如何在Linux中读取文件并获取特定行 [英] How to read a file and get certain lines in Linux

问题描述

我有一个文件，它的内容是这样的:

I have a file and it has content like this:

abc.com.        IN A            10.10.40.94 ;10.120.40.61    ;10.60.3.135
def.com.            IN CNAME        some-domain.com.

xyz.com.        IN A            10.123.40.32    ;10.145.40.133    ;1.240.3.215
one.com.                IN CNAME        some-other-domain.com.

我想做的是，用IN A获取所有行并将其打印到另一个文件中，以便最终文件如下所示:

What I want to do is, get all the lines with IN A and print them to another file so the final file would look like below:

我尝试了awk，如下所示:

cat dev-mytaxi.com.zone | awk '{print $3}'

但是我如何写文件以及下面提到的最终结果?

but how can I write to a file and the final outcome as mentioned below?

写入新文件的最终输出应类似于:

final output written to a new file should look like:

abc.com 10.10.40.94 10.120.40.61 10.60.3.135
xyz.com 10.123.40.32 10.145.40.133 1.240.3.215

推荐答案

您可以使用自定义FS尝试更简单的awk:

You may try this simpler awk with custom FS:

awk -F '[.[:blank:]]+IN A[[:blank:]]+' 'NF > 1 {
gsub(/[;[:blank:]]+/, " ", $2); print $1, $2}' file

abc.com 10.10.40.94 10.120.40.61 10.60.3.135
xyz.com 10.123.40.32 10.145.40.133 1.240.3.215

工作方式:

• 在这里，我们使用正则表达式[.[:blank:]]+IN A[[:blank:]]+作为输入字段分隔符，该记录将每条记录都用IN A文本分隔，两边都用空格包围.
• NF > 1仅对具有IN A记录的行为真
• gsub转换多个空格，;转换IP地址之间的单个空格

• Here we are using a regex [.[:blank:]]+IN A[[:blank:]]+ as input field separator that splits each record by a IN A text surrounded by whitespaces on either side.
• NF > 1 will be true only for lines with IN A record
• gsub converts multiple spaces and ; with a single space between ip addresses

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