Question

我正在尝试使用django创建一个字符串对象

我的模特：

lines = []
with open('file.txt', 'r+') as f:
    # You can actually process the file line by line without saving it into a list:
    for line in f:
        # For clearer code, you can index the split function here:
        line = line.split(' ', 1)[0]
        lines.append(line)

创建代码：

class Example(models.Model):
    user_id = models.IntegerField(blank=True, null=True)
    field_1 = models.DecimalField(max_digits=15, decimal_places=2, blank=True, null=True)
    field_2 = models.DecimalField(max_digits=15, decimal_places=2, blank=True, null=True)
    field_3 = models.DecimalField(max_digits=15, decimal_places=2, blank=True, null=True)

如何创建此。..谢谢.. !!

Answer 1

 real_field = "field_"+str(level)
 example = Example(
                user_id=user_id,
                ......
 )
example.setattr(real_field,value_1)
example.save()

setattr是您要寻找的东西

或

 data = {}
 data["field_"+str(level)] = value1
 example = Example(
                user_id=user_id,
                ......,
                **data
 )
 example.save()

Answer 2

您可以将字典传递给模型构造函数：

example = Example(**{
                'user_id': user_id,
                real_field: value_1,
                ......
 })
 example.save()

Django查询创建-字段由字符串确定

2 个答案: