本文主要介绍了python中“TypeError: Can’t convert ‘int’ object to str implicitly”报错的解决办法。
原文出处：
TypeError: Can’t convert ‘int’ object to str implicitly
问题
我在尝试着写一个文字游戏，遇到了一个函数错误，这个函数实现的功能是：在你输入完字符后，就会消耗你的技能分。刚开始时报错信息显示我在试图用一个整数减去 一个字符，对应代码为“balance - strength”，这个错误很明显，因此我将其改为“strength = int(strength)”修复了… 但是现在我遇到了一个以前从未见过的 错误(o(╯□╰)o我是一个新手)，我不知道它试图在告诉我什么以及如何修复它。
以下为该函数对应的代码：
def attributeSelection():
balance = 25
print("Your SP balance is currently 25.")
strength = input("How much SP do you want to put into strength?")
strength = int(strength)
balanceAfterStrength = balance - strength
if balanceAfterStrength == 0:
print("Your SP balance is now 0.")
attributeConfirmation()
elif strength < 0:
print("That is an invalid input. Restarting attribute selection. Keep an eye on your balance this time!")
attributeSelection()
elif strength > balance:
print("That is an invalid input. Restarting attribute selection. Keep an eye on your balance this time!")
attributeSelection()
elif balanceAfterStrength > 0 and balanceAfterStrength < 26:
print("Ok. You're balance is now at " + balanceAfterStrength + " skill points.")
else:
print("That is an invalid input. Restarting attribute selection.")
attributeSelection()
以下为运行此部分代码后的报错信息：
Your SP balance is currently 25.
How much SP do you want to put into strength?5
Traceback (most recent call last):
File "C:\Python32\APOCALYPSE GAME LIBRARY\apocalypseGame.py", line 205, in <module>
gender()
File "C:\Python32\APOCALYPSE GAME LIBRARY\apocalypseGame.py", line 22, in gender
customizationMan()
File "C:\Python32\APOCALYPSE GAME LIBRARY\apocalypseGame.py", line 54, in customizationMan
characterConfirmation()
File "C:\Python32\APOCALYPSE GAME LIBRARY\apocalypseGame.py", line 93, in characterConfirmation
characterConfirmation()
File "C:\Python32\APOCALYPSE GAME LIBRARY\apocalypseGame.py", line 85, in characterConfirmation
attributeSelection()
File "C:\Python32\APOCALYPSE GAME LIBRARY\apocalypseGame.py", line 143, in attributeSelection
print("Ok. You're balance is now at " + balanceAfterStrength + " skill points.")
TypeError: Can't convert 'int' object to str implicitly
有人知道如何解决这个问题吗？先行感谢。
(译者注：提问者报错信息中涉及较多，部分为其项目代码文件。为缩短报错信息，我将提问者所提到的函数部分代码粘贴到本机后，运行完对应的报错信息如下)
Your SP balance is currently 25.
How much SP do you want to put into strength?5
Traceback (most recent call last):
File "test.py", line 26, in <module>
attributeSelection()
File "test.py", line 20, in attributeSelection
print("Ok. You're balance is now at " + balanceAfterStrength + " skill points.")
TypeError: cannot concatenate 'str' and 'int' objects
答案
你不能将整型(int)与字符串(string)连在一起。你需要使用’str’函数将整型(int)转换为字符型(string)，或者使用’formatting’格式化输出。
将
print("Ok. Your balance is now at " + balanceAfterStrength + " skill points.")
改为：
({} .format方式)
print("Ok. Your balance is now at {} skill points.".format(balanceAfterStrength))
或改为：
(使用str函数转换类型)
print("Ok. Your balance is now at " + str(balanceAfterStrength) + " skill points.")
或按照下面的一条评论所提及的那样做，使用’,’将不同的字符串传递给print函数，而不是使用’+’连接。
(涉及的评论为：你不能使用’,’连接字符串；你可以用’,’将参数分开传递给print函数，这些参数会以空格分割，一个接一个的打印出来)
print("Ok. Your balance is now at ", balanceAfterStrength, " skill points.")
总结
当同时打印字符及整型变量时，有以下几种方式来避免“TypeError”报错。
假设变量temp = 3，要输出的内容为the number you input is 3.
1.使用str强制将整型转换为字符型
print 'the nume you input is ' + str(temp)
2.使用格式化输出(python2中适用，“format % values”形式)，详细使用方法可参考官方文档：string-formatting
print 'the nume you input is %s' % temp
3.使用” str.format()”(python2.6以上)格式化输出，详细使用方法可参考官方文档：string-formatting
print 'the nume you input is {}' .format(temp)
4.使用逗号将变量和字符串分隔
print 'the nume you input is' , temp
% 及 .format() 两种格式化输出对比
更多实例对比请参考：
https://pyformat.info/ https://github.com/ulope/pyformat.info
基本输出
Old '%s %s' % ('one', 'two')
New '{} {}'.format('one', 'two')
Output one two
Old '%d %d' % (1, 2)
New '{} {}'.format(1, 2)
Output 1 2
#右对齐
Old '%10s' % ('test',)
New '{:>10}'.format('test')
Output test #test左边有六个空格
#左对齐
Old '%-10s' % ('test',)
New '{:<10}'.format('test')
Output test #test右边有六个空格
#字典
person = {'first': 'Jean-Luc', 'last': 'Picard'}
New '{p[first]} {p[last]}'.format(p=person)
Output Jean-Luc Picard
#列表
data = [4, 8, 15, 16, 23, 42]
New '{d[4]} {d[5]}'.format(d=data)
Output 23 42
#Accessing arguments by position:
>>> '{0}, {1}, {2}'.format('a', 'b', 'c')
'a, b, c'
>>> '{}, {}, {}'.format('a', 'b', 'c') # 3.1+ only
'a, b, c'
>>> '{2}, {1}, {0}'.format('a', 'b', 'c')
'c, b, a'
>>> '{2}, {1}, {0}'.format(*'abc') # unpacking argument sequence
'c, b, a'
>>> '{0}{1}{0}'.format('abra', 'cad') # arguments' indices can be repeated
'abracadabra'
#Accessing arguments by name:
>>> 'Coordinates: {latitude}, {longitude}'.format(latitude='37.24N', longitude='-115.81W')
'Coordinates: 37.24N, -115.81W'
>>> coord = {'latitude': '37.24N', 'longitude': '-115.81W'}
>>> 'Coordinates: {latitude}, {longitude}'.format(**coord)
'Coordinates: 37.24N, -115.81W'