Question

class Line(models.Model):
    _name = "line"

line1.quantity = 5
line2.quantity = 8
line3.quantity = 3

def count_line(self):
    for line in self:
        # i need to count all lines and print that line in the end #that has least quantity.

它更像是伪代码，我只想知道如何做到这一点。

Answer 1

如果您有一个包含所有行的列表，则可以执行以下操作：

line_list.sort(key=lambda x: x.quantity, reverse=False)
less_quantity = line_list[0].quantity

Answer 2

由于对quantity属性没有任何限制，因此此代码应处理边缘情况。请注意，它根据需要打印quantity最小的行，并返回总行数。如果没有行，则打印输出为None，并返回数字0。如果quantity有一个下限，则可以简化代码。

def count_line(self):
    """Count all lines, print that line in the end that has least
    quantity, and return the number of lines."""
    counter = 0
    result = None
    for counter, line in enumerate(self):
        if result is None or result.quantity > line.quantity:
            result = line
    print(result)
    return counter

计算行数并减少打印量

2 个答案: