SED to find and replace exact word beginning with $

Question

My php file has some defined variables with $downtime_hosts, what i need is a single command to just find and replace entire $downtime_hosts = 8; into $downtime_hosts = 15; and without affecting the other $downtime_hosts used several time within the file.

here my number 8,15,16 may change anytime, what i need is find the line beginning with$downtime_hosts = anyinteger and simply replace with my new line $downtime_hosts = newinteger . please note anyinteger / newinteger=2,3,4,15or any

$downtime_hosts = 8;

$total_hosts = $all_hosts - $downtime_hosts;

if ($host_up == $total_hosts )

Hosts under downtime $downtime_hosts `

Any inputs are really welcomed !

Answer 1

sed 's/$downtime_hosts = 8;/$downtime_hosts = 15;/' file.php

The $ poses no problem since it won't act as an anchor unless it's found at the end of the pattern. The sed script does need to be single-quoted though a the shell otherwise would try to expand $downtime_hosts as a shell variable.

To only match the pattern in the beginning of a line:

sed 's/^$downtime_hosts = 8;/$downtime_hosts = 15;/' file.php

If the integer 8 could be any integer:

sed 's/^$downtime_hosts = [0-9]*;/$downtime_hosts = 15;/' file.php

To replace the integer with an integer held by a shell variable $newint:

sed "s/^\$downtime_hosts = [0-9]*;/\$downtime_hosts = $newint;/" file.php

Notice that we now have to use double quotes around the sed editing script for the shell to expand the $newint variable. This also means that we are forced to escape the two existing $ from the shell.