如何解析JSON模式并创建默认的json对象或生成Java类? [英] How can I parse a JSON schema and create a default json object or generate a java class?

问题描述

我确定有人做过此事:一位同事经历了制作JSON模式(可用于验证JSON对象)的麻烦；但是，我很懒，只想从架构中生成一个默认"空JSON对象.有没有简单的方法可以做到这一点?

I'm sure someone has done this before: A co-worker went through the trouble of making a JSON schema which can be used to validate a JSON object; however, I am lazy, and just want to generate a "default" empty JSON object from the schema. Is there an easy way to do this?

我想用Java做到这一点.

I would like to do this in Java.

失败了，有没有办法基于此模式生成Java类(在运行时不生成)?

Failing this, is there a way to generate a Java class (NOT at run-time) based on this schema?

格拉西亚斯(Muchas gracias)，朋友.

Muchas gracias, amigos.

推荐答案

" jsonschema2pojo "怎么样?

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