错误：'int'对象不是可订阅的

我正在尝试一段简单的代码，获取某人的姓名和年龄，并让他/她知道他们何时21岁......不考虑负面信息和所有这些，只是随机。错误：'int'对象不是可订阅的

我一直收到这个'int' object is not subscriptable错误。

name1 = raw_input("What's your name? ") 
age1 = raw_input ("how old are you? ") 
x = 0 
int([x[age1]]) 
twentyone = 21 - x 
print "Hi, " + name1+ " you will be 21 in: " + twentyone + " years."

来源

2011-11-22 Gal Appelbaum

为什么它说，它不会代替标化说这是不是可转位的？ – nog642

的问题是在该行，

int([x[age1]])

你想要的是

x = int(age1)

您还需要为int转换为字符串输出...

print "Hi, " + name1+ " you will be 21 in: " + str(twentyone) + " years."

完整的脚本看起来像，

name1 = raw_input("What's your name? ") 
age1 = raw_input ("how old are you? ") 
x = 0 
x = int(age1) 
twentyone = 21 - x 
print "Hi, " + name1+ " you will be 21 in: " + str(twentyone) + " years."

来源

2011-11-22 01:05:15 Charles

当你键入x = 0是创建一个新的int变量（名称），并为其分配一个零。

当您键入试图访问age1'th条目的x[age1]时，就好像x是一个数组。

来源

2011-11-22 01:05:44

实际解释发生了什么的唯一答案应该在顶部。 – Caelum

当你写x = 0，x是一个int ...所以你不能这样做，因为x[age1]是xint

来源

2011-11-22 01:05:58 TJD

什么是你想在这里做的：int([x[age1]])？这个不成立。

你只需要投年龄输入为int：

name1 = raw_input("What's your name? ") 
age1 = raw_input ("how old are you? ") 
twentyone = 21 - int(age1) 
print "Hi, %s you will be 21 in: %d years." % (name1, twentyone)

来源

2011-11-22 01:06:11 juliomalegria

你需要先转换AGE-1转换成int，所以它可以做负。之后把结果返回字符串显示：

name1 = raw_input("What's your name? ") 
age1 = raw_input ("how old are you? ") 
twentyone = str(21 - int(age1)) 
print "Hi, " + name1+ " you will be 21 in: " + twentyone + " years."

来源

2011-11-22 01:07:46

name1 = input("What's your name? ") 
age1 = int(input ("how old are you? ")) 
twentyone = str(21 - int(age1)) 

if age1<21: 
    print ("Hi, " + name1+ " you will be 21 in: " + twentyone + " years.") 

else: 
    print("You are over the age of 21")

来源

2017-01-27 14:59:47

错误：'int'对象不是可订阅的

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