我有下面的代码
test = "have it break."
selectiveEscape = "Print percent % in sentence and not %s" % test
print(selectiveEscape)
我想要得到输出:
Print percent % in sentence and not have it break.
实际发生了什么:
selectiveEscape = "Use percent % in sentence and not %s" % test
TypeError: %d format: a number is required, not str
当前回答
>>> test = "have it break."
>>> selectiveEscape = "Print percent %% in sentence and not %s" % test
>>> print selectiveEscape
Print percent % in sentence and not have it break.
其他回答
>>> test = "have it break."
>>> selectiveEscape = "Print percent %% in sentence and not %s" % test
>>> print selectiveEscape
Print percent % in sentence and not have it break.
或者，从Python 2.6开始，你可以使用新的字符串格式(在PEP 3101中描述):
'Print percent % in sentence and not {0}'.format(test)
当你的字符串变得更复杂时，这尤其方便。
您不能有选择地转义%，因为%总是有一个特殊的含义，这取决于后面的字符。
在Python文档中，在该部分的第二个表的底部，它声明:
'%' No argument is converted, results in a '%' character in the result.
因此，你应该使用:
selectiveEscape = "Print percent %% in sentence and not %s" % (test, )
(请注意将tuple的参数明确地更改为%)
如果不知道这些，我可能会这样做:
selectiveEscape = "Print percent %s in sentence and not %s" % ('%', test)
用你已经掌握的知识。
如果你正在使用Python 3.6或更新版本，你可以使用f-string:
>>> test = "have it break."
>>> selectiveEscape = f"Print percent % in sentence and not {test}"
>>> print(selectiveEscape)
... Print percent % in sentence and not have it break.
尝试使用%%打印%符号。
