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提示:

$ 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">>

我没有看到告诉 grep not 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。

最佳答案

使用 fgrep(已弃用)、grep -F 或 grep --fixed-strings，使其将模式视为列表固定字符串，而不是正则表达式。

作为引用，文档提到(摘录):

-F --fixed-strings Interpret 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.)

fgrep is the same as grep -F. Direct invocation as fgrep is deprecated, but is provided to allow historical applications that rely on them to run unmodified.

如需完整引用，请查看: https://www.gnu.org/savannah-checkouts/gnu/grep/manual/grep.html

https://stackoverflow.com/questions/9416390/