UNIX：使用egrep或sed查找第一次出现字符串的行？

Question

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.

Answer 1

用这个：

grep -m1 auir myfile.txt

Answer 2

This sed command

sed -n '/auir/p' myfile.txt | head -1

solves your problem.

Answer 3

This might work for you:

sed '/auir/!d;q' file

or

sed -n '/auir/{p;q}' file

Answer 4

sed -n -e'4s / auir / auir / p'文件

Answer 5

或者它可以像这样简单

grep auir myFile.txt|head -1