我找到了解决方案。我使用了Steve Gentile，jQuery和ASP.NETMVC的解决方案-将JSON发送到Action-Revisited。
我的ASP.NET MVC视图代码如下所示：
function getplaceholders() {
var placeholders = $('.ui-sortable');
var results = new Array();
placeholders.each(function() {
var ph = $(this).attr('id');
var sections = $(this).find('.sort');
var section;
sections.each(function(i, item) {
var sid = $(item).attr('id');
var o = { 'SectionId': sid, 'Placeholder': ph, 'Position': i };
results.push(o);
});
});
var postData = { widgets: results };
var widgets = results;
$.ajax({
url: '/portal/Designer.mvc/SaveOrUpdate',
type: 'POST',
dataType: 'json',
data: $.toJSON(widgets),
contentType: 'application/json; charset=utf-8',
success: function(result) {
alert(result.Result);
}
});
};
[JsonFilter(Param = "widgets", JsonDataType = typeof(List<PageDesignWidget>))]
public JsonResult SaveOrUpdate(List<PageDesignWidget> widgets
由于链接断开，因此这是JsonFilterAttribute的代码
public class JsonFilter : ActionFilterAttribute
{
public string Param { get; set; }
public Type JsonDataType { get; set; }
public override void OnActionExecuting(ActionExecutingContext filterContext)
{
if (filterContext.HttpContext.Request.ContentType.Contains("application/json"))
{
string inputContent;
using (var sr = new StreamReader(filterContext.HttpContext.Request.InputStream))
{
inputContent = sr.ReadToEnd();
}
var result = JsonConvert.DeserializeObject(inputContent, JsonDataType);
filterContext.ActionParameters[Param] = result;
}
}
}
JsonConvert.DeserializeObject来自Json.NET
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