Shell Programming and Scripting

Might not understand this correctly, but what keeps you from doing:

Code:

awk -F= '{print $NF}' <yourinputfile>

btw, cat is not a character.
By characters I meant special ones used in unix commands like $,^,|,/ etc.
If you want to use any of these characters, not for its feature, you have to escate it using backslash.

Eg: $ is used to represent end of a line in unix.
But if you want to represent it as a normal character in a string you have to escape it with backslash, telling the shell interpreter that

"Hello Shell, Please dont give '$' any special meaning as you would;but its a normal character in my string"

Let me give you an example.

There is an input file like this.

XX|$$|YY
YY|$$|ZZ

%s/$/AA/g

This command will change the file as below
XX|$$|YYAA
YY|$$|ZZAA

because shell interpreted $ as end of line.
======================================================
%s/\$\$/AA/g
This command will change the file as below
XX|AA|YY
YY|AA|ZZ

because, you escaped $ . So shell will not interpret $ as end of file, but as a normal character.

Let me know if you have more doubts.

No, it isn't. It is used by some commands to represent the end of a line. It may also represent the end of a file.

Or it may represent nothing but itself, as in:

Code:

echo $

Or it may be the beginning of some form of expansion, as in:

Code:

printf "%s\n" "$HOME"  "$(( ${x:-3} + ${y:-2} ))"  $'\t\n'

In some situations you might have to do that.

That is not a Unix command. It may be a command in some Unix application, but I have never seen it (perhaps it's vi?).

Without the percent sign, it could be a sed command.

No, an individual command may interpret $ as end of line. Many will not.

$ by itself has many meanings based on the context of usage.
I gave the meanings, based on the context of my example I provided.

btw, %s// is used in vi

By itself it means only '$'.

In other contexts it means one of several different things.

Mr Johnson,

Do you see anything wrong in the example I provided?