I know the typical sed editing statements, such as
sed "s/substitutethis/withthis/g" file
However, when I searched for how to extract lines between two phrases, I found this:
sed -n "/startphrase/,/stopphrase/p" file
Could someone explain, please? The man page does not help me.
I know that -n suppresses echoing of output. Why do they use -n and p? Leaving out both does not work. What does the comma stand for?
This:
"/startphrase/,/stopphrase/p"
means "from the line containing startphrase to the line containing stopphrase, print".
There are many sed tutorials that talk about "range of lines".
Related
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.
Please bear with me as I'm new to the forums and tried to do my research before posting this. What I'm trying to do is to use sed to look through multiple lines of a file and any line that contains the words 'CPU Usage" I want it to comment out that line and also 19 lines immediately after that.
Example file.txt
This is some random text CPU USAGE more random text
Line2
Line3
Line4
Line5
etc.
I want sed to find the string of text CPU usage and comment out the line and the 19 lines following
#This is some random text CPU USAGE more random text
#Line2
#Line3
#Line4
#Line5
#etc.
This is what I've been trying but obviously it is not working since I'm posting on here asking for help
sed '/\/(CPU Usage)s/^/#/+18 > File_name
sed: -e expression #1, char 17: unknown command: `^'
I'd like to be able to use this on multiple files. Any help you can provide is much appreciated!
GNU sed has a non-standard extension (okay, it has many non-standard extensions, but there's one that's relevant here) of permitting /pattern/,+N to mean from the line matching pattern to that line plus N.
I'm not quite sure what you expected your sed command to do with the \/ part of the pattern, and you're missing a single quote in what you show, but this does the trick:
sed '/CPU Usage/,+19 s/^/#/'
If you want to overwrite the original files, add -i .bak (or just -i if you don't mind losing your originals).
If you don't have GNU sed, now might be a good time to install it.
This can easily be done with awk
awk '/CPU Usage/ {f=20} f && f-- {$0="#"$0}1' file
When CPU Usage is found, set flag f=20
If flag f is true, decrements until 0 and for every time, add # in front of the line and print it.
Think this should work, cant test it, if anyone finds something wrong just let me know :)
awk '/CPU Usage/{t=1}t{x++;$0="#"$0}x==19{t=0;x=0}1' file
This should be extremely simple, but for the life of me I just can't get gnu-sed to do it this afternoon.
The file in question has lines that look like this:
PART NUMBER PART NUMBER QUANTITY WEIGHT -999 -4,999 -9,999
w/ UL APPROVAL
MIN-3
I need to prepend every line like the "MIN-3" line with a ">" character, and the only thing specifically differentiating those lines from the others are two things:
The first character is a space " ".
The lines do not contain a comma.
I've tried mostly things like any of the following:
/^ +[^,]+$/ s/^/>/
/^ +[\w\-]+$/ s/^/>/
/^ +(\w|\-)+$/ s/^/>/
I will admit, I am somewhat new to sed. :)
Edit: Answers that use perl, or awk could also be appreciated, though my initial target is sed.
try this:
sed '/^ [^,]*$/s/^/>/'
the output is, only the line with MIN-3 with leading >
sed default uses basic regex. so the + should be \+ in your script. I think that could be the problem killing your time. You could add -r however, to let sed use extended-regex.
According to your description this should do:
sed 's/^\([ ][^,]*\)$/> \1/' input
which matches the complete line if the line starts with a space and then contains anything but a comma until the end.
Here is a simple answer:
sed 's/^ [^,]*$/>&/'
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
I have many lines of the form
ko04062 ko:CXCR3
ko04062 ko:CX3CR1
ko04062 ko:CCL3
ko04062 ko:CCL5
ko04080 ko:GZMA
and would dearly like to get rid of the ko: bit of the right-hand column. I'm trying to use sed, as follows:
echo "ko05414 ko:ITGA4" | sed 's/\(^ko\d{5}\)\tko:\(.*$\)/\1\2/'
which simply outputs the original string I echo'd. I'm very new to command line scripting, sed, pipes etc, so please don't be too angry if/when I'm doing something extremely dumb.
The main thing that is confusing me is that the same thing happens if I reverse the \1\2 bit to read \2\1 or just use one group. This, I guess, implies that I'm missing something about the mechanics of piping the output of echo into sed, or that my regexp is wrong or that I'm using sed wrong or that sed isn't printing the results of the substitution.
Any help would be greatly appreciated!
sed is outputting its input because the substitution isn't matching. Since you're probably using GNU sed, try this:
echo "ko05414 ko:ITGA4" | sed 's/\(^ko[0-9]\{5\}\)\tko:\(.*$\)/\1\2/'
\d -> [0-9] since GNU sed doesn't recognize \d
{} -> \{\} since GNU sed by default uses basic regular expressions.
This should do it. You can also skip the last group and simply use, \1 instead, but since you're learning sed and regex this is good stuff. I wanted to use a non-capturing group in the middle (:? ) but I could not get that to play with sed for whatever reason, perhaps it's not supported.
sed --posix 's/\(^ko[0-9]\{5\}\)\( ko:\)\(.*$\)/\1 \3/g' file > result
And ofcourse you can use
sed --posix 's/ko://'
You don't need sed for this
Here is how you can do it with bash:
var="ko05414 ko:ITGA4"
echo ${var//"ko:"}
${var//"ko:"} replaces all "ko:" with ""
See Manipulating Strings for more info
#OP, if you just want to get rid of "ko:", then
$ cat file
ko04062 ko:CXCR3
ko04062 ko:CX3CR1
ko04062 ko:CCL3
ko04062 ko:CCL5
some text with a legit ko: this ko: will be deleted if you use gsub.
ko04080 ko:GZMA
$ awk '{sub("ko:","",$2)}1' file
ko04062 CXCR3
ko04062 CX3CR1
ko04062 CCL3
ko04062 CCL5
some text with a legit ko: this ko: will be deleted if you use gsub.
ko04080 GZMA
Jsut a note. While you can use pure bash string substitution, its only more efficient when you are changing a single string. If you have a file, especially a big file, using bash's while read loop is still slower than using sed or awk.