Python 格式抛出 KeyError [英] Python format throws KeyError

问题描述

以下代码片段:

template = "\                                                                                
function routes(app, model){\                                                                
  app.get('/preNew{className}', function(req, res){\                                         
    res.render('{className}'.ejs, {});\                                                      
  });\                                                                                       
});".format(className=className)

抛出一个KeyError:

throws a KeyError:

Traceback (most recent call last):   File "createController.py", line 31, in <module>
    });".format(className=className) KeyError: '  app'

有人知道为什么吗?

推荐答案

该代码中有许多未转义的大括号.Python 将所有大括号视为占位符，并试图将它们全部替换.但是，您只提供了一个值.

You have a number of unescaped braces in that code. Python considers all braces to be placeholders and is trying to substitute them all. However, you have only supplied one value.

我希望您不希望所有大括号都成为占位符，因此您应该将不想替换的大括号加倍.如:

I expect that you don't want all your braces to be placeholders, so you should double the ones that you don't want substituted. Such as:

template = """                                                                  
function routes(app, model){{
  app.get('/preNew{className}', function(req, res){{
    res.render('{className}'.ejs, {{}});                                           
  }};                                                      
}});""".format(className=className)

我还冒昧地为字符串文字使用了三重引号，因此您不需要在每行末尾使用反斜杠.

I also took the liberty of using triple quotes for the string literal so you don't need the backslashes at the end of each line.

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