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Top Forums Shell Programming and Scripting Building JSON command with bash script Post 303039232 by psysc0rpi0n on Thursday 26th of September 2019 05:02:08 PM
Old 09-26-2019
Quote:
Originally Posted by Chubler_XL

...

These escaped quotes are simply to protect them from the shell the actual argument values that will react the bitcoin-cli process will be
Ok, I'm going to try your examples. This is not trivial for a starter. I've just been reading but not practicing much, so this is not clear at first sight!

--- Post updated at 10:02 PM ---

Quote:
Originally Posted by Chubler_XL
The reason for this is that quote removal only occurs on the original input words, not the result of expansions. To demonstrate:

Code:
$ echo "Hello, world"
Hello, world
$ VAR='echo "Hello, world"'
$ $VAR
"Hello, world"

It's also worth noting that quote matching is done before any expansion, for example:
Code:
$ QT='"'
$ echo "$QT"
"
$ echo $QT
"

If you save this code a pparam(or make it a function in your script) you can use it to debug what the shell is doing to your parameters before they are passed on to scripts/functions:

Code:
for((i=1;i<=${#};i++))
do
   echo Param $i: ${!i}
done

Code:
$ VAR='"one" "two"'
$ ./pparam $VAR
Param 1: "one"
Param 2: "two"
$ ./pparam "$VAR"
Param 1: "one" "two"
$ ./pparam "one" "two"
Param 1: one
Param 2: two

I don't fully understand what you mean by:
Quote:
that quote removal only occurs on the original input words, not the result of expansions.
and
Quote:
It's also worth noting that quote matching is done before any expansion, for example:
.

I need to read about expansions!
 

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