Building JSON command with bash script


 
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Old 09-20-2019
Correct Usage of quoting ;-)

Hello Psy,
  • Please don't full quote forum posts. Please only quote the short passage of text you're relating to. Refrain from full quoting because it adds redundant information and makes the thread harder to read.
  • regarding quoting #1

    Code:
    bitcoin-cli -testnet sendmany $backcomp $json_obj $min_conf $comment $json_lst $replc $min_conf $est_mde

    It's always a good idea to quote parameters. Else the shell will split your parameters at whitespace. For sure it splits your json-object - which is not what you want. Do it that way:

    Code:
    bitcoin-cli -testnet sendmany "$backcomp" "$json_obj" "$min_conf" "$comment" "$json_lst" "$replc" "$min_conf" "$est_mde"

  • regarding quoting #2

    This(single quotes) '$comment' does not replace $comment with the value contained in $comment. This(double quotes) "$comment" does.
  • regarding quoting #3

    Quote:
    You mean to add double quotes when I run the command? Because I have already added double quotes when I assigned the variable due to the existence of 2 separate words. Won't "$comment" expand to ""Periodic payments""? Or should I remove the double quotes at the assignment of the comment variable?
    You misunderstood the concept of quoting a bit. If you assign a variable like that: variable="my value", the content of $variable is not "my value" it is my value(without the quotes). The quotes just protect the content from word splitting. If you leave the quotes out, it would be a complete different command. If you run this command: variable=my value this means assign my to variable for the next program call and call the program named value.

    You can only get quotes as data into some variable if you quote it. ;-) like this quotes_variable='"' or like this: quotes_variable="\"".
  • I recommend to read a bit about quoting to fully understand the concept, as it is really important!
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