如何访问深JSON在Python属性

Question

这个PHP代码工作（以下JSON）：

$usaData = json_decode(stripslashes(file_get_contents('usaJson.txt')),true); 
foreach($usaData['USA']['States'] as $state){ 
    foreach($state['Cities'] as $city){ 
      $zipCode = $city['zipcode']; 
     } 
    } 
} 

我试图做同样的蟒蛇，但它给了我<type 'exceptions.TypeError'>: string indices must be integers

usaData = json.loads(get_file('usaJson.txt')) 
for state in usaData['USA']['States']: 
    for city in state['Cities']: 
     zipCode = city['zipcode'] 

我的数据结构是这样的：

{ 
    "USA":{ 
    "States":{ 
     "AL":{ 
     "Cities":[ 
      { 
      "city":"auburn", 
      "zipcode":"36830" 
      }, 
      { 
      "city":"birmingham", 
      "zipcode":"35201" 
      } 
     ] 
     }, 
     "AK":{ 
     "Cities":[ 
      { 
      "city":"anchorage", 
      "zipcode":"99501" 
      }, 
      { 
      "city":"fairbanks", 
      "zipcode":"99701" 
      } 
     ] 
     } 
    } 
    } 
} 

那么如何访问python中的zipcode？

Answer 1

for state in usaData.USA.States: 

，或者可能是，这取决于结构：

for state in usaData.USA.States.all(): 

编辑：不漂亮，可以使用清理，但这应该工作：

for state in usaData['USA']['States'].keys(): 
    for cities in usaData['USA']['States'][state]['Cities']: 
     print cities['zipcode'] 

Answer 2

您的问题是迭代通过字典给你的关键，而不是在字典中的价值。所以，你需要的是这样的：

for state_name in usaData['USA']['States'] 
    for city in usaData['USA']['States'][state_name]['Cities'] 
    print city['zipcode'] 

这是非常丑陋的

一个更好的版本是使用在字典中的items功能，它看起来像这样：

for state_name, state in usaData['USA']['States'].items(): 
    for city in state['Cities'] 
    print city['zipcode']