replace a line containing pattern only in the first occurrence after another line that contains anot-CodePudding

I want to replace a line that contains the matching pattern ForwardX11 in a file with sed, but only want to replace it when the line containing it is the first line found after another line that contains another pattern.

Example:

Host *
#    ForwardAgent no
#    ForwardX11 no
#    ForwardX11Trusted no
#    AnotherLine no

Host localhost
    ForwardX11 yes

In the example above I only want to replace the line # ForwardX11 no with Forward yes because this line is the first one after the line Host *. But I want to keep the line Forward yes after the line Host localhost unchanged.

So far I tried using sed like this:

sed -i "/ForwardX11Trusted/! s/.*ForwardX11.*/    ForwardX11 yes/" /etc/ssh/ssh_config

But it changes the second line too (ForwardX11 after Host localhost).

EDIT 1: It is supposed that I don't know if the line containing the pattern ForwardX11 has # at beginning or not, and also I dont know if ForwardX11 is set to yes or no.

EDIT 2: It does not have to be done with sed.

CodePudding user response：

Does it have to be sed? Easier (IMO) with awk

awk '
    $1 == "Host" {in_block = $2 == "*"}
    in_block && $2 == "ForwardX11" {print "    ForwardX11 yes"; next}
    1
' ssh_config

Host *
#    ForwardAgent no
    ForwardX11 yes
#    ForwardX11Trusted no
#    AnotherLine no

Host localhost
    ForwardX11 yes

CodePudding user response：

Using sed

$ sed -i.bak '/^Host \*/,/ForwardX11/ s/#\?\(.*X11\) no/\1 yes/' /etc/ssh/sshd_config
Host *
#    ForwardAgent no
    ForwardX11 yes
#    ForwardX11Trusted no
#    AnotherLine no

Host localhost
    ForwardX11 yes