Need some help regex please

sysmicuser · 05-28-2017, 08:06 AM

Hi Guys,

I am have a yaml file and it has the following text:

Code:

  /data/app0:
     server: x.y.z.p
     share: /data/app0
     mount: /data/app/0

And I want to have the following output:

Code:

  /data/app0:
     server: x.y.z.p
     share: /data/app0
     mount: /data/app/0
     options: _netdev,noatime,nodiratime

I tried the following which is a simple regex expression used -i and a but it is throwing an error:

Code:

[user@jump nodes]$ cat test.yaml
  /data/app0:
     server: x.y.z.p
     share: /data/app0
     mount: /data/apps/0
  /data/app1:
     server: x.y.z.p
     share: /data/app1
     mount: /data/apps/1
  /data/app2:
     server: x.y.z.p
     share: /data/app2
     mount: /data/apps/2
  /data/app3:
     server: x.y.z.p
     share: /data/app3
     mount: /data/apps/3

[user@jump nodes]$ sed -i '/mount: /data/app/*/a \
options: _netdev,noatime,nodiratime \' test.yaml
sed: -e expression #1, char 11: extra characters after command
[user@jump nodes]$

No idea why it is going wrong ?

grail · 05-28-2017, 09:03 AM

sed will see all of the / as arguments so more than 2 without the use of 's', as in s///, and it is incorrect . You will need to escape all the ones that are part of the regex.

Also, do not confuse globbing with regex. /data/app/*/ -- this is looking for zero or more / after the string 'app' followed by a single / (also it is probably meant to be 'apps'.
Try using a digit qualifier, [0-9]

hydrurga · 05-28-2017, 09:07 AM

I'm not an expert on sed but:

i) The second forward slash (before data) will be interpreted as the end of the sed clause. You need to escape all literal forward slashes.

ii) What is that second backslash (\) character doing there? It will escape the final single quote.

iii) You're looking for /data/app/... but the data in your data file contains /data/apps/...

Edit: Oops, beaten to it...