Question

我想从Python KeyError例外获取密钥名称：

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]

我已经尝试过这个

print e
print e.args
print e.message

我的代码在django视图里面！

如果我使用ipython例如并尝试e.arg或e.message它工作正常。但后来我在django视图中尝试它，我得到了这个结果：

"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']}>

虽然我只想要'key2'

Answer 1

您可以使用e.args：

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

来自docs：

except子句可以在异常名称后面指定一个变量（或者元组）。该变量与一个异常实例绑定存储在instance.args
中的参数

Answer 2

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 3

异常对象将是缺少的键

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

Answer 4

对于KeyErrors，sys.exc_info()的第二个元素（索引1）保存密钥：

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

输出：

KeyError，违规密钥：'key2'

Answer 5

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

在这种情况下，

将打印key2

已编辑以添加...

不熟悉django但是如果e.message是一致的（并且始终是字符串），则可以使用普通的字符串方法提取密钥，例如。

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

或类似的正则表达式？

Answer 6

由于您的字典实际上是QueryDict，因此返回不同的错误消息，您需要解析异常消息以返回单引号中的第一个单词：

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

从Python KeyError异常中获取密钥名称

6 个答案: