解码Django POST请求正文

时间:2018-12-10 21:38:17

标签: python jquery json django post

我正在使用cordova构建地图应用,并发出发布请求以发送以下JSON(功能)

{
"type": "Feature",
"geometry": {
    "type": "Point",
    "coordinates": [
        -6.6865857,
        53.2906136
    ]
},
"properties": {
    "amenity": "pub",
    "name": "The Parade Ring"
}

}

这是发送请求的JQuery代码

function savePub(feature){

$.ajax({
    type: "POST",
    headers: {"csrfmiddlewaretoken": csrftoken},
    url: HOST + URLS["savePub"],
    data: {
        pub_feature: JSON.stringify(feature)
    },
    contentType:"application/json; charset=utf-8"
}).done(function (data, status, xhr) {
    console.log(data + " " + status);
    pubDialogAlert("Pub saved",feature);
}).fail(function (xhr, status, error) {
    showOkAlert(error);
    console.log(status + " " + error);
    console.log(xhr);
}).always(function () {
    $.mobile.navigate("#map-page");
});

}

在Django后端收到请求时,我不确定为什么当我打印请求正文时看起来像这样,

b'pub_feature=%22%7B%5C%22type%5C%22%3A%5C%22Feature%5C%22%2C%5C%22geometry%5C%22%3A%7B%5C%22type%5C%22%3A%5C%22Point%5C%22%2C%5C%22coordinates%5C%22%3A%5B-6.6865857%2C53.2906136%5D%7D%2C%5C%22properties%5C%22%3A%7B%5C%22amenity%5C%22%3A%5C%22pub%5C%22%2C%5C%22name%5C%22%3A%5C%22The+Parade+Ring%5C%22%7D%7D%22'

,当我尝试对其进行解码然后使用json.loads()时,会引发此错误

@api_view(['POST'])
def save_pub(request):
if request.method == "POST":

    data = request.body.decode('utf-8')
    received_json_data = json.loads(data)

    return Response(str(received_json_data) + " written to db", status=status.HTTP_200_OK)


JSONDecodeError at /savepub/
Expecting value: line 1 column 1 (char 0)

我假设是因为一旦解码了二进制字符串,由于这些字符%22等而无法将其转换为有效的JSON,但是我不知道解决方案是什么。

任何帮助将不胜感激。 谢谢

1 个答案:

答案 0 :(得分:1)

您在这里混淆了两件事,即表单编码和JSON格式。您所拥有的是带有一个键pub_feature的表单编码帖子,其值是JSON对象。

相反,您应该直接发布JSON:

data: JSON.stringify(feature),

然后您应该能够像已经完成的那样加载它-尽管请注意,确实应该让DRF为您处理