python - 按模型创建记录

Question

假设我有这样的模型:

class Recipe (models.Model):
    par_recipe = models.CharField(max_length=200)

class Line (models.Model):
    par_machine = models.CharField(max_length=200)

class Measurements (models.Model):
    par_value = models.IntegerField(default=0)
    id_line = models.ForeignKey(Line)
    id_recipe = models.ForeignKey(Recipe)

我是否理解正确，通过这种方式我有一个 1:1 关系，并且添加条目 id 将自动创建 id_line,id_recipe。

我将添加例如:

for row in ws.iter_rows(row_offset=1):
        recipe =Recipe()
        line = line()
        measurements = Measurements()

        recipe.par_recipe = row[1].value
        line.par_machine = row[2].value
        measurements.par_value = row[8].value

关于测量的小问题被设想为所有辅助键都应该去它，现在它正确实现了吗？

Answer 1

不是那样的，你必须把它们绑在一起:

for row in ws.iter_rows(row_offset=1):
    recipe =Recipe.objects.create(par_recipe=row[1].value)
    line = Line.objects.create(par_machine=row[2].value)
    measurements = Measurements.objects.create(
        par_value=row[8].value, 
        id_line=line, 
        id_recipe=recipe
    )

这些都不是数据库优化的，您可以使用事务来优化数据库写入。

如果有很多行，你可以通过使用事务使它更快:

from django.db import transaction

with transaction.atomic():
    for row in ws.iter_rows(row_offset=1):
        recipe =Recipe.objects.create(par_recipe=row[1].value)
        line = Line.objects.create(par_machine=row[2].value)
        measurements = Measurements.objects.create(
            par_value=row[8].value, 
            id_line=line, 
            id_recipe=recipe
        )

这将创建一个事务并写入一个而不是每次。但它也会因错误而导致整个交易失败。

参见 Django Database Transactions

您可以通过计算记录数并每 1000 条记录写入一次来获得更多创意，例如:

from django.db import transaction

with transaction.atomic():
    for idx, row in enumerate(ws.iter_rows(row_offset=1)):
        recipe =Recipe.objects.create(par_recipe=row[1].value)
        line = Line.objects.create(par_machine=row[2].value)
        measurements = Measurements.objects.create(
            par_value=row[8].value, 
            id_line=line, 
            id_recipe=recipe
        )
        # every 1000 records, commmit the transaction
        if idx % 1000 == 0:
           transaction.commit()