我的问题是从这个http请求中调用了第一个Get方法: http://localhost:56690/api/testelements/?name=aeg
我不希望我的这个打到第二个(字符串重载)方法,因为它有一个字符串参数?如何让它达到第二个(字符串重载)方法?
这是我在c#
中的web api控制器// GET api/<controller>
public IEnumerable<TestElement> Get()
{
return testelements;
}
// GET api/<controller>/searchString
//[HttpGet, Route("api/testelements/{name=name}")]
public IEnumerable<TestElement> Get(string searchString)
{
return Array.FindAll(testelements, s => s.name.Contains(searchString));
}
对于我的角度服务我有这个打字稿代码
search(term: string): Observable<testelement[]> {
return this.http
.get(`api/testelements/?name=${term}`)
.map(response => response.json() as testelement[]);
}
我的WepApiConfig.cs
public static class WebApiConfig
{
public static void Register(HttpConfiguration config)
{
// Web API configuration and services
// Web API routes
config.MapHttpAttributeRoutes();
config.Routes.MapHttpRoute(
name: "DefaultApi",
routeTemplate: "api/{controller}/{id}",
defaults: new { id = RouteParameter.Optional }
);
}
}
我不希望它覆盖此功能!:
public IHttpActionResult Get(int id)
{
var testelement = testelements.FirstOrDefault((p) => p.id == id);
if (testelement == null)
{
return NotFound();
}
return Ok(testelement);
}
答案 0 :(得分:2)
试试这个,
[HttpGet, Route("api/testelements/{searchString}")]
public IEnumerable<TestElement> Get(string searchString)
{
return Array.FindAll(testelements, s => s.name.Contains(searchString));
}
search(term: string): Observable<testelement[]> {
return this.http
.get(`api/testelements/` + ${term}) //append your search term at end of your Url.
.map(response => response.json() as testelement[]);
}
要从浏览器进行测试的网址:http:yourUrl/api/testelements/123
<强>更新强>
我不希望它覆盖此功能!:
public IHttpActionResult Get(int id)
{
var testelement = testelements.FirstOrDefault((p) => p.id == id);
if (testelement == null)
{
return NotFound();
}
return Ok(testelement);
}
如果您不想覆盖上述方法,则需要更改此功能的属性路由,
[HttpGet, Route("api/test/{searchString}")] // previous route [HttpGet, Route("api/testelements/{searchString}")]
public IEnumerable<TestElement> Get(string searchString)
{
return Array.FindAll(testelements, s => s.name.Contains(searchString));
}
如果您需要有关路由的更多信息,请Check this out