bash UNIX：使用 egrep 或 sed 查找第一次出现字符串的行？
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UNIX: Using egrep or sed to find the line with the first occurrence of a string?
提问by tf.rz
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:
我在 bash shell 中工作，我试图只打印字符串第一次出现的行。例如，对于字符串 ' auir'，如果我有文件 myfile.txt 并且它包含：
123
asdf
4wirajw
forauir somethingelse
starcraft
mylifeforauir
auir
something else
tf.rzauir
I want to output "forauir somethingelse"
我想输出“ forauir somethingelse”
So far, I use the command
到目前为止，我使用命令
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.
这给了我这个字符串的所有出现。我怎样才能只得到“ auir”出现的第一行？如果它只是一个命令或命令管道，那就太好了。
Any insight is greatly appreciated.
任何见解都非常感谢。
采纳答案by beerbajay
This sedcommand
这个sed命令
sed -n '/auir/p' myfile.txt | head -1
sed -n '/auir/p' myfile.txt | head -1
solves your problem.
解决你的问题。
回答by the paul
Use this:
用这个：
grep -m1 auir myfile.txt
回答by potong
This might work for you:
这可能对你有用：
sed '/auir/!d;q' file
or
或者
sed -n '/auir/{p;q}' file
回答by loganaayahee
sed -n -e '4s/auir/auir/p' file
sed -n -e '4s/auir/auir/p' 文件
回答by mawia
Or it can be as simple as this
或者它可以像这样简单
grep auir myFile.txt|head -1
