Question

我不知道我是否针对我的问题找到了正确的问题。

在我的查询中，我想动态更改注释名称的名称取决于其危险程度。但我似乎无法得到它。它只返回名称本身，即＆＃34; hazard_level＆＃34;。

这一行问题.annotate(hazard_level=Count('brgy_locat')))。如何获取这些危险等级，然后将其用作注释查询的名称。所以JSON将是这样的：

[
        {
            "Low": 2,
            "brgy_locat": "Barangay 9",
            "municipali": "Cabadbaran City"
        },
        {
            "High": 5,
            "brgy_locat": "Comagascas",
            "municipali": "Cabadbaran City"
        }
        ...
]

这是我的代码：

    if request.method == "GET":
    # create a list
    to_json = []

    # this code is results a messy JSON data that need underscore.js to manipulate
    # in order for us to use datatables
    hazard_levels = ['High', 'Medium', 'Low']
    for hazard_level in hazard_levels:
        reference = FloodHazard.objects.filter(hazard=hazard_level)
        ids = reference.values_list('id', flat=True)
        for myid in ids:
            getgeom = FloodHazard.objects.get(id=myid).geom
            response = list( PolyStructures
                        .objects
                        .filter(geom__within=getgeom)
                        .values('brgy_locat', 'municipali')
                        .annotate(hazard_level=Count('brgy_locat'))
                       )
            to_json.append(response)    

    return HttpResponse(list(json.dumps(to_json)), 
                        content_type='application/json')

Answer 1

这是一种无代码方法：

    response = list( PolyStructures
                        .objects
                        .filter(geom__within=getgeom)
                        .values('brgy_locat', 'municipali')
                        .annotate(hazard_level=Count('brgy_locat'))
                       )
    response_cooked = [ { ( "high" if x.hazard_level >= 5 
                                   else "low" 
                          ) : x.hazard_level,                                     
                            "brgy_locat": x."brgy_locat", 
                            "municipali": x.municipali  
                         }
                         for x in response 
                       ]
    to_json.append(response_cooked)

如何在for循环中更改注释查询的变量名？

1 个答案: