python - 为什么我不能在 raw_input 期间捕获 KeyboardInterrupt？

Question

这是一个测试用例。

try:
    targ = raw_input("Please enter target: ")
except KeyboardInterrupt:
    print "Cancelled"
print targ

当我按 ctrl+c- 时，我的输出如下

NameError: name 'targ' is not defined

我的意图是“取消”输出。当我尝试在 raw_input 期间捕获 KeyboardInterrupt 时为什么会发生这种情况的任何想法？

谢谢！

Answer 1

在上面的代码中，当引发异常时，没有定义 targ。您应该仅在未引发异常时打印。

try:
    targ = raw_input("Please enter target: ")
    print targ
except KeyboardInterrupt:
    print "Cancelled"

Answer 2

发生错误是因为如果KeyboardInterrupt引发 a ，则变量targ永远不会被初始化。

try:
    targ = raw_input("Please enter target: ")
except KeyboardInterrupt:
    print "Cancelled"

Please enter target: 
Cancelled

>>> targ

Traceback (most recent call last):
  File "<pyshell#19>", line 1, in <module>
    targ
NameError: name 'targ' is not defined

当它不发生时，

try:
    targ = raw_input("Please enter target: ")
except KeyboardInterrupt:
    print "Cancelled"


Please enter target: abc

>>> targ
'abc'

targ如果未引发异常，您可以更改代码以打印，通过在try语句中打印它，请参阅以下演示。

try:
    targ = raw_input("Please enter target: ")
    print targ
except KeyboardInterrupt:
    print "Cancelled"


Please enter target: abc
abc

try:
    targ = raw_input("Please enter target: ")
    print targ
except KeyboardInterrupt:
    print "Cancelled"


Please enter target: 
Cancelled