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我有一个这样的搜索Ajax请求:如何将数据从View传递给对象使用ajax的控制器?
$.ajax({
type: 'POST',
data: { FirstName: firstname, LastName: lastname},
contentType: "application/json; charset=utf-8",
url: 'GetPeople',
dataType: 'json',
}
});
在GetPeole动作我可以得到我的参数(名字,姓氏)
public virtual JsonResult GetPeople(string FirstName,string LastName)
{
....
}
如果我改变像
$.ajax({
type: 'POST',
data: { FirstName: firstname, LastName: lastname,Age=age},
contentType: "application/json; charset=utf-8",
url: 'GetPeople',
dataType: 'json',
}
});
我的Ajax请求
我必须改变我的GetPeople
public virtual JsonResult GetPeople(string FirstName,string LastName,int Age)
{
....
}
我希望得到我的searchparameters(名字,姓氏,年龄),如Getpeople对象这样
public virtual JsonResult GetPeople(searchParam)
{
.....
}