bash - parsing a log file with awk -

I am trying to parse a log file with the following code:

  If [[$ line-match "=====. *"]]]]; So awk $ 2 = $ vFiler; Repeat "$ vFiler";  

Explanation:

The lines starting with a pattern such as ===== or 'vol / ...' are one Name will find all the lines that start with ===== or'vol / ... 'and set the string which is equal to one variable, but I actually I am thinking about an order with if awk (I know that -Mail is not real command).

The input looks like this:

  ===== vfiler0 / vol / vol0 Vfiler1 / vol / vol1  

output Should be:

Can you help me? I am trying to do this in a bass script, as if I tried it above, "if ..." is possible to get it?

I will use the following:

  awk -v RS = "=====" -V OFS = "," BEGIN {print} vFiler "," type "} NF {print $ 1, $ 2} 'file  

:

  vFiler, type vfiler0, / vol / vol0 vFiler1, / vol / vol1  

We record the separator ===== . In this way, a block of ==== will be handled every time. Then, whenever there is at least one ( NF ), we print the first and second fields Are there.