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我正在构建一个在数据库中查找每个电话号码的应用程序。如果有任何重复，我想获取找到的第一个电话号码作为该电话号码的主记录，然后对于重复信息（姓名，位置），获取这些字段中的每一个，并将其添加到主记录电话数字字段（名称、位置），以分号分隔。

检查发现的主电话号码记录的重复信息后，结果如下所示：

Name                      Location               Phone number
Helene,Sandra             New Yok, Boston        000-000

请在下面找到我的模型：

class Document(models.Model):
    name = models.CharField(null=True, max_length=254, blank=True)
    location = models.CharField(null=True, max_length=254, blank=True)
    phone_number = models.CharField(null=True, max_length=254, blank=True)

我对实现上述目标有点迷失。任何帮助将非常感激。

以下是我迄今为止尝试过的：（不工作）

 from django.shortcuts import render
    from .models import Document

    def index(request):
        search_number = list(Document.objects.order_by('-created').values("phone_number").distinct().order_by()) # Dictionary list of all numbers sorted by creation data without duplicate

        for x in search_number:
            try:
                look_up = Document.objects.values("phone_number")
                list_in_dba = look_up.phone_number
                x in list_in_dba['phone_number']
                print("Yes")
            except:
                print("No")

        return render(request, 'snippets/index.html')
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1 回答 1

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我会从这样的事情开始。

## this will get you all document records that have a duplicate phone-number 
## and also group them by phone-number.
duplicate_phone_numbers = Document.objects.values('phone_number').\
    annotate(total_items=Count('phone_number')).order_by('-total_items').filter(total_items__gt=1)

for entry in duplicate_phone_numbers:
    records = Document.objects.filter(phone_number=entry.get('phone_number')
    ## unsure whether you want to just output the info here or 
    ## update the actual record
    all_names = ''
    all_locations = ''
    for x in records:
        all_names += x.name + ";"
        all_locations += x.location + ";"
    print all_names, all_locations, entry.get('phone_number')
    # to update the actual record
    record = records[0]
    record.name = all_names
    record.location = all_locations
    record.save()
于 2019-02-22T05:13:53.360 回答