Question

我的库存计数有N个位置，这个位置需要计算N次，所以我有一个模型用于“位置标题”，另一个模型用于每个标题的项目列表。

我需要链接，排序和获取N个查询集中项目的唯一结果

我有这个：

loc_id = request.POST['loc_id'] # the Id of my location pivot
inv_location = InventoryLocations.objects.get(pk=loc_id) # get the location count pivot
inv_locations = InventoryLocations.objects.filter(location=inv_location.location,
                inventory=inv_location.inventory) #get all related locations counts

# At this point i can have N inv_locations

count_items = [] # list of items in all inventory counts 
for l in inv_locations:
    items = InventoryDetails.objects.filter(inventory_location = l) # get items of every count 
    count_items.append(items)

# Now I have all the items counted in the counts_items array, I need to get from this a single
# list of items Ordered and not repeated    

all_items = chain(count_items) <<< IS THIS CORRECT??
sorted_items = sorted(all_items,key=lambda item: item.epc) << THIS GIVE ME ERROR
unique_items = ???

我的模特是：

class InventoryCount(models.Model):
    ...nothing important

class InventoryLocation(models.Model):
    inventory= models.ForeignKey(InventoryCount)
    location= models.ForeignKey(Location)
    ...

class InventoryDetails(models.Model):
    inventory_location= models.ForeignKey(InventoryLocations)
    epc = models.CharField(max_length=25, null=True, blank=True)
    item= models.ForeignKey(Item)
    ...

基本上，我需要一个按epc排序的数组中所有inventoryDetails计算的所有项目的列表，而不是重复

我被困在这里，我不知道连锁店是否做得对，排序功能给我一个错误，说明该项目没有'epc'属性。

帮助PLZ！

Answer 1

解决您的直接问题 - 假设itertools.chain，chain需要多次迭代。使用chain(*count_items)展开您的查询集列表。

但是使用InventoryDetails.objects.filter(inventory_location__in=inv_locations).order_by('epc').distinct()可以省去一些麻烦 - 在数据库中进行排序和统一，而不是在视图中进行排序。

Django / Python链和排序查询集

1 个答案: