查找特定的图案并使用awk或sed打印完整的文本块 [英] Find specific pattern and print complete text block using awk or sed

问题描述

如何在文本块中找到特定的数字并打印以关键字"BEGIN"开始并以"END"结尾的 complete 文本块?基本上，这就是我的文件的样子:

How can find a specific number in a text block and print the complete text block beginning with the key word "BEGIN" and ending with "END"? Basically this is what my file looks like:

BEGIN
A: abc
B: 12345
C: def
END

BEGIN
A: xyz
B: 56789
C: abc
END

BEGIN
A: ghi
B: 56712
C: pqr
END

[...]

如果我正在寻找'^B: 567'，我想获得以下输出:

If I was looking for '^B: 567', I would like to get this output:

BEGIN
A: xyz
B: 56789
C: abc
END

BEGIN
A: ghi
B: 56712
C: pqr
END

我可以在这里使用grep( grep -E -B2 -A2 "^B: 567" file )，但我想获得更通用的解决方案.我猜 awk 或 sed 也许可以做到这一点！?

I could use grep here (grep -E -B2 -A2 "^B: 567" file), but I would like to get a more general solution. I guess awk or sed might be able to do this!?

谢谢！ :)

推荐答案

$ awk -v RS= -v ORS='\n\n' '/\nB: 567/' file
BEGIN
A: xyz
B: 56789
C: abc
END

BEGIN
A: ghi
B: 56712
C: pqr
END

请注意在B之前的\n以确保它出现在行的开头，这代替了您最初使用的^字符串开始字符，因为现在每行都不是自己的细绳.您需要在上面设置ORS，以便在记录之间重新插入空白行.

Note the \n before B to ensure it occurs at the start of a line.This is in place of the ^ start-of-string character you had originally since now each line isn't it's own string. You need to set ORS above to re-insert the blank line between records.

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