使用Umbraco Base返回Json而不是XML

Question

我有一个.NET方法,为我的数据库添加一个新成员.它通过AJAX请求完成此操作.我有这个工作正常,但我在返回正确的响应消息时遇到问题,因此我可以向用户打印正确的消息.

我的方法目前看起来像这样:

    public static string MemberRegister(int process)
    {

        //here we find form values posted to the current page
        HttpRequest post = HttpContext.Current.Request;

        //get values from ajax URL
        var name = post["name"];
        var email = post["email"];
        var username = post["username"];
        var password = post["password"];

        //check if email exists
        if (Member.GetMemberFromEmail(email) == null)
        {

            MemberType userMemberType = new MemberType(1111); //id of membertype 'demo'
            Member newMember = Member.MakeNew(name, userMemberType, new umbraco.BusinessLogic.User(0));

            newMember.AddGroup(MemberGroup.GetByName("Active").Id);
            newMember.Email = email;
            newMember.Password = password;
            newMember.LoginName = username;

            newMember.Save();

            return "success";
        }
        else
        {
            return "emailError";
        }        

    }

我的Ajax代码如下:

        // submit
    $registerForm.submit(function() {

            $loader.show();

            jQuery.ajax({
                url: "/processform.aspx",
                type: "POST",
                data: $(this).serialize()

            }).complete(function( response ) {

            alert(response.responseText);

                if( response.responseText === "success" ) {
                    $registerSuccess.fadeIn(); 
                } elseif( response.responseText === "emailError" ) {
                    $registerEmailError.fadeIn(); 
                } else {
                    $registerError.slideDown();                        
                }

                $loader.hide();

            });



            return false;

    });

例如,如果所有成员都存在,则返回响应:

<value>emailError</value> 

我只是想让它返回emailError而不是值标签.我该怎么做呢？

提琴手(原始):

 HTTP/1.1 200 OK
 Content-Type: text/xml; charset=utf-8
 Vary: Accept-Encoding
 Server: Microsoft-IIS/7.5
 X-Powered-By: ASP.NET
 Date: Sun, 25 Mar 2012 20:59:54 GMT
 Content-Length: 25

 <value>emailError</value>


    POST http://domain.com/base/Forms/MemberRegister/process.aspx HTTP/1.1
    Host: domain.com
    Connection: keep-alive
    Content-Length: 125
    Origin: http://domain.com
    X-Requested-With: XMLHttpRequest
    User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/535.11 (KHTML, like Gecko) Chrome/17.0.963.79 Safari/535.11
    Content-Type: application/json; charset=UTF-8
    Accept: application/json, text/javascript, */*; q=0.01
    Referer: http://domain/register.aspx
    Accept-Encoding: gzip,deflate,sdch
    Accept-Language: en-GB,en-US;q=0.8,en;q=0.6
    Accept-Charset: ISO-8859-1,utf-8;q=0.7,*;q=0.3
    Cookie: ASP.NET_SessionId=ys4mmhsn2mpqcpja1iyjg04m; UMB_UPDCHK=1; __utma=256732567.15732944.1331581910.1332617890.1332627641.11; __utmc=256732567; __utmz=256732567.1331581910.1.1.utmcsr=(direct)|utmccn=(direct)|utmcmd=(none); UMB_UCONTEXT=12621b40-16fe-4422-a027-cf4fa68fe03d; __utma=176230262.1311679778.1332368941.1332694687.1332708163.12; __utmb=176230262.3.10.1332708163; __utmc=176230262; __utmz=176230262.1332368941.1.1.utmcsr=(direct)|utmccn=(direct)|utmcmd=(none)

    __VIEWSTATE=%2FwEPDwUENTM4MWRkEytUM47m6NUA6dpfOudOh9t51j6okfG%2BQhu4Em%2B26KU%3D&name=ppp&email=ppp&username=ppp&username=ppp

谢谢罗伯特

Answer 1

以下是如何使用Umbraco Base返回JSON对象的一个​​很好的示例:

/* Be sure to add References to:
 * 
 * umbraco.dll
 * System.Web.dll
 * System.Web.Extensions.dll
 */

using System.Web;
using System.Web.Script.Serialization;

using umbraco.presentation.umbracobase;

namespace CoB.Umb.Base.Example
{
    [RestExtension("Example")]
    public class Example
    {
        [RestExtensionMethod(returnXml = false, allowAll = true)]
        public static void Get()
        {
            string json = "";

            var person = new
            {
                firstName = "John",
                lastName = "Doe"
            };

            json = new JavaScriptSerializer().Serialize(person);

            HttpContext.Current.Response.ContentType = "application/json";
            HttpContext.Current.Response.Write(json);
        }
    }
}

和javascript:

$.getJSON('/base/Example/Get', function (data) {
    alert("Hey, " + data.firstName + " " + data.lastName);
});

Answer 2

您可能需要装饰该方法以返回 Json

[WebMethod]
[ScriptMethod(ResponseFormat = ResponseFormat.Json)]
 public static string MemberRegister(int process)


complete(function( response ) {
   if (response.d == "success") {
     //
    }
}

编辑

 jQuery.ajax({
                url: "/processform.aspx",
                type: "POST",
                contentType: "application/json; charset=utf-8",
                dataType: "json",
                data: $(this).serialize()

            })..