UNIX for Advanced & Expert Users

Hi,

I have a command in one of my shell script as below.

sed -n "/^$a/,\$p" $home/abc.log

$a and $home are variables. Not sure about $p.

Can you please let me know what this command is exactly trying to do.

Thanks in advance
LM

Lijju,
The "sed" command: sed -n "/^$a/,\$p" $home/abc.log
Will replace all records in the file "$home/abc.log" that start with the value of the
variable "$a" by a three character string ",$p" (comma-dollar-p).
Because there is a backslash "\", the dollar sign "$" is treated literally, as opposed
to a prefix of a shell variable.

don't think it does replacement. there should be 's' modifier for replacement.

Yes it is not replacing. It doing some other thing

You are right, ghostdog74 -- my eyes didn't catch this one.

Shell_life,
I dont think so. It will print all lines of the file between the line in the file that begins with the value of $a and the "$" - Last line of the file.

Simply once the sed finds out the line which begins with the value of $a,it prints all lines from there to the end of the file.

Example:
Assuming the variable "a" is set to "the" ie a=the
Please correct me of am wrong.

Thanks
Nagarajan Ganesan

ennstate, you are correct.

The $ is escaped because of the use of double quotes. The shell would otherwise attempt to perform substitution.

assuming a=foo

Without variables, one would normall write this as:

Code:

sed '/^foo/,$p'