I am working in a bash shell and I am trying to print only the line of the first occurrence of the string. For example, for the string ' auir
', if I have the file myfile.txt and it contains:
123
asdf
4wirajw
forauir somethingelse
starcraft
mylifeforauir
auir
something else
tf.rzauir
I want to output " forauir somethingelse
"
So far, I use the command
sed -n '/auir/p' myfile.txt
which gives me all the occurrences of this string. How can I only get the first line that ' auir
' occurs on? It'd be great if it was just a single command or pipeline of commands.
Any insight is greatly appreciated.
用这个:
grep -m1 auir myfile.txt
This sed
command
sed -n '/auir/p' myfile.txt | head -1
solves your problem.
This might work for you:
sed '/auir/!d;q' file
or
sed -n '/auir/{p;q}' file
sed -n -e'4s / auir / auir / p'文件
或者它可以像这样简单
grep auir myFile.txt|head -1
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