bash 匹配后在线打印所有内容
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Print everything on line after match
提问by twk789
I have an large textfile that contains an unique string in the middle. What i want to do is to print everything AFTER the string by using grep.
我有一个大文本文件，中间包含一个唯一的字符串。我想要做的是使用 grep 在字符串之后打印所有内容。
cat textfile | grep "target_string"
This highlights target_string but prints the whole file
cat textfile | grep -o "target_string"
This prints only target_string
cat textfile | grep -o "target_string*"
This prints only target_string
How can i print everything after target_string and nothing before?
如何在 target_string 之后打印所有内容而之前没有任何内容？
采纳答案by ysdx
You forgot the '.':
你忘记了“.”：
cat textfile | grep -o "target_string.*"
回答by chimeric
回答by geekosaur
With GNU grep, try -B0 -A999999999or similar. A better choice might be awk:
使用 GNU grep，尝试-B0 -A999999999或类似。更好的选择可能是awk：
awk '/target_string/ {seen = 1}
seen {print}'
If (your problem specification is slightly unclear) you don't also need to print the matching line, sedis even shorter:
如果（你的问题说明有点不清楚）你也不需要打印匹配的行，sed甚至更短：
sed '1,/target_string/d'
回答by Chris Koknat
This will print everything after each match, on that same line only:
这将在每次匹配后打印所有内容，仅在同一行上：
perl -lne 'print if /target_string(.*)/' textfile
This will do the same, except it will also print all subsequent lines:
这将执行相同的操作，除了它还会打印所有后续行：
perl -lne 'if ($found){print} else{if (/target_string(.*)/){print ; $found++}}' textfile
