Is there is a way to use sed to edit it the way it looks like
ajohn
bmary
ccharles
And also
john_a
mary_b
charles_c
etc...
Can anyone assist me into getting the commands to do so, any explanation will be awesome as well. I would like to understand how it works if possible. I usually get confused when I see things like 's/\.(.*.... without knowing what some of those mean... anyway thanks in advance.
EDIT ... I change the username
sed s/\(user\)\(.\)/\2\1/
Breakdown:
sed s/string/replacement/ will replace all instances of string with replacement.
Then, string in that sed expression is \(user\)\(.\). This can be broken down into two
parts: \(user\) and \(.\). Each of these is a capture group - bracketed by \( \). That means that once we've matched something with them, we can reuse it in the replacement string.
\(user\) matches, surprisingly enough, the user part of the string. \(.\) matches any single character - that's what the . means. Then, you have two captured groups - user and a (or b or c).
The replacement part just uses these to recreate the pattern a little differently. \2\1 says "print the second capture group, then the first capture group". Which in this case, will print out auser - since we matched user and a with each group.
ex:
$ echo "usera
> userb
> userc" | sed "s/\(user\)\(.\)/\2\1/"
auser
buser
cuser
You can change the \2\1 to use any string you want - ie. \2_\1 will give a_user, b_user, c_user.
Also, in order to match any preceding string (not just "user"), just replace the \(user\) with \(.*\). Ex:
$ echo "marya
> johnb
> alfredc" | sed "s/\(.*\)\(.\)/\2\1/"
amary
bjohn
calfred
here's a partial answer to what is probably the easy part. To use sed to change usera to user_a you could use:
sed 's/user/user_/' temp
where temp is the name of the file that contains your initial list of usernames. How this works: It is finding the first instance of "user" on each line and replacing it with "user_"
Similarly for your dot example:
sed 's/user/user./' temp
will replace the first instance of "user" on each line with "user."
Sed does not offer non-greedy regex, so I suggest perl:
perl -pe 's/(.*?)(.)$/$2$1/g' file
ajohn
bmary
ccharles
perl -pe 's/(.*?)(.)$/$1_$2/g' file
john_a
mary_b
charles_c
That way you don't need to know the username before hand.
Simple solution using awk
awk '{a=$NF;$NF="";$0=a$0}1' FS="" OFS="" file
ajohn
bmary
ccharles
and
awk '{a=$NF;$NF="";$0=$0"_" a}1' FS="" OFS="" file
john_a
mary_b
charles_c
By setting FS to nothing, every letter is a field in awk. You can then easy manipulate it.
And no need to using capturing groups etc, just plain field swapping.
This might work for you (GNU sed):
sed -r 's/^([^_]*)_?(.)$/\2\1/' file
This matches any charactes other than underscores (in the first back reference (\1)), a possible underscore and the last character (in the second back reference (\2)) and swaps them around.
This is a simple question, I'm not sure if i'm able to do this with sed/awk
How can I make sed search for these 3 lines and replace with a line with a determined string?
<Blarg>
<Bllarg>
<Blllarg>
replace with
<test>
I tried with sed "s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/g" But it just don't seem to find these lines. Probably something with my break line character (?) \n. Am I missing something?
Because sed usually handles only one line at a time, your pattern will never match. Try this:
sed '1N;$!N;s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/;P;D' filename
This might work for you:
sed '/<Blarg>/ {N;N;s/<Blarg>\n<Bllarg>\n<Blllarg>/<test>/}' <filename>
It works as follows:
Search the file till <Blarg> is found
Then append the two following lines to the current pattern space using N;N;
Check if the current pattern space matches <Blarg>\n<Bllarg>\n<Blllarg>
If so, then substitute it with <test>
You can use range addresses with regular expressions an the c command, which does exactly what you are asking for:
sed '/<Blarg>/,/<Blllarg>/c<test>' filename