正则表达式匹配特定模式

Question

I have

[root@centos64 ~]# cat /tmp/out
[
  "i-b7a82af5",
  "i-9d78f4df",
  "i-92ea58d0",
  "i-fa4acab8"
]

I would like to pipe though sed or grep to match the format "x-xxxxxxxx" ie a mix of az 0-9 always in 1-[8 chars length], and omit everything else

[root@centos64 ~]# cat /tmp/out| sed s/x-xxxxxxxx/
i-b7a82af5
i-9d78f4df
i-92ea58d0
i-fa4acab8

I know this is basic, but I can only find examples of text substitution.

Answer 1

grep -Eo '[a-z0-9]-[a-z0-9]{8}' file

The -E option makes it recognize extended regular expressions, so it can use {8} to match 8 repetitions.

The -o option makes it only print the part of the line that matches the regexp.

Answer 2

Why not just print whatever's between the quotes:

$ sed -n 's/[^"]*"\([^"]*\).*/\1/p' file
i-b7a82af5
i-9d78f4df
i-92ea58d0
i-fa4acab8

$ awk -F\" 'NF>1{print $2}' file                     
i-b7a82af5
i-9d78f4df
i-92ea58d0
i-fa4acab8

Answer 3

Through GNU sed,

$ sed -nr 's/.*([a-z0-9]-[a-z0-9]{8}).*/\1/p' file
i-b7a82af5
i-9d78f4df
i-92ea58d0
i-fa4acab8

Answer 4

I think this is all you need: [0-9a-zA-Z]-[0-9a-zA-Z]{8} . Try it out here .

Answer 5

这应该可以工作^[a-z0-9]-[a-zA-Z0-9]{8}$