CODEWITHSUNDEEP
wcfjsonpostget
petko_stankoski
petko_stankoski

Reputation: 10713

Wcf service RESTful

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?

Upvotes: 0

Views: 165

Answers (1)

Jeff Ogata
Jeff Ogata

Reputation: 57793

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.

Upvotes: 1

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