Question

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/她知道他们何时年满21岁......不考虑否定和所有这些，只是随机。

我不断收到此'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."

Answer 1

当您键入正在创建新x = 0变量（名称）并为其指定零的int时。

当您输入试图访问x[age1]条目的age1时，就好像x是一个数组一样。

Answer 2

问题在于，

int([x[age1]])

你想要的是

x = int(age1)

您还需要将int转换为输出的字符串...

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

完整的脚本如下，

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."

Answer 3

当您撰写x = 0时，x是一个整数...因此您无法x[age1]，因为x是int

Answer 4

你想在这做什么：int([x[age1]]) ??这毫无意义。

您只需将年龄输入转换为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)

Answer 5

这样做会简单得多；

name = input("What's your name? ")
age = int(input("How old are you? "))
print ("Hi,{0} you will be 21 in {1} years.".format(name, 21 - age))`

Answer 6

您需要先将age1转换为int，因此可以执行减号。之后将结果返回到字符串以显示：

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."

Answer 7

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")

Answer 8

嗯，所有这些答案都是正确的，但这是一种更现代的方法！

router.post('/order', auth.check.bind(auth), (req, res) => {
const paramsProdOrderId = [req.body.product_id, req.body.product_ordered];
db.query(sqlProdOrder, paramsProdOrderId).then((results) => {
    if (results.rowCount === 0) {
        res.redirect('/products');
        return;
    }
    res.redirect('/');
})

.catch((err) => {
    res.redirect('/error/404')
});

Answer 9

在Python中，

'int'对象不可下标是TypeError。为了更好地理解此错误的发生方式，让我们考虑以下示例：

list1 = [1, 2, 3]
print(list1[0][0])

如果运行代码，您将在Python3中收到相同的TypeError。

TypeError: 'int' object is not subscriptable

此处列表的索引超出范围。如果将代码修改为：

print(list1[0])

现在列表的索引在范围内，输出将为1（因为Python列表中的索引从零开始）。

运行代码（与问题一起提供）时，发生TypeError并指向代码的第4行：

int([x[age1]])

其目的可能是创建一个整数列表（尽管完全不需要为单个数字创建列表）。只需将输入（然后转换为整数）分配给变量即可。

因此，最好以这种方式进行编码：

name = input("What's your name? ")
age = int(input('How old are you? '))
twentyOne = 21 - age
if(twentyOne < 0):
    print('Hi {0}, you are above 21 years' .format(name))
elif(twentyOne == 0):
    print('Hi {0}, you are 21 years old' .format(name))
else:
    print('Hi {0}, you will be 21 years in {1} year(s)' .format(name, twentyOne))

输出：

What's your name? Steve
How old are you? 21
Hi Steve, you are 21 years old

Answer 10

x已经是整数（x = 0），并且您再次尝试使x再次成为整数，并且您还给出了索引限制，因为x已经只有一个索引（0），并且您尝试给与x相同的索引年龄，这就是为什么您会收到此错误。用这个简单的代码

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

错误：＆＃39; int＆＃39; object不是可下载的 - Python

10 个答案: