Posts by Diego Szmidtke • 1 point
2 posts
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answer581
viewsA: Deserialize JSON with dynamic field names
I was able to solve it this way: JSONObject req = new JSONObject(json); JSONArray locs = req.getJSONArray("Pedido"); //JSONArray recs = locs.getJSONArray("record"); for (int i = 0; i <…
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0
votes1
answer581
viewsQ: Deserialize JSON with dynamic field names
I have a Json like this, and within the entity "purchase", I receive an object with the purchase id, and then the information I need to access, I have already generated the classes, however my…