将多个数据部分解析为单独的文件

Question

我有一个包含多个日志的串联日志文件，我试图将其解析为单个日志文件。我稍后会将它们重命名为每个的日期/时间。每个日志以“--- LOG REPORT ---”分隔。

到目前为止，我有：

sed -n '/--- LOG REPORT ---/,/--- LOG REPORT ---/p' logname.log > test.out

但是，您可以想象，这仅输出模式的第一个实例。我查看了 sed 的手册页，但我不相信它可以输出多个文件。也许我可以继续从文件中提取直到它为空，但这似乎工作量太大。我怎么能做到这一点？也许我应该改用 awk ？

输入文件 filename.log 的示例

--- LOG REPORT ---
Mary
Had
A
Little
Lamb
--- LOG REPORT ---
Her
Fleece
Was
White
As
Snow

期望的输出：

在 filename_1.log 中

--- LOG REPORT ---
Mary
Had
A
Little
Lamb

在 filename_2.log 中

--- LOG REPORT ---
Her
Fleece
Was
White
As
Snow

Answer 1

这个命令怎么样？

csplit logname.log /---\ LOG\ REPORT\ ---/ {*}

测试

cat logname.log
--- LOG REPORT ---
Mary
Had
A
Little
Lamb
--- LOG REPORT ---
Her
Fleece
Was
White
As
Snow

运行上述命令后，我得到的输出是，

cat xx01
--- LOG REPORT ---
Mary
Had
A
Little
Lamb
cat xx02
--- LOG REPORT ---
Her
Fleece
Was
White
As
Snow

Answer 2

怎么样的东西

awk '/--- LOG REPORT ---/ {n++;next} {print > "test"n".out"}' logname.log