can't output dialog value on bash

Question

I have a dynamic menu on dialog (the items are provided from an array) but the menu is not displaying the selected option

gp_options=()
for i in `find ~ -type d -name .git`; do
    gp_options+=("" "$i")
done

gp_dialog=(dialog --stdout --extra-button --help-button \
--ok-label 'Access repository' \
--extra-label 'Create repository' \
--cancel-label 'Remove repository'
--help-label 'Cancel' \
--backtitle "Welcome to Git Bash `whoami`!" \
--title ' Manage repositories ' \
--menu 'Manage repositories' \
0 0 0 \
"${gp_options[@]}")

dialog --stdout --msgbox "$manage_repositories" 0 0

(1) What is dialog supposed to write on stdout? (2) Did you try tor run with set -x so that you can see whether the parameters are correct? — user1934428
– user1934428, Commented Mar 20, 2018 at 14:23
dialog is supposed to write the option selected by the user on the menu (gp_dialog) for example: /home/user/.git (and how do use set -x?) — henriquehbr
– henriquehbr, Commented Mar 20, 2018 at 14:31
You must have made a mistake copying your example. Is gp_dialog supposed to be an array, or the result of running dialog in which case you need =$(...). What is in var manage_repositories? — meuh
– meuh, Commented Mar 20, 2018 at 18:13
The gp_dialog is the actual menu dialog, and you're right, i forgot to put the =$(), thank you @meuh — henriquehbr
– henriquehbr, Commented Mar 20, 2018 at 18:42

henriquehbr · Accepted Answer · 2018-03-20 18:40:50Z

After hours of research, i finally found a answer:

repositorios=() ; i=0
while read -r line; do
    let i=$i+1
    repositorios+=($i "$line")
done < <( find ~ -type d -name .git )

gerenciar_repositorios=$(dialog --stdout --extra-button --help-button \
--ok-label "Acessar repositório" \
--extra-label 'Criar repositório' \
--cancel-label 'Remover repositório' \
--help-label 'Cancelar' \
--backtitle "Bem vindo ao Git Bash `whoami`!" \
--title ' Gerenciar repositórios ' \
--menu 'Gerenciar repositórios' 0 0 0 \
${repositorios[@]})

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