Python django管理列表显示从多到多的排序

请帮助我解决一个问题，我尝试了很多，但不幸的是找不到解决方案

我的密码在下面

型号.py

    class MeasuredController(MeasuredObject):
        .....
        tempraturemeasure = models.ManyToManyField(MeasuredTemperature, blank=True, null=True)
        .....


    class MeasuredTemperature(MeasuredObject):
        ......
        temperature = models.DecimalField(max_digits=6,decimal_places=1,verbose_name='Temperature[C]')         .....

    class MeasuredControllerAdmin( admin.ModelAdmin):
        list_display = ('tem0', 'tem1', 'tem2')

        def tem2(self, obj): 
            ...............   
            return obj.tempraturemeasure.filter(deviceindex="2").latest("senddate").temperature
            .......

        def tem1(self, obj): 
            ...............   
            return obj.tempraturemeasure.filter(deviceindex="1").latest("senddate").temperature
            .......

        def tem0(self, obj): 
            ...............   
            return obj.tempraturemeasure.filter(deviceindex="0").latest("senddate").temperature
            .......

admin.py

    class MeasuredController(MeasuredObject):
        .....
        tempraturemeasure = models.ManyToManyField(MeasuredTemperature, blank=True, null=True)
        .....


    class MeasuredTemperature(MeasuredObject):
        ......
        temperature = models.DecimalField(max_digits=6,decimal_places=1,verbose_name='Temperature[C]')         .....

    class MeasuredControllerAdmin( admin.ModelAdmin):
        list_display = ('tem0', 'tem1', 'tem2')

        def tem2(self, obj): 
            ...............   
            return obj.tempraturemeasure.filter(deviceindex="2").latest("senddate").temperature
            .......

        def tem1(self, obj): 
            ...............   
            return obj.tempraturemeasure.filter(deviceindex="1").latest("senddate").temperature
            .......

        def tem0(self, obj): 
            ...............   
            return obj.tempraturemeasure.filter(deviceindex="0").latest("senddate").temperature
            .......

---

我的问题是，是否可以对管理中的

tem0

，

tem1

，

tem2

列进行排序

您可以通过创建子查询来实现这一点，尝试如下

class MeasuredControllerAdmin( admin.ModelAdmin):
    list_display = ('tem0', 'tem1', 'tem2')

    def tem2(self, obj): 
        ...............   
        return obj.tempraturemeasure.filter(deviceindex="2").latest("senddate").temperature
        .......

    def tem1(self, obj): 
        ...............   
        return obj.tempraturemeasure.filter(deviceindex="1").latest("senddate").temperature
        .......

    def tem0(self, obj): 
        ...............   
        return obj.tempraturemeasure.filter(deviceindex="0").latest("senddate").temperature
        .......

    def get_queryset(self, request):
        qs = super(MeasuredControllerAdmin, self).get_queryset(request)
        qs = qs.extra(
        select = { 
            "tem2" : "<sub query for selecting the latest temp2>"),
            "tem1" : "<sub query for selecting the latest temp1>"),
sub query for query for selecting the latest temp1"),
            "temp0:"<sub query for selecting the latest temp0>")})

tem2.admin_order_field = 'tem2'
tem1.admin_order_field = 'tem1'
tem0.admin_order_field = 'tem0'

class MeasuredControllerAdmin（admin.ModelAdmin）：
列表显示=（'tem0'、'tem1'、'tem2'）
def tem2（自身、obj）：
...............   
返回对象温度测量过滤器（deviceindex=“2”）.最新（“发送日期”）.温度
.......
def tem1（自身、obj）：
...............   
返回对象温度测量过滤器（deviceindex=“1”）.最新（“发送日期”）.温度
.......
def tem0（自身、obj）：
...............   
返回对象温度测量过滤器（deviceindex=“0”）.最新（“发送日期”）.温度
.......
def get_queryset（自我，请求）：
qs=super（MeasuredControllerAdmin，self）.get\u queryset（请求）
qs=qs.extra(
选择={
“tem2”：“），
“tem1”：“），
用于选择最新时间段的查询的子查询1“，
“temp0:”）}）
tem2.admin\u order\u字段='tem2'
tem1.admin\u order\u字段='tem1'
tem0.admin\u order\u字段='tem0'