bash 以特定字符串开头的 grep 行
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原文地址: http://stackoverflow.com/questions/43579802/
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grep lines that start with a specific string
提问by Jonathan Matthews
I want to find all the lines in a file that start with a specific string. The problem is, I don't know what's in the string beforehand. The value is stored in a variable.
我想查找文件中以特定字符串开头的所有行。问题是，我事先不知道字符串中的内容。该值存储在一个变量中。
The na?ve solution would be the following:
天真的解决方案如下：
grep "^${my_string}" file.txt;
Because if the Bash variable my_stringcontains ANY regular expression special characters, grepwill cry, and everyone will have a bad day.
因为如果 Bash 变量my_string包含 ANY 正则表达式特殊字符，grep会哭，每个人都会有不好的一天。
You don't want to make grepcry, do you?
你不grep想哭，是吗？
采纳答案by anubhava
You should use awkinstead of grepfor non-regex search using indexfunction:
您应该使用awk而不是grep使用index函数进行非正则表达式搜索：
awk -v s="$my_string" 'index($ cat file
crying now
no cry
$ var="n.*"
$ echo "$var"
n.*
$ grep "^$var" file
no cry
, s) == 1' file
index($0, s) == 1ensures search string is found only at start.
index($0, s) == 1确保仅在开始时找到搜索字符串。
回答by James Brown
What do you mean by if the Bash variable my_stringcontains ANY regular expression special characters, grepwill cry
如果 Bash 变量包含任何正则表达式特殊字符，你是什么意思，会哭my_stringgrep
$ cat file
OMG it's full of
*s
$ var="*"
$ grep "^.\{"${#var}"\}" file|grep -F "$var"
*s
Now that @anubhava spoon-fed me the problem (thank you sir), using grepfor the job does seem to need at least two greps and since you want to grepthe regex characters literally, use -Fin the latter (or fgrep):
现在@anubhava 用勺子给我解决了这个问题（谢谢先生），grep用于这项工作似乎确实需要至少两个greps，并且由于您想要grep字面上的正则表达式字符，请-F在后者（或fgrep）中使用：
${#var}returns varlength in Bash and that amount of chars we extract from the beginning of file to be examined with the latter grep.
${#var}返回varBash 中的长度以及我们从文件开头提取的字符数量，以供后者检查grep。
(Quote from 2001)
（引自2001 年）
