python：从函数返回异常

Question

假设我有以下功能：

def test():

  ...

  if x['error']:

    raise

无论是否x['error']定义，这都会引发异常。

相反，如果我尝试这个，它不会抛出任何异常：

def test():

  ...

  try:

    if x['error']:

      raise

  except:

    return

如何测试特定值并在定义时返回异常，如果未定义则成功返回？

Answer 1

def test():  ...  if x.get(‘error’):    raise您可以使用字典的内置get函数避免无意中引发错误。None如果指定键处的值不存在，则Get 将返回而不是抛出异常。

Answer 2

如果要将错误作为字符串返回：>>> def test():&nbsp; &nbsp; try:&nbsp; &nbsp; &nbsp; &nbsp; if x['error']:raise&nbsp; &nbsp; except Exception as err:&nbsp; &nbsp; &nbsp; &nbsp; return err>>> test()NameError("name 'x' is not defined",)如果要发生错误：>>> def test():&nbsp; &nbsp; try:&nbsp; &nbsp; &nbsp; &nbsp; if x['error']:raise&nbsp; &nbsp; except:&nbsp; &nbsp; &nbsp; &nbsp; raise>>> test()Traceback (most recent call last):&nbsp; File "<pyshell#20>", line 1, in <module>&nbsp; &nbsp; test()&nbsp; File "<pyshell#19>", line 3, in test&nbsp; &nbsp; if x['error']:raiseNameError: name 'x' is not defined

Answer 3

试试这个def check_not_exist(d,k):&nbsp; &nbsp;#if keys exists in dict,raise it&nbsp; &nbsp;if k in d:&nbsp; &nbsp; &nbsp;raise&nbsp; &nbsp;else:&nbsp; &nbsp; &nbsp;return True