使用带有可变数据的 Jquery Ajax 渲染局部视图
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Render Partial View using Jquery Ajax with variable data
提问by dan_vitch
I have already read this post, but I am not sure know to make it work taking data from the user. Here is the ajax jquery I am using. I know (or at least think) that this cant render a partial. But it works all the way until the render fails. I thought it may be helpful to have.
我已经阅读了这篇文章，但我不确定如何从用户那里获取数据。这是我正在使用的 ajax jquery。我知道（或至少认为）这不能呈现部分。但它一直有效，直到渲染失败。我认为拥有它可能会有所帮助。
$.ajax(
{
type: 'POST',
contentType: 'application/json; charset=utf-8',
data: "{'test':" + "'" + dateText + "'}",
dataType: 'json',
url: 'Site/Grab/',
success: function (result) {
alert('Success');
},
error: function (error) {
alert('Fail');
}
});
Here is my controller
这是我的控制器
[HttpPost]
public ActionResult Grab(string test)
{
DateTime lineDate= Convert.ToDateTime(test);
List<Info> myInfo= GameCache.Data(lineDate);
return PartialView("_PartialView", myInfo);
}
回答by RPM1984
Okay, couple of things to try:
好的，有几件事要尝试：
1) dataTypeis the expected result of the ajax call. In your case, your sendingJSON, but receivingHTML. The content-typeparameter specifies the request, which you have (and what you have is correct). So the data type should be:
1)dataType是ajax调用的预期结果。在您的情况下，您发送JSON，但接收HTML。该content-type参数指定了您拥有的请求（以及您拥有的请求是正确的）。所以数据类型应该是：
dataType: 'html',
2) You need to serialize the JSON. Try grabbing the lightweight JSONlibrary and stringify'ing:
2）您需要序列化JSON。尝试获取轻量级JSON库并进行字符串化：
var test = { test: 'testvalue' };
$.ajax {
...
data: JSON.stringify(test),
...
});
Much easier than trying to coerce a JSON string with quoatations. Create a regular JS variable, then stringify it.
比尝试使用引号强制转换 JSON 字符串要容易得多。创建一个常规 JS 变量，然后对其进行字符串化。
The rest of your code looks fine.
您的其余代码看起来不错。
If it's a problem with the HTML/markup of the partial view itself, run in debug mode and Visual Studio should stop on the line in the markup that is causing the problem.
如果部分视图本身的 HTML/标记有问题，请在调试模式下运行，Visual Studio 应该在导致问题的标记行上停止。
Bonus Hint: ASP.NET MVC 3 includes built-in JSON model binding. So you can create a basic POCO that matches the fields of your JSON object, then accept it as a strongly-typed object in the action method:
额外提示：ASP.NET MVC 3 包括内置的 JSON 模型绑定。因此，您可以创建一个与 JSON 对象的字段匹配的基本 POCO，然后在 action 方法中将其作为强类型对象接受：
[HttpPost]
public ActionResult Grab(MyJsonObject obj)
{
DateTime lineDate= Convert.ToDateTime(obj.test);
List<Info> myInfo= GameCache.Data(lineDate);
return PartialView("_PartialView", myInfo);
}
Since your only sending one parameter, it's overkill - but if you have more than 2 then it's worthwhile using a JSON POCO.
由于您只发送一个参数，因此有点矫枉过正 - 但如果您有 2 个以上，那么使用 JSON POCO 是值得的。
回答by Rashid Hussain Siddiqui
Change your controller code to:
将您的控制器代码更改为：
public ActionResult Grab(string test) {
DateTime lineDate= Convert.ToDateTime(test);
List<Info> myInfo= GameCache.Data(lineDate);
return Json(new { data = this.RenderPartialViewToString("_PartialView", myInfo) });
}
