Question

可能重复：
Syntax error on print with Python 3

我知道Python应该是解释语言之神......它的简单性，缺乏冗余，等等等等。但由于我已经习惯了C，C ++，Java，Javascript和Php，我必须承认这很烦人。这是我的代码：

#!/usr/bin/python3.1
def shit(texture, weight):
    if textura is "green":
        texturaTxt = "Shit is green"
    elif textura is "blue":
        texturaTxt = "Shit is blue"
    elif textura is "purple":
        texturaTxt = "Shit is purple"
    else:
        print "Incorrect color"
        break
    if weight < 20:
        pesoTxt = " and weights so few"
    elif weight >= 20 and peso <=50:
        pesoTxt = " and weights mid"
    elif weight > 50:
        pesoTxt = " and weights a damn lot"
    else:
        print "Incorrect weight"
    return texturaTxt + pesoTxt

c = input("Introduce crappy color: ")
e = input("Introduce stupid weight: ")
r = shit(c, w)
print r

我正在努力学习Python以及我想要实现的目标是：

...
function shit(texture, weight) {
    string textureTxt = "Shit is ", pesoTxt = " and weights ";
    switch(texture) {
        case "green": textureTxt .= "green as an applee"; break;
        case "blue": textureTxt .= "blue as the sky"; break;
        case "purple": textureTxt .= "purple as Barny"; break;
        default: cout<<"Incorrect texture, try again later" <<endl; exit;
    }
    //etc
}
string color = "";
int weight = 0;
cout<<"Introduce color: ";
cin>>color;
//etc 
cout<<shit(color, weight);
...

但是我放弃了，我不能让它发挥作用，它会抛出我所有的错误。希望有一些C ++或php或C到python转换器。

感谢您的帮助

Answer 1

Python3不再支持print作为特殊形式，其参数位于print关键字后面（如Python 2.x中的print x）。相反，print现在是一个函数，需要一个参数列表，如： print(x)。请参阅http://docs.python.org/release/3.0.1/whatsnew/3.0.html

中的“打印是一项功能”

此外，break语句不能在循环外发生。除了C switch语句之外，if不支持break。因为if语句没有直接逻辑，所以不需要它。为了停止执行该函数并返回给调用者，请使用return。

使用等于运算符is而不是==运算符。 is测试对象是否相同，这是一个比平等更严格的测试。可以找到更多详细信息here。

此外，你将体重作为一个字符串。您可能希望使用函数int(weight)将字符串转换为整数以与其他整数值进行比较。

还有许多其他小错误：

当您尝试在函数调用中使用未定义的变量名e时，您将权重的用户输入分配给w
函数中的第一个参数名为texture，但您在函数体中使用textura。
您在一个实例中使用peso而不是weight

这是一个（不那么冒犯）重写，消除了这些错误：

def stuff(color, weight):

    color_txt = "Your stuff is "
    if color == "green":
        color_txt += "green as an apple"
    elif color == "blue":
        color_txt += "blue as the sky"
    elif color == "purple":
        color_txt += "purple as an eggplant"
    else:
        print("Invalid color!")
        return

    w = 0
    # converting a string to an integer might
    # throw an exception if the input is invalid
    try:
        w = int(weight)
    except ValueError:
        print("Invalid weight!")
        return

    weight_txt = ""
    if w<20:
        weight_txt = " and weighs so few"
    elif 20 <= w <= 50:
        weight_txt = " and weighs mid"
    elif w > 50:
        weight_txt = " and weighs a lot"
    else:
        print("Invalid weight!")

    return color_txt + weight_txt

c = input("Input color: ")
w = input("Input weight: ")
r = stuff(c, w)
print(r)

基本Python 3问题的语法错误

1 个答案: