过滤复杂文件中的特定模式

Question

我有这个日志文件，我试图避免打印包含模式“APLHA”的文本。

   +++ skdfhahjsahsdjk >
** ALPHA EDFJDJFKLJDKFJKSDLFJL
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

   +++ skdfhahjsahsdjk >
* KJENRFKES DFJKLSDFJEDFJDJFKLJDKFJKSDLFJL
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

   +++ skdfhahjsahsdjk >
CL mesukww juwaehdiearfa
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

   +++ skdfhahjsahsdjk >
*C KJENRFKESDFJ ksludhieokdaewmdp
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

 +++ skdfhahjsahsdjk >
** BETA EDFJDJFKLJDKFJKSDLFJL
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

我尝试通过这种方式但没有 successful.It 打印其余文本，"ALPHA"

除外

cat 日志文件 |grep -v "ALPHA"
结果：

+++ skdfhahjsahsdjk >
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

我期望的是包含 "ALPHA" 的整个日志未打印，如下所示：

 +++ skdfhahjsahsdjk >
* KJENRFKES DFJKLSDFJEDFJDJFKLJDKFJKSDLFJL
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

   +++ skdfhahjsahsdjk >
CL mesukww juwaehdiearfa
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

   +++ skdfhahjsahsdjk >
*C KJENRFKESDFJ ksludhieokdaewmdp
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

 +++ skdfhahjsahsdjk >
** BETA EDFJDJFKLJDKFJKSDLFJL
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

能否就此案例提供其他解决方案？

此致Dom

Answer 1

试一试：

awk 'BEGIN{RS=ORS="\n\n"}!/ALPHA/' file

Answer 2

使用 perl 的 slurp 模式（-00 启用）的替代解决方案。这将告诉 perl 考虑段落是一个记录而不是行。稍后很清楚，仅当 ALPHA 不存在时才打印。

perl -00 -ne  'print if  !/ALPHA/' inputfile
   +++ skdfhahjsahsdjk >
* KJENRFKES DFJKLSDFJEDFJDJFKLJDKFJKSDLFJL
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

   +++ skdfhahjsahsdjk >
CL mesukww juwaehdiearfa
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

   +++ skdfhahjsahsdjk >
*C KJENRFKESDFJ ksludhieokdaewmdp
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

 +++ skdfhahjsahsdjk >
** BETA EDFJDJFKLJDKFJKSDLFJL
   SDFSDGDRGRTG
   WEFETTFYRT #168113++-

过滤复杂文件中的特定模式

Filter specifik patern in a complex file

grep

uniq