python - 如何在 Python 中获取解析为列表/字典的 JSON 服务器响应的详细信息

Question

我是 Python 新手。我一直在尝试解析作为函数参数发送的响应。

我一直在尝试将函数从 Perl 转换为 Python。

Perl 块看起来像这样：

sub fetchId_byusername
{
     my ($self,$resString,$name) =@_;
     my $my_id;
     my @arr = @{$json->allow_nonref->utf8->decode($resString)};
     foreach(@arr)
     {
          my %hash = %{$_};
          foreach my $keys (keys %hash)
          {
               $my_id = $hash{id} if($hash{name} eq $name);
          }
     }
     print "Fetched Id is : $my_id\n";
     return $my_id;

解析 JSON 数据的部分让我很困扰。我如何在 python3 中编写这个。

我尝试了类似的东西

    def fetchID_byUsername(self, resString, name):
        arr = []
        user_id = 0
        arr = resString.content.decode('utf-8', errors="replace")
        for item in arr:
            temp_hash = {}
            temp_hash = item
            for index in temp_hash.keys():
                if temp_hash[name] == name:
                    user_id = temp_hash[id]
        print("Fetched  ID is: {}".format(user_id))
        return user_id

现在我不确定这是否是正确的方法。

json 输入类似于：

[{"id":12345,"name":"11","email":"[e-mail]","groups":[{"id":6967,"name":"Test1"},{"id":123456,"name":"E1"}],"department":{"id":3863,"name":"Department1"},"comments":"111","adminUser":false},{"id":123457,"name":"1234567","email":"[e-mail]","groups":[{"id":1657,"name":"mytest"},{"id":58881,"name":"Service Admin"}],"department":{"id":182,"name":"Service Admin"},"comments":"12345000","adminUser":true}]

提前致谢。

标签： pythonjsonpython-3.xcode-conversion

Answer 1

首先导入 json 库并使用 json.loads() 像：

import json

x = json.loads(json_feed) #This converts the json feed to a python dictionary
print(x["key"]) #values to "key"