这是从视图到控制器的值的基本传递,但这并不寻求工作。当我单击用于更新数据库中的记录的更新按钮时,来自视图的值未正确传递值到控制器。将调试器放入javascript后,每个变量都能够正确获取其值并将对象存储在它们存储的位置。
这个问题可能是什么原因?
这是Javascript中的按钮onclick事件代码。
$('#updatePrescription').click(function () {
debugger;
ValidateFields();
var drugListIsEmpty = CheckDrugList();
var error = $(".text-danger").length;
if (error == 0 && !drugListIsEmpty) {
debugger;
var prescription = [];
var template = {};
template.templateName = $("#prescriptionTemplateName").val();
template.templateTypeId = $('input[name=templateTypeId]:checked').val();
template.prescriptionTemplateItemList = [];
template.instructionId = $('.instruction').val();
template.frequencyId = $('.frequency').val();
template.day = $('.inputDays').val();
template.quantity = $('.inputQuantity').val();
template.dispenseLocationId = $('.selectDispenseLocation').val();
template.statusId = $('.status').val();
//template.categoryId = $('.templateCategory').filter(":visible").last().val();
template.templateId = $('#prescriptionTemplateId').val();
//if (template.categoryId == null) {
// template.categoryId = 0;
//}
var x = 0;
$('#tblPrescriptionSaveTemplateBody tr').each(function (key, value) {
debugger;
var row = $(this).closest('tr');
var next_row = $(row).next();
var drugId = $(value).find('.drugId').val();
var dosage = $(value).find('.inputDosage').val();
var dosageUnitId = $(value).find('.selectUnitId').val();
var statusId = "41";
var remarks = $(value).find('.inputDescription').val();
var groupId = $(value).find('.inputGroupNo').val();
var unit = $(value).find('.selectUnitId').val();
var prescriptionTemplateItemId = $(value).find('.prescriptionTemplateItemId').val();
x++;
var obj = {
// templateId: prescriptionTemplateId,
prescriptionTemplateId: template.templateId,
prescriptionTemplateItemId: prescriptionTemplateItemId,
drugId: drugId,
dosage: dosage,
dosageUnitId: dosageUnitId,
instructionId: template.instructionId,
frequencyId: template.frequencyId,
day: template.day,
quanitity: template.quantity,
unit: unit,
remarks: remarks,
dispenseLocationId: template.dispenseLocationId,
groupId: groupId,
statusId: template.statusId
}
template.prescriptionTemplateItemList.push(obj);
//prescription.push(obj)
})
$.ajax({
type: 'POST',
url: '/WestMedicinePrescriptionTemplate/UpdateTemplate',
dataType: 'json',
contentType: 'application/json',
data: JSON.stringify(template),
success: function (data) {
ShowNotificationMessage(data.notification);
window.location.href = '/WestMedicinePrescriptionTemplate/Index';
}
});
}
});
预计会在控制器的参数“newtemplate”中传递模型的结果,但结果为null
public ActionResult UpdateTemplate([FromBody] PrescriptionTemplateVM newtemplate)
{
int empId = Convert.ToInt32(HttpContext.Session.GetInt32("EmployeeId"));
var notif = "Update Failed.";
try
{
if (ModelState.IsValid)
{
bool updateSuccessful = _prescription.UpdatePrescriptionTemplateAndItems(newtemplate, empId);
if (updateSuccessful)
{
notif = "Update Successful.";
}
}
}
catch (Exception ex)
{
notif = ex.Message;
}
return Json(new { notification = notif });
}
代码中的问题可能是什么
答案 0 :(得分:1)
这样做:
[HttpPost]
public ActionResult UpdateTemplate(PrescriptionTemplateVM newtemplate)
您需要确保使用在PrescriptionTemplateVM中定义的相同变量名称
并且不要将数据转换为Json。这样做:
$.ajax({
type: 'POST',
url: '/WestMedicinePrescriptionTemplate/UpdateTemplate',
dataType: 'json',
contentType: 'application/json',
data: {newtemplate: template},
success: function (data) {
ShowNotificationMessage(data.notification);
window.location.href =
'/WestMedicinePrescriptionTemplate/Index';
}
});