Question

在学校为作业分配字符打印机时遇到了一些问题，尽管在StackOverflow上到处都看不到如何解决它。我试图使用户能够输入他们想要的前两行顶行和底行的符号，并根据字长将其两个文本的正反斜线居中。但是，这已被证明是一个问题。

我试图更改变量以及创建将文本长度封装为整数的变量，但事实证明这是不成功的。

Repeats = input("How many times should the symbol repeat (1-30 CHARACTERS)?:")
insertText = input("Please enter text:")
insertText = str(insertText)
insertMoreText =input("Please enter text (again):")
insertMoreText= str(insertMoreText)
Repeats = int(Repeats)

#Determining text lengths/stored variables:
text1length = len(insertText)
text2length = len(insertMoreText)
forwardSlash = str("/")
backSlash = str("\\")
symbolsbeforeText = "*"
if(Repeats <= 30):
    print(firstSymbol * Repeats)
    print(secondSymbol * Repeats)
    print(symbolsbeforeText * Repeats)
    print (forwardSlash * text1length) & (insertText.upper()) & (forwardSlash 
    * text1length)
    print (backSlash * text2length) + (insertMoreText.lower()) + (backSlash * 
    text2length)
    print(symbolsbeforeText * Repeats)
    print(secondSymbol * Repeats)
    print(firstSymbol * Repeats)    
else:
    print("You have reached the repetition threshold, please try again.")

错误：

Traceback (most recent call last):
 File "C:\Users\colby\trainingTime.py", line 28, in <module>
 print (forwardSlash * text1length) & (insertText.upper()) & 
 (forwardSlash * text1length)
 TypeError: unsupported operand type(s) for &: 'NoneType' and 'str'

Process terminated with an exit code of 1

Answer 1

#!/bin/bash function debug() { echo "it is debugging" "$@" } function first() { echo "doing some first stuff" } function second() { echo "doing some second stuff" } debug first debug second echo "doing main stuff"函数始终返回print，并且您正在执行返回值None和print（这是一个字符串）的按位与运算。导致上述错误。

您应该使用整个括号内的整个表达式作为参数来调用(insertText.upper())：

print

Answer 2

您的问题是在Python 3+中，print是一个返回None的函数，所以：

print(forwardSlash * text1length)

返回None和

insertText.upper()

返回一个str。因此，您无法在&和None上执行逐位str操作。

解决此问题的一种非常好的方法是使用f字符串，这是自Python 3.6起可用的Python功能：

print(f"{backSlash * text2length} {insertMoreText.lower()} {backSlash * text2length}")

这将使您的代码更具可读性。看到您不需要使用+连接字符串，这可能会引起您的困惑。

TypeError：＆：不支持的操作数类型：“ NoneType”和“ str”

2 个答案: