Question

try:
    f = int(factor)
    if (factor == 0):
        print("No factors present to derive");
    elif (factor != int):
        print ("Numbers only please!");
    elif (factor >> 4):
        print("Four maximum factors!");
    else:
        f = int(factor);
if f == 1:
    coefficientone = raw_input("What is the coefficient on the first term?")
        try:
        coef1 = int(coefficientone)
        if (coef1 == 0):
            print "No coefficient present, please re-enter polynomial"
        elif (coef1 != int)
            print "Numbers only please!"
        else:
            coef1 = int(coefficientone)

这会在if f == 1：行返回语法错误。请帮忙！从我在这个网站上看到的和大多数其他网站看起来是正确的。我也很感激代码的任何其他部分的任何帮助，因为这是我第一次使用python。提前致谢。

Answer 1

如果您要添加try块，则必须使用匹配的except来处理您正在捕获的异常。

如果你不知道为什么要添加try块，最好只是删除它，以免掩盖潜在的错误。最好只有except个特定的错误类型。

这是固定代码。这些变化都附有评论。

try:
    f = int(factor)
    if (factor == 0):
        print("No factors present to derive");
    elif (factor != int):
        print ("Numbers only please!");
    elif (factor >> 4):
        print("Four maximum factors!");
    else:
        f = int(factor);
except Exception as e:                            # Add this
    print 'ERROR: {0}'.format(e)                  #

if f == 1:
    coefficientone = raw_input("What is the coefficient on the first term?")
    try:                                          # Un-indent this
        coef1 = int(coefficientone)
        if (coef1 == 0):
            print "No coefficient present, please re-enter polynomial"
        elif (coef1 != int)
            print "Numbers only please!"
        else:
            coef1 = int(coefficientone)
    except Exception as e:                        # Add this
        print 'ERROR: {0}'.format(e)              #

最后，您似乎计划向用户询问他指定的系数。为此，您可能需要考虑使用for循环来处理输入，并使用list来存储值。我将把实施细节留给您。那里有很多资源。

Answer 2

#an error that I am seeing is that factor != int should be:
elif type(factor) != int:
#and adding to the try I think that you will also need a except ValueError:

如果语句语法错误？

2 个答案: