Google Firebase. Cloud Firestore Structure

How do I get Firebase to return a real value, it says >>Structure<<

Jogo


You’ve made 3 posts about this, please do not do it again in the future as spam is not allowed on this forum.

Firebase already returns the correct full value. Your variable in firebase is a structure variable therefore it is a structure variable in GDevelop. VariableString will show the string contents of a variable, however a structure does not have 1 value - it inherently has many values since it is a container for multiple variables, and therefore it doesn’t has a string content to show. Thus trying to get a string value of a structure variable will always yield [Structure] in GDevelop.

Access the children value directly if you want to show them, or use ToJSON to generate a text representation of the structure instead of VariableString.

2 Likes

Firstly thanks for saying about spam, I hadn’t noticed

I understood the structure part that will always return to me [Structure]

Now about how to access the child value directly if I want to display it or use ToJSON to generate a text representation of the structure instead of VariableString. How do I do this? Because you actually understand what I need? I need to check in the Letters derivative if the word I want exists in any of the fields, which in this case will be several