如何使用python从嵌套的json响应中提取多个json值

Question

我的测试 url 返回以下 json 响应： 一个={ “版本”：“1.3.0”， “服务”： [ { “版本”：“1.0.0”， "serviceGun": "com.xx.version.1", “描述”：“测试服务”， “链接”：[ { "rel": "能力", "href": "/abc/test", “提示”：[ { “方法”：“获取” } ]

    },
{
    "version": "1.0.0",
    "serviceGun": "com.xx.abx.version.1",
    "description": "mode",
    "links": [ 
    {
        "rel": "mediaConfiguration",
        "href": "/abc/rest",
        "hints": [ 
        {
            "method": "get"

        },
        {
            "method": "put"
        } ]

    },
    {
        "rel": "modeConfiguration",
        "href": "/abc/mode",
        "hints": [ 
        {
            "method": "get"

        },
        {
            "method": "put"
        } ]

    },

{
    "version": "1.0.0",
    "serviceGun": "com.xx.abc.version.1",
    "description": "controlPanel",
    "links": [ 
    {
        "rel": "configuration",
        "href": "/abc/controlPanel",
        "hints": [ 
        {
            "method": "get"

        },
        {
            "method": "put"
        } ]

    },
    {
        "rel": "state",
        "href": "/abc/awc",
        "state": "unavailable",
        "stateReason": "inactivated",
        "hints": [ 
        {
            "method": "get"
        } ]
    } ]

},
 ]

}

通过以下函数，我可以从 json 中检索单个值（例如 href、rel 等）。 但我希望该函数返回一个列表，其中包含所有“href”以及每个href的“提示”中的支持方法。例如：[(href1,hints1),(href1,hints1),(href1,hints1)]

预期输出：[('/abc/test',['get','put']),('/abc/rest',['get','put']),...]

def extract_values(obj, key): """从嵌套的 JSON 中拉取指定键的所有值。""" arr = []

def extract(obj, arr, key):
    """Recursively search for values of key in JSON tree."""
    if isinstance(obj, dict):
        for k, v in obj.items():
            if isinstance(v, (dict, list)):
                extract(v, arr, key)
            elif k == key:
                arr.append(v)
    elif isinstance(obj, list):
        for item in obj:
            extract(item, arr, key)
    return arr

results = extract(obj, arr, key)
return results

运行：test_ref = extract_values(a,'href') 打印 test_ref 输出：['/abc/test','/abc/rest',','/abc/mode','/abc/controlPanel','abc/awc']

预期输出：[('/abc/test',['get','put']),('/abc/rest',['get','put']),...]

Answer 1

我认为您的递归试图解决一个不存在的问题。您的 JSON 具有清晰可见的层次结构。

1. a['services'] 是服务列表
2. a['services'][i]['links'] 是ith 服务的链接列表
3. a['services'][i]['links'][j]['href'] 是ith 服务的jth 链接的URL

所以我的函数应该是这样的：

def parse_links_and_methods(json_result):
    for service in json_result['services']:
        for link in service['links']:
            for hint in link['hints']:
                if 'method' in hint:
                    yield link['href'], hint['method']

这基本上只是一个巨大的列表理解。您可以将其用作：

print list(parse_links_and_methods(a))

并获取[('/abc/test', 'get'), ('/abc/rest', 'get'), ('/abc/rest', 'put'), ('/abc/mode', 'get'), ('/abc/mode', 'put'), ...]

编辑我误读了您的预期输出。以你想要的方式使用它

def parse_links_and_methods(json_result):
    for service in json_result['services']:
        for link in service['links']:
            methods = [hint['method'] for hint in link['hints']
                       if 'method' in hint]
            yield link['href'], methods