python - 如何从Python中的字典列表中获取所有一个键？

Question

我有一个很大的 JSON 文件，如下所示:

{
  "data" : [
    {"album": "I Look to You", "writer": "Leon Russell", "artist": "Whitney Houston", "year": "2009", "title": "\"A Song for You\""},
    {"album": "Michael Zager Band", "writer": "Michael Zager", "artist": "Whitney Houston", "year": "1983", "title": "\"Life's a Party\""},
    {"album": "Paul Jabara & Friends", "writer": "Paul Jabara", "artist": "Whitney Houston", "year": "1978", "title": "\"Eternal Love\""},
    ...

...我正在尝试制作一个非常简单的 API 来获取不同的值。现在，我可以相当轻松地获取 localhost/data/1/title 例如，获取第一个标题值，但我想通过执行 来获取所有 标题>localhost/titles 之类的。我如何修改此处的 do_GET 方法来添加此类功能？

def do_GET(self):
    self.send_response(200)
    self.send_header('Content-type', 'application/json')
    self.end_headers()

    path = self.path[1:]
    components = string.split(path, '/')

    node = content
    for component in components:
        if len(component) == 0 or component == "favicon.ico":
            continue

        if type(node) == dict:
            node = node[component]

        elif type(node) == list:
            node = node[int(component)]

    self.wfile.write(json.dumps(node))

    return

Answer 1

这里的答案将遵循您当前的动态 URL 模式，无需重大架构更改或要求。

在这里，我使用“all”代替您的网址模式中给定的数字索引，因为我觉得这更好地代表了您的data/[item(s)]/[attribute]范例

以下是一些 URL 和示例输出:

1. /data/1/album =>“迈克尔·扎格乐队”
2. /data/0/title => “一首给你的歌”
3. /data/all/title => [“为你献上一首歌”、“生活就是一场派对”、“永恒的爱”]
4. /data/all/year => ["2009", "1983", "1978"]
5. /data/1 => {"album": "Michael Zager Band", "title": "Life's a Party", "writer": "Michael Zager", "year": “1983”，“艺术家”:“惠特尼·休斯顿”}

PS - 我稍微改变了架构，使用递归，我认为这更好地遵循你想要做的事情。

def do_GET(self):
    self.send_response(200)
    self.send_header('Content-type', 'application/json')
    self.end_headers()

    path = self.path[1:]
    components = string.split(path, '/')

    node = parse_node(content, components)

    self.wfile.write(json.dumps(node))

    return

def parse_node(node, components):
    # For a valid node and component list:
    if node and len(components) and components[0] != "favicon.ico":
        # Dicts will return parse_node of the top-level node component found, 
        # reducing the component list by 1
        if type(node) == dict:
            return parse_node(node.get(components[0], None), components[1:])

        elif type(node) == list:
            # A list with an "all" argument will return a full list of sub-nodes matching the rest of the URL criteria
            if components[0] == "all":
                return [parse_node(n, components[1:]) for n in node]
            # A normal list node request will work as it did previously
            else:
                return parse_node(node[int(components[0])], components[1:])
    else:
        return node

    # Handle bad URL
    return None