获取键名

Question

我想从Python KeyError例外拿到钥匙名称：

例如：

myDict = {'key1':'value1'} 
try: 
    x1 = myDict['key1'] 
    x2 = myDict['key2'] 
except KeyError as e: 
    # here i want to use the name of the key that was missing which is 'key2' in this example 
    print error_msg[missing_key] 

我已经尝试过这种

print e 
print e.args 
print e.message 

我的代码在Django视图中！

如果我使用ipython为例，并尝试e.arg或e.message它工作正常。 但后来我尝试它，而Django视图中我得到这个结果：

"Key 'key2' not found in <QueryDict: {u'key1': [u'value']}>" 
("Key 'key2' not found in <QueryDict: {u'key1': [u'value']}>",) 
Key 'key2' not found in <QueryDict: {u'key1': [u'value']}> 

，而我只是想在这种情况下，“密钥2”

Answer 1

至于你的字典实际上是QueryDict，因此，返回不同的错误信息，则需要分析异常消息，以在单引号返回的第一个字：

import re 
#... 
except KeyError as e: 
    m = re.search("'([^']*)'", e.message) 
    key = m.group(1) 

Answer 2

myDict = {'key1':'value1'} 
try: 
    x1 = myDict['key1'] 
    x2 = myDict['key2'] 
except KeyError as e: 
    print e.message 

将打印key2

编辑添加...

不熟悉django，但如果e.message是一致的（并且始终是一个字符串），您可以使用普通字符串方法提取密钥，如。

my_non_key = e.message.split(" ")[1] 

或类似的正则表达式的东西？

Answer 3

myDict = {'key1':'value1'} 
try: 
    x1 = myDict['key1'] 
    x2 = myDict['key2'] 
except KeyError as e: 
    # here i want to use the name of the key that was missing which is 'key2' in this example 
    print "missing key in mydict:", e.message 

Answer 4

您可以使用e.args：

[53]: try: 
    x2 = myDict['key2'] 
except KeyError as e:  
    print e.args[0] 
    ....:  
key2 

从docs：

的except子句中的名称（或 元组）后，可以指定一个变量。变量绑定到一个异常实例与存储在instance.args

Answer 5

的异常对象的 参数将是缺少关键

a={"a":1} 
try: 
    print(a["b"]) 
except KeyError as e: 
    print(e)  #prints "b" 

Answer 6

对于KeyErrors，所述第二元件的sys.exc_info()（索引1）成立的关键：

import sys 
try: 
    x1 = myDict['key1'] 
    x2 = myDict['key2'] 
except KeyError: 
    print 'KeyError, offending key:', sys.exc_info()[1] 

输出：

KeyError异常，得罪键：“键2”