无法从配置中迭代字典并将其格式化以打印

Question

大约三周前，我开始自学 Python，并且一直在为我的 Ark 服务器为 Red discord bot 开发 RCON cog。我有一个正在尝试格式化的配置，以便我可以使用命令轻松查看服务器的所有设置。

这是我目前为服务器创建的配置

{
   "clusters": {
      "pvp": {
         "joinchannel": 739270285450018927,
         "leavechannel": 851819367439400960,
         "adminlogchannel": 848933429164507156,
         "globalchatchannel": 841118994399494164,
         "servers": {
            "rag": {
               "ip": "192.168.1.201",
               "port": 27020,
               "password": "passwd",
               "chatchannel": 770891102601084928
            },
            "val": {
               "ip": "192.168.1.203",
               "port": 27024,
               "password": "passwd",
               "chatchannel": 770891102601084928
            },
            "island": {
               "ip": "192.168.1.205",
               "port": 27021,
               "password": "passwd",
               "chatchannel": 770891102601084928
            }
         }
      },
      "pve": {
         "joinchannel": 739270285450018927,
         "leavechannel": 851819367439400960,
         "adminlogchannel": 848933429164507156,
         "globalchatchannel": 841118994399494164,
         "servers": {
            "rag": {
               "ip": "192.168.1.202",
               "port": 27022,
               "password": "passwd",
               "chatchannel": 770891102601084928
            },
            "val": {
               "ip": "192.168.1.204",
               "port": 27023,
               "password": "passwd",
               "chatchannel": 770891102601084928
            },
            "island": {
               "ip": "192.168.1.206",
               "port": 27025,
               "password": "passwd",
               "chatchannel": 770891102601084928
            }
         }
      }
   }
}

如果我要打印“服务器设置”，我希望它看起来像这样

PvP 集群

map: rag
ip: 192.168.1.201
port: 27020
password: psswrd
chatchannel: 770891102601084928

map: val
ip: 192.168.1.203
port: 27024
password: psswrd
chatchannel: 770891102601084928

map: island
ip: 192.168.1.205
port: 27021
password: psswrd
chatchannel: 770891102601084928

PvE 集群

map: rag
ip: 192.168.1.202
port: 27022
password: psswrd
chatchannel: 770891102601084928

map: val
ip: 192.168.1.204
port: 27023
password: psswrd
chatchannel: 770891102601084928

map: island
ip: 192.168.1.206
port: 27025
password: psswrd
chatchannel: 770891102601084928

到目前为止我已经尝试过

        serverlist = []
        settings = await self.config.guild(ctx.guild).all()
        for cluster in settings["clusters"]:
            for name, server in cluster["servers"].items():
                serverlist.extend(settings[cluster]["servers"])
        await ctx.send(serverlist)

还有一些其他可能是菜鸟的逻辑，让它按照我想要的方式格式化，但是有这么多嵌套的字典，我迷路了。

最终我会把它做成一个嵌入。但现在我只是想更好地了解如何从我的配置中操作字典以更好地在不和谐中显示它。

我也想一次性打印所有这些作为一条消息。

如果有人能在这方面牵着我的手，并使用我的配置展示一个示例，那将不胜感激:)

Answer 1

此代码输出您想要的内容，但我怀疑您正在以结构化形式转发信息：

types = ['pvp','pve']
maps = ['rag','val','island']
items = ['ip','port','password','chatchannel']

for t in types:
    servers = settings['clusters'][t]['servers']
    print(t,'Cluster')
    for m in maps:
        s = servers[m]
        for i in items:
            print(i,s[i])
        print()
    print()

按要求输出

Answer 2

这应该可行：

print_all = []
for pv in settings["clusters"]:
    print_all.append(f'\n{pv.upper()} Cluster\n')
    for server in settings["clusters"][pv]["servers"]:
        print_all.append(f'map: {server}')
        for k,v in settings["clusters"][pv]["servers"][server].items():
            print_all.append(f'{k}: {v}')
        print_all.append('')
print_all = '\n'.join(print_all) # a single string
print(print_all)