WebWelcome to the Treehouse Community. Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? WebOct 1, 2014 · Here is a working solution. public class ResponseCapture : IDisposable { private readonly HttpResponseBase response; private readonly TextWriter originalWriter; private StringWriter localWriter; public ResponseCapture (HttpResponseBase response) { this.response = response; originalWriter = response.Output; localWriter = new …
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Web8 Answers. You can create your own IHttpActionResult class instance to return the JSON and a method in your controller or base controller class to utilize it. Create the IHttpActionResult instance that sets the content and status code: public class JsonTextActionResult : IHttpActionResult { public HttpRequestMessage Request { get; … WebMar 4, 2016 · You can try this. [HttpGet (" {}")] public IActionResult getUser () { Benutzer benutzerData = _context.benutzer.FirstOrDefault (); return Ok (benutzerData); } you cannot store json string to model class object instead you can store it in string and you can pass the string to the model.And also use cannot call controller in model class. dutch harbor fishing charters
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WebOct 3, 2016 · 1172. You can just use the ContentResult to return a plain string: public ActionResult Temp () { return Content ("Hi there!"); } ContentResult by default returns a text/plain as its contentType. This is overloadable so you can also do: return Content ("This is poorly formatted xml.", "text/xml"); Share. WebJun 14, 2024 · I've changed my IActionResult to ActionResult, and removed the Ok() and just return the T object now. ... public async Task GetUsersInPool(string poolId) { // do work to get users return Ok(pool.Users); } ... CS0029 Cannot implicitly convert type 'System.Collections.Generic.ICollection' to … Web2 Answers. Sorted by: 9. You'll need to return a class that inherits from ActionResult. Probably you want something like: public ActionResult ViewQuery () { DBController dBController = new DBController (); return View (dBController.usp_xxxxx ()); } This returns the view with the result passed in as Model. If you want to return JSON use: dutch harbor webcam