1

我正在尝试在我的 ajax 保存函数中发送多个数据数组。

我可以单独做每个数组data:hardwarePayload,它会起作用。如果我这样做,{hardware: hardwarePayload, service:servicePayload}我会得到非常奇怪的 JSON 输出。看起来像:

硬件=%5B%7B%22hardwareName%22%3A%221%22%2C%22hardwareQuantity%22%3A%22%22%2C%22hardwareBYOD%22%3A%22%22%7D%5D&服务=%5B%7B% 22serviceName%22%3A%223%22%2C%22serviceQuantity%22%3A%22%22%7D%5D

我真的需要两个阵列,一个硬件和一个服务,所以我可以单独获取每个阵列。

我的代码看起来像这样..

self.save = function (form) {
    var hardwareModel = [];
    var serviceModel = [];
    ko.utils.arrayForEach(self.services(), function (service) {
        serviceModel.push(ko.toJS(service));
    });
    ko.utils.arrayForEach(self.hardwares(), function (hardware) {
        hardwareModel.push(ko.toJS(hardware));
    }); 
  //allModel.push({accountId: ko.toJS(account)});
    var hardwarePayload = JSON.stringify(hardwareModel);
    var servicePayload = JSON.stringify(serviceModel);
  //alert(JSON.stringify(serviceModel) +JSON.stringify(allModel));
    $.ajax({
        url: '/orders/add',
        type: 'post',
        data: {hardware: hardwarePayload, service:servicePayload}, //            data:hardwarePayload,
        contentType: 'application/json',
        success: function (result) {
            alert(result);
        }
    });
};
4

2 回答 2

1

你应该试试这个

var hardwarePayload = hardwareModel;
var servicePayload = serviceModel;

var postData = {'hardware': hardwarePayload, 'service':servicePayload};

var postData = JSON.stringify(postData);

alert(postData);

$.ajax({
    url: '/orders/add',
    type: 'post',
    data: postData,
    contentType: 'application/json',
    success: function (result) {
        alert(result);
    }
});
于 2013-03-21T05:02:28.140 回答
0

我认为如果您不对数据进行字符串化处理会更好:

$.ajax({
    url: '/orders/add',
    type: 'post',
    data: {hardware: hardwareModel, service:serviceModel}, //            data:hardwarePayload,
    contentType: 'application/json',
    success: function (result) {
        alert(result);
    }
});

(请注意,我使用的是字符串化的hardwareModelserviceModel

这样,您可以让 jQuery 处理请求的 (json) 数据。

于 2013-03-20T23:07:33.160 回答