Question

我有一个包含许多块的文件，如下所示：

==9673== 
==9673== HEAP SUMMARY:
==9673==     in use at exit: 0 bytes in 0 blocks
==9673==   total heap usage: 75,308 allocs, 75,308 frees, 7,099,382 bytes allocated
==9673== 
==9673== All heap blocks were freed -- no leaks are possible
==9673== 
==9673== For counts of detected and suppressed errors, rerun with: -v
==9673== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)
....
....
....
....

==9655== 
==9655== HEAP SUMMARY:
==9655==     in use at exit: 0 bytes in 0 blocks
==9655==   total heap usage: 75,308 allocs, 75,308 frees, 7,099,382 bytes allocated
==9655== 
==9655== All heap blocks were freed -- no leaks are possible
==9655== 
==9655== For counts of detected and suppressed errors, rerun with: -v
==9655== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

....
....
....

==9699== 
==9699== HEAP SUMMARY:
==9699==     in use at exit: 0 bytes in 0 blocks
==9699==   total heap usage: 75,308 allocs, 75,308 frees, 7,099,382 bytes allocated
==9699== 
==9699== All heap blocks were freed -- no leaks are possible
==9699== 
==9699== For counts of detected and suppressed errors, rerun with: -v
==9699== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

我想从line：

开始提取最后一个块

 ==XXXX== HEAP SUMMARY:

所以在我的例子中，我想只提取最后一个集团：

==9699== HEAP SUMMARY:
==9699==     in use at exit: 0 bytes in 0 blocks
==9699==   total heap usage: 75,308 allocs, 75,308 frees, 7,099,382 bytes allocated
==9699== 
==9699== All heap blocks were freed -- no leaks are possible
==9699== 
==9699== For counts of detected and suppressed errors, rerun with: -v
==9699== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

我怎么能用bash做到这一点？

Answer 1

使用grep -zoP和负前瞻正则表达式：

grep -zoP '==\w{4}== HEAP SUMMARY:(?![\s\S]*==\w{4}== HEAP SUMMARY:)[\s\S]*\z' file

==9699== HEAP SUMMARY:
==9699==     in use at exit: 0 bytes in 0 blocks
==9699==   total heap usage: 75,308 allocs, 75,308 frees, 7,099,382 bytes allocated
==9699==
==9699== All heap blocks were freed -- no leaks are possible
==9699==
==9699== For counts of detected and suppressed errors, rerun with: -v
==9699== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

-z将文件数据视为空终止而不是新行终止
(?![\s\S]*==\w{4}== HEAP SUMMARY:)是负面预测，声称我们在下面的文件中没有相同的其他实例。

RegEx Demo

Answer 2

如果你有tac，这可能是最简单的

$ tac file | awk '1; /==....== HEAP SUMMARY/{exit}' | tac

Answer 3

如果你知道这些块总是9行，你只需使用tail：

tail -n9 file

Answer 4

使用sed：

$ sed -n '/HEAP SUMMARY/{:a;/ERROR SUMMARY/bb;N;ba;:b;$p;d}' infile
==9699== HEAP SUMMARY:
==9699==     in use at exit: 0 bytes in 0 blocks
==9699==   total heap usage: 75,308 allocs, 75,308 frees, 7,099,382 bytes allocated
==9699== 
==9699== All heap blocks were freed -- no leaks are possible
==9699== 
==9699== For counts of detected and suppressed errors, rerun with: -v
==9699== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

这是如何工作的：

sed -n '                   # Do not print lines at end of each cycle
    /HEAP SUMMARY/ {       # If line matches "HEAP SUMMARY"
        :a                 # Label to jump back to
        /ERROR SUMMARY/bb  # If line matches "ERROR SUMMARY", jump to :b
        N                  # Append next line to pattern space
        ba                 # Jump to :a
        :b                 # Label to jump forward to
        $p                 # If we are on the last line, print pattern space
        d                  # Delete pattern space
    }
' infile

每次遇到HEAP SUMMARY时，它会将所有直到下一个ERROR SUMMARY的行读入模式空间。然后，它检查是否已到达最后一行;如果是，则打印图案空间，否则将被删除。

Answer 5

如果文件的最后一行也有块号，这将获得快速的块号（不读取整个文件以找到哪个号码）：

n="$(tail -n1 infile | awk '{print $1}')"

然后我们可以从最终选择具有此类块号的所有行：

tac infile | awk -vn="$n" '!($1~n){exit};1'| tac

Answer 6

这可能适合你（GNU sed）：

sed '/HEAP SUMMARY:/h;//!H;$!d;x' file

当遇到HEAP SUMMARY:时，用当前行替换保留空间（HS）中的任何内容。对于任何其他模式附加到HS的线。当模式空间（PS）与HS交换并打印出PS时，删除除最后一行之外的所有行。

Answer 7

使用数据前面的数字作为ID /组号：

id=$(tail -n1 file | grep -Po '(?<=\=\=)[0-9]*') && grep "$id" file |tail -n+2

如何从文件中提取最后一个块

7 个答案: