On Tue, Dec 20, 2011 at 2:30 AM, Bob Proulx <[email protected]> wrote: > lina wrote: >> Bob Proulx wrote: >> > lina wrote: >> >> aaa >> >> model 0 >> >> bbb >> >> ddd >> >> model 1 >> >> ccc >> >> >> >> I want to print out the parts which match the "model 0" and ends with >> >> match "model 1" >> >> >> >> for the final expected output is: >> >> >> >> model 0 >> >> bbb >> >> ddd >> > >> > Try this: >> > >> > sed -n '/^model 1/q;/^model 0/,$p' >> >> Just realize the sed -n '/model 0/,/model 1/'p can also do that. (so >> newbie I was/am). > > Not quite. You said you wanted the output to include the model 0 line > but NOT the model 1 line. Doing /model 0/,/model 1/p prints both of > the model lines including the model 1 line. The way I did it won't. Thanks, Yesterday I got some time, so started doing some sed tutorial. then I thought the questions I asked days ago, so back to this thread, but forget I was looking for the model 0 till meet model 1.
I did not know the q means quit. now I understand. cool. (sometimes always run for other things before one thing really fixed,) Thanks for both of you, Best regards, > >> just still don't understand above sentence. sed -n '/^model 1/q;/^model >> 0/,$p' > > By default sed will print lines at the end of the processing loop. > The -n option tells sed not to print lines by default. Otherwise > every line would be printed. Then for the lines you want to print > there is the explicit 'p' option to print that line.[1] > > Using /model 0/,/model 1/ means print the lines starting on the first > pattern and ending on the second pattern. This prints the lines > between along with the starting and ending lines. > > To avoid printing the ending pattern I did it differently. Let's look > at the second part first. /^model 0/,$p says to print from the model > line all of the way to the end of file. The end of file is designated > by the '$' which is a line number and $ means the last line in the > file when sed reads EOF from the file. I don't know how long the file > will be so just print all lines starting with /^model 0/. > > Then to stop the printing where I want I use /^model 1/q which tells > sed that when it sees the model 1 line that it should quit (exit) and > when it exits it will of course no longer be printing. > > The exit clause must come before the printing clause so that it won't > print the line that you didn't want printed. If they were in the > other order then it would print the closing pattern line and then > execute the exit clause. > > Normally sed will recycle the pattern. If you have multiple strings > then it will reset and print the next set. > > $ printf "a\nb\nCCC\nd\ne\na\nb\nCCC\nd\ne\n" > a > b > CCC > d > e > a > b > CCC > d > e > > $ printf "a\nb\nCCC\nd\ne\na\nb\nCCC\nd\ne\n" | sed -n '/b/,/d/p' > b > CCC > d > b > CCC > d > > See how it reset above? But if you only want to print the first set > then you would still want to 'q' to quit the program early. > > $ printf "a\nb\nCCC\nd\ne\na\nb\nCCC\nd\ne\n" | sed -n '/b/,/d/p;/d/q' > b > CCC > d > > Hope that helps, > Bob > > [1] GNU sed has a bug making it slightly incompatible with the classic > Unix sed program. In Unix sed if you use 'p' without -n the line is > only printed once. > > On HP-UX (System V sed) for example: > > $ echo hello | sed p > hello > > In GNU sed it by default always prints and if there is a command to > print then it prints the line again. > > $ echo hello | sed p > hello > hello > > For portable use never use p without also using -n to avoid duplicated > lines and use that would not be the same on a traditional Unix system. -- To UNSUBSCRIBE, email to [email protected] with a subject of "unsubscribe". Trouble? Contact [email protected] Archive: http://lists.debian.org/CAG9cJmkKEqCAfK-iCajQmp2A6O_cExPquk6rrky7FP3szjCÚ@mail.gmail.com
