How to grep 7 lines after match and replace text?

Question

I'm trying grep 7 lines after match and replace a context in those 7 lines and save it in the same file.

For Example:

<IfModule prefork.c>
StartServers       8
MinSpareServers    5
MaxSpareServers   20
ServerLimit      256
MaxClients       256
MaxRequestsPerChild  4000
</IfModule>

i would like to replace the value of MaxRequestsPerChild with 5000.

I have tried with:

grep "IfModule prefork.c" httpd.conf -A7 | /bin/sed -nE "/\bMaxRequestsPerChild\b/ s/[0-9]+/50/"

But no luck.

Answer 1

Here is a SED solution.

sed '/<IfModule prefork.c>/,/MaxRequestsPerChild/s/MaxRequestsPerChild.*/MaxRequestsPerChild 5000/' apacheconf

It works by matching the Module name then replacing the next "MaxRequestsPerChild" line with the 5000 value. It will ignore the "MaxRequestsPerChild" of all other modules.

Answer 2

With GNU sed (not sure if syntax/feature is okay for other versions)

To change within 6 lines after matching line:

sed -E '/IfModule prefork\.c/{N;N;N;N;N;N;s/(MaxRequestsPerChild +)[0-9]+/\15000/}'

here, N will add a newline to the pattern space, then append the next line of input to the pattern space

If it is always 6th line:

sed -E '/IfModule prefork\.c/{n;n;n;n;n;n;s/[0-9]+/5000/}'

here n will replace the pattern space with the next line of input

Or, simpler syntax with relative addressing

sed -E '/IfModule prefork\.c/,+6 s/(MaxRequestsPerChild +)[0-9]+/\15000/'

Answer 3

You could put sed's range operator , to good use here in case you are indifferent to line numbers:

sed -e '
   /<IfModule prefork.c>/,\|</IfModule>|!b
   H;\|</IfModule>|!d;g;s/.//;
   s/\(\(\n\)MaxRequestsPerChild[[:blank:]]\{1,\}\)[1-9][0-9]*\n/\15000\2/
' config-file

1. Store the range in the hold space.
2. When the range end is detected, retrieve the range from the hold area, and
3. Perform the desired substitution.
4. Lines not falling in the range are left to themselves and passed on to stdout.