比较一个嵌套较多的Json，并打印出结构和值的差异

自动化: 可以自动遍历和比较复杂的嵌套结构。
详细输出: 能够提供具体的路径和差异值，便于定位问题。
灵活性: 可以处理不同类型的JSON结构，包括对象和数组。

要比较两个嵌套较多的JSON对象并打印出结构和值的差异，可以使用递归函数来遍历JSON对象的每一个节点。以下是一个示例代码，展示了如何实现这一功能：

import json

def compare_json(obj1, obj2, path=""):
    if isinstance(obj1, dict) and isinstance(obj2, dict):
        for key in set(obj1.keys()) | set(obj2.keys()):
            new_path = f"{path}.{key}" if path else key
            if key not in obj1:
                print(f"Key '{new_path}' missing in first JSON")
            elif key not in obj2:
                print(f"Key '{new_path}' missing in second JSON")
            else:
                compare_json(obj1[key], obj2[key], new_path)
    elif isinstance(obj1, list) and isinstance(obj2, list):
        if len(obj1) != len(obj2):
            print(f"List length mismatch at '{path}': {len(obj1)} vs {len(obj2)}")
        for i, (item1, item2) in enumerate(zip(obj1, obj2)):
            compare_json(item1, item2, f"{path}[{i}]")
    else:
        if obj1 != obj2:
            print(f"Value mismatch at '{path}': {obj1} vs {obj2}")

# 示例JSON对象
json1 = {
    "name": "Alice",
    "age": 30,
    "address": {
        "city": "New York",
        "zip": "10001"
    },
    "skills": ["Python", "Java"]
}

json2 = {
    "name": "Alice",
    "age": 31,
    "address": {
        "city": "New York",
        "zip": "10002"
    },
    "skills": ["Python", "C++"]
}

# 将JSON字符串转换为Python对象
obj1 = json.loads(json.dumps(json1))
obj2 = json.loads(json.dumps(json2))

# 比较两个JSON对象
compare_json(obj1, obj2)