int()参数必须是字符串，类似字节的对象或数字，而不是“方法" [英] int() argument must be a string, a bytes-like object or a number, not 'method'

问题描述

在django 1.8/python 3.4

In django 1.8 /python 3.4

我该怎么办int的方法?

How can I do method to int？

示例:

models.py

models.py

class Bid(models.Model):
    bid_Price = models.IntegerField(default=1)

    def __str__(self):
        return self.bid_Price

    def topPrice(self):
        return 100

view.py

def BidView(request, bid_id):
    bid_data = Bid.objects.get(id=bid_id)
    if int(bid_data.topPrice_text) < 100):

推荐答案

方法的编写方法已经返回一个int，因此您实际上需要调用该方法以使用其返回值:

The way it's written your method is already returning an int, so what you would have to do is actually call the method to use its return value:

if bid_data.topPrice()<100:

我不得不说我不太了解您要在这里做什么，但是 topPrice 可能无论如何都不应该是一种方法.

I do have to say that I don't quite know what you're trying to do here, but topPrice probably shouldn't be a method anyways.

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