Dynamically decide which java class to create instance of using yaml/json

Asked Sep 30 '22 at 08:17

Active Sep 30 '22 at 08:17

Viewed 131 times

I have an interesting problem in with one of my projects. I have ZipInputeStream from uploaded zip in spring boot application with several yaml file in it, which can have any name and dynamic content which we know and have java POJO objects for them. For e.g. Below yaml corresponds to BaseModel.java

---
baseModel:
    other contents

class BaseModel{}

and below yaml will correspond to TopModel.java

---
topModel:
    other contents

class TopModel{}

Now the Question is how do i decide which java class to use for object creation at runtime based on the yaml file string. I am using jackson and tried converting yaml string to JSON but that requires a type in JSON which we dont have in the file and we cant get it added as well.

Any help on this would really be appreciated. Thanks.

asked Sep 30 '22 at 08:17

ADITYA PRAKASH

I assume a conditional statement would be nice, but since the names match .... what is there to 'decide'? – Stultuske Sep 30 '22 at 08:21
If the names don't always match, how about a simple map from name to class (or to factory method)? – ewramner Sep 30 '22 at 08:22
I dont want to have a complex conditional or switch block. Is there any cleaner solution? – ADITYA PRAKASH Sep 30 '22 at 08:22
Any example would be nice to have here please – ADITYA PRAKASH Sep 30 '22 at 08:23
an example of your actual problem as well. as you posted it, all you need to do is check the name of the yaml file to know the type – Stultuske Sep 30 '22 at 08:23
@Stultuske Filename can be anything, Decision has to be made on basis of content , specially the name of starting node – ADITYA PRAKASH Sep 30 '22 at 08:36

Dynamically decide which java class to create instance of using yaml/json

0 Answers0