Linux Grep 不是正则表达式
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Grep not as a regular expression
提问by dotancohen
I need to search for a PHP variable $someVar. However, Grep thinks that I am trying to run a regex and is complaining:
我需要搜索一个 PHP 变量$someVar。但是，Grep 认为我正在尝试运行正则表达式并抱怨：
$ grep -ir "Something Here" * | grep $someVar
Usage: grep [OPTION]... PATTERN [FILE]...
Try `grep --help' for more information.
$ grep -ir "Something Here" * | grep "$someVar"
<<Here it returns all rows with "someVar", not only those with "$someVar">>
I don't see an option for telling grep notto interpret the string as a regex, but to include the $as just another string character.
我没有看到告诉 grep不要将字符串解释为正则表达式的选项，而是将 包含$为另一个字符串字符。
采纳答案by Mr Lister
Escape the $by putting a \in front of it.
$通过\在它前面放一个来逃避它。
回答by Hasturkun
Use fgrep(deprecated), grep -For grep --fixed-strings, to make it treat the pattern as a list of fixed strings, instead of a regex.
使用fgrep(deprecated), grep -Forgrep --fixed-strings使其将模式视为固定字符串列表，而不是正则表达式。
For reference, the documentation mentions (excerpts):
作为参考，文档提到（摘录）：
-F--fixed-stringsInterpret the pattern as a list of fixed strings (instead of regular expressions), separated by newlines, any of which is to be matched. (-F is specified by POSIX.)
fgrepis the same asgrep -F. Direct invocation as fgrep is deprecated, but is provided to allow historical applications that rely on them to run unmodified.
-F--fixed-strings将模式解释为由换行符分隔的固定字符串列表（而不是正则表达式），其中任何一个都将被匹配。（-F 由 POSIX 指定。）
fgrep与 相同grep -F。作为 fgrep 的直接调用已被弃用，但提供它是为了允许依赖它们的历史应用程序未经修改地运行。
For the complete reference, check: https://www.gnu.org/savannah-checkouts/gnu/grep/manual/grep.html
如需完整参考，请查看：https: //www.gnu.org/savannah-checkouts/gnu/grep/manual/grep.html
回答by rkhayrov
grep -Fis a standard way to tell grepto interpret argument as a fixed string, not a pattern.
grep -F是grep将参数解释为固定字符串而不是模式的标准方法。
回答by huelbois
You have to tell grep you use a fixed-string, instead of a pattern, using '-F' :
您必须使用 '-F' 告诉 grep 您使用的是固定字符串，而不是模式：
grep -ir "Something Here" * | grep -F $somevar
回答by Christophe Drevet-Droguet
In this question, the main issue is not about grepinterpreting $as a regex. It's about the shell substituting $someVarwith the value of the environment variable someVar, likely the empty string.
在这个问题中，主要问题不在于grep解释$为正则表达式。这是关于 shell 替换$someVar环境变量的值someVar，可能是空字符串。
So in the first example, it's like calling grepwithout any argument, and that's why it gives you a usageoutput. The second example should not return all rows containing someVarbut all lines, because the empty string is in all lines.
所以在第一个例子中，这就像grep不带任何参数的调用，这就是为什么它会给你一个usage输出。第二个示例不应返回包含someVar但所有行的所有行，因为空字符串在所有行中。
To tell the shell to not substitute, you have to use '$someVar'or \$someVar. Then you'll have to deal with the grep interpretation of the $character, hence the grep -Foption given in many other answers.
要告诉外壳不要替换，您必须使用'$someVar'或\$someVar。然后你将不得不处理$字符的 grep 解释，因此grep -F在许多其他答案中给出了选项。
So one valid answer would be:
所以一个有效的答案是：
grep -ir "Something Here" * | grep '$someVar'
回答by Alo?ké Go
+1 for the -Foption, it shall be the accepted answer.
Also, I had a "strange" behaviour while searching for the -I..pattern in my files, as the -Iwas considered as an option of grep; to avoid such kind of errors, we can explicitly specify the end of the arguments of the command using --.
+1为-F选项，它应该是被接受的答案。此外，-I..在我的文件中搜索模式时，我有一个“奇怪”的行为，因为它-I被认为是grep; 为避免此类错误，我们可以使用 明确指定命令参数的结尾--。
Example:
例子：
grep -HnrF -- <pattern> <files>
Hope that'll help someone.
希望这会帮助某人。
