awk将字段数据复制到新字段

Question

如何使用/作为字段分隔符值复制到行尾？使用？作为分隔符？

我有一个类似这样的文件：

/folder/subfolder1/user1/1234/file.jpg,2014-11-24 11:49:55.6590447890 +0200
/folder/subfolder1/user2/ddaa/file.jpg,2014-11-24 11:49:55.6590447890 +0200
/folder/subfolder1/user2/sssffd/file.jpg,2014-11-24 11:43:03.6438296240 +0200
/folder/subfolder1/user3/882ssa/file.jpg,2014-11-24 11:43:00.6918628210 +0200

我需要的输出是：

/folder/subfolder1/user1/1234/file.jpg,2014-11-24 11:49:55.6590447890 +0200,1234,1234 2014-11-24 11:49:55.6590447890 +0200,user1
/folder/subfolder1/user2/ddaa/file.jpg,2014-11-24 11:49:55.6590447890 +0200,ddaa,ddaa 2014-11-24 11:49:55.6590447890 +0200,user2
/folder/subfolder1/user2/sssffd/file.jpg,2014-11-24 11:43:03.6438296240 +0200,sssffd,sssffd 2014-11-24 11:49:55.6590447890 +0200,user2
/folder/subfolder1/user3/882ssa/file.jpg,2014-11-24 11:43:00.6918628210 +0200,882ssa,882ssa 2014-11-24 11:49:55.6590447890 +0200,user3

所以我需要将该行的某些部分复制到它的末尾，并将几个字段组合在一起。

Answer 1

怎么样

$ awk -F[/,] '{$0=$0","$5","$5" "$7","$4}1' input
/folder/subfolder1/user1/1234/file.jpg,2014-11-24 11:49:55.6590447890 +0200,1234,1234 2014-11-24 11:49:55.6590447890 +0200,user1
/folder/subfolder1/user2/ddaa/file.jpg,2014-11-24 11:49:55.6590447890 +0200,ddaa,ddaa 2014-11-24 11:49:55.6590447890 +0200,user2
/folder/subfolder1/user2/sssffd/file.jpg,2014-11-24 11:43:03.6438296240 +0200,sssffd,sssffd 2014-11-24 11:43:03.6438296240 +0200,user2
/folder/subfolder1/user3/882ssa/file.jpg,2014-11-24 11:43:00.6918628210 +0200,882ssa,882ssa 2014-11-24 11:43:00.6918628210 +0200,user3