Wcf服务RESTful [英] Wcf service RESTful

查看:133
本文介绍了Wcf服务RESTful的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

  [OperationContract] 
[WebInvoke(Method =POST ,
ResponseFormat = WebMessageFormat.Json,
RequestFormat = WebMessageFormat.Json,
odyStyle = WebMessageBodyStyle.Bare)]
List< Human> GetHuman(UserEnteredName,人名);

UserEnteredName类只有一个属性 - 字符串。



它有效。但是,我需要使它得到,而不是后。

我试过这个:

  [WebInvoke(Method =GET,UriTemplate =GetHuman?username = {John},
ResponseFormat = WebMessageFormat.Json,
RequestFormat = WebMessageFormat.Json)]

但它不起作用。根据你的 UriTemplate ,我需要改变什么?

解决方案你的方法必须看起来像

 人类GetHuman(字符串约翰)

我怀疑你错误地将一个可能的参数值放在 UriTemplate 中。尝试像

  [WebInvoke(Method =GET,UriTemplate =GetHuman?username = {userName},
ResponseFormat = WebMessageFormat.Json,
RequestFormat = WebMessageFormat.Json)]
人类GetHuman(字符串用户名)

另外,对于 GET ,您可以使用 WebGetAttribute ,它稍微干净一点。






我会改变你的方法来取一个 string 参数并构造 UserEnteredName方法体中的实例。如果它使用 TypeConverterAttribute ,可能可以使用 UserEnteredName 类型作为参数,但我从来没有这样做过,所以我不能说它有多容易(或不)。请参阅 WCF Web HTTP编程模型概述,特别是 UriTemplate查询字符串参数和网址部分。


I made my method with post like this:

[OperationContract]
[WebInvoke(Method = "POST",
           ResponseFormat = WebMessageFormat.Json,
           RequestFormat = WebMessageFormat.Json,
           BodyStyle = WebMessageBodyStyle.Bare)]
List<Human> GetHuman(UserEnteredName humanName);

The UserEnteredName class has just one property - string.

And it works. But, I need to make it to be get, not post.

I tried with this:

[WebInvoke(Method= "GET", UriTemplate = "GetHuman?username={John}", 
           ResponseFormat = WebMessageFormat.Json, 
           RequestFormat = WebMessageFormat.Json)]

But it doesn't work. What do I need to change?

解决方案

According to your UriTemplate, your method would have to look something like

Human GetHuman(string John)

I suspect you are mistakenly putting a possible parameter value in your UriTemplate. Try something like

[WebInvoke(Method= "GET", UriTemplate = "GetHuman?username={userName}", 
           ResponseFormat = WebMessageFormat.Json, 
           RequestFormat = WebMessageFormat.Json)]
Human GetHuman(string userName)

Also, for GET, you can use the WebGetAttribute, which is slightly cleaner.


I would change your method to take a string parameter and construct the UserEnteredName instance in the method body. It may be possible to use your UserEnteredName type as a parameter if it uses the TypeConverterAttribute, but I have never done this, so I can't say how easy (or not) it is. See the WCF Web HTTP Programming Model Overview, specifically the UriTemplate Query String Parameters and URLs section.

这篇关于Wcf服务RESTful的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆