AWK：显示匹配正则表达式（多） [英] awk: Display matched regex (multiline)

问题描述

我有一个看起来像这样的文件：

I have a file that looks like this:

SF:/home/developer/project/test/resources/somefile.js
DA:1,2
DA:3,2
end_of_record
SF:/home/developer/project/src/resources/otherfile.js
DA:9,2
DA:15,2
DA:22,2
end_of_record

...some more SF:/home/xxx and end_of_record lines...

该文件包括与开始块SF的：... 和 end_of_record 结束。请注意，行的计数其间（ DA，则x ）可以是不同的。我想打印在第一行字符串测试的所有块（如SF：/家庭/开发/ 测试 /资源/ ...在这里）。在这个例子中，我想要的输出将是：

The file consists of blocks beginning with SF: ... and ending with end_of_record. Note that the count of the lines inbetween (DA:x,x) can be different. I want to print all blocks that have the String "test" in its first line (like "SF:/home/developer/test/resources/..." here). For this example, my wanted output would be:

SF:/home/developer/project/test/resources/somefile.js
DA:1,2
DA:3,2
end_of_record

我想这样做在的Linux 环境。

我的第一次尝试是这样使用sed的命令的事，但一些研究似乎像AWK后做多行操作的更合适的工具。

My first try was to do this using the "sed" command, but after some research it seemed like "awk" is a more suitable tool for doing multiline operations.

使用AWK和正则表达式，这是我到目前为止的命令：

Using awk and Regex, this is the command that I have so far:

awk '/SF[:\/a-zA-Z0-9]*test[\/A-Za-z0-9.,:\n]*end_of_record/ {print}' FS="\n" RS="" examplefile

但它输出完整的 examplefile ，不仅包含测试在第一线的块。我不知道如果我的正则表达式是错误的，或者如果我错过在我的 AWK 电话什么的。

But it outputs the full examplefile, not only the blocks containing "test" in the first line. I am not sure if my Regex is wrong or if I am missing something in my awk call.

我怎样才能得到只有阻滞在第一行测试？

How can I get only the block with "test" in the first line?

推荐答案

您需要一个标志：

awk '/^SF.*test.*/{f=1}f;/end_of_record/{f=0}' yourFile

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