多个 greps 和 sed 在文件中使用管道

Question

我想在文件中查找一个字符串并用某物替换，但是下面的命令不起作用。

$ grep "CUDA_ARCH" lib/gpu/Makefile.linux | grep -v "#CUDA_ARCH" | grep -v " CUDA_ARCH " | xargs sed -i 's/sm_.*/sm_86/g'
sed: invalid option -- 'a'
Usage: sed [OPTION]... {script-only-if-no-other-script} [input-file]...

以下命令正确

$ grep "CUDA_ARCH" lib/gpu/Makefile.linux | grep -v "#CUDA_ARCH" | grep -v " CUDA_ARCH "
CUDA_ARCH = -arch=sm_50
$ grep "CUDA_ARCH" lib/gpu/Makefile.linux | grep -v "#CUDA_ARCH" | grep -v " CUDA_ARCH " | sed 's/sm_.*/sm_86/g'
CUDA_ARCH = -arch=sm_86
$ grep "CUDA_ARCH" lib/gpu/Makefile.linux
#     - Change CUDA_ARCH for your GPU
#CUDA_ARCH = -arch=sm_30
#CUDA_ARCH = -arch=sm_32
#CUDA_ARCH = -arch=sm_35
#CUDA_ARCH = -arch=sm_37
CUDA_ARCH = -arch=sm_50
#CUDA_ARCH = -arch=sm_52

我该如何解决这个问题？

Answer 1

这是您的错误的简化示例：

$ echo 'abc -a xyz' | xargs sed 's/abc/123/'
sed: invalid option -- 'a'

# if - isn't present in the stdin data, you might get file errors
$ echo 'abc a xyz' | xargs sed 's/abc/123/'
sed: can't read abc: No such file or directory
sed: can't read a: No such file or directory
sed: can't read xyz: No such file or directory

# you can remove xargs, but then you'll not be able to use -i
$ echo 'abc -a xyz' | sed 's/abc/123/'
123 -a xyz

试试这个（将ip.txt 替换为lib/gpu/Makefile.linux）。如果可行，请使用-i 选项。

$ cat ip.txt
#CUDA_ARCH = -arch=sm_50
 CUDA_ARCH = -arch=sm_50
CUDA_ARCH = -arch=sm_50

$ sed -E '/(^|[^# ])CUDA_ARCH/ s/sm_.*/sm_86/' ip.txt
#CUDA_ARCH = -arch=sm_50
 CUDA_ARCH = -arch=sm_50
CUDA_ARCH = -arch=sm_86

sed支持过滤，这里不用grep。

/(^|[^# ])CUDA_ARCH/ 将匹配行首的CUDA_ARCH，或者如果它前面没有# 或空格字符。

对于这样的行，将应用替换命令。

如果单个正则表达式不起作用并且您需要多个条件，则可以使用 /regexp1/ { /regexp2/ {stmts} } 之类的内容。使用/regexp/! 反转条件。