CODEWITHSUNDEEP
javascriptjqueryrazormodel-view-controller
user3872094
user3872094

Reputation: 3351

how to access a controller return value in javascript

I've a java script function that is used to save a webcam Image and the code is as below.

onCapture: function () {
                webcam.save("@Url.Content("~/Image/ImageCapture")/");
                MatchFindFunction();
            }

and in my controller the code is as below.

public void ImageCapture() //Capturing the image and save to folder
        {

            if (!Directory.Exists(Server.MapPath(@"~/Temp")))
            {
                Directory.CreateDirectory(Server.MapPath(@"~/Temp"));
            }

            string sessionIdValue = getSessionID();
            tempPath = "~/Temp/" + sessionIdValue + ".jpg";
            System.Web.HttpContext.Current.Session["tempImagePath"] = sessionIdValue;

            CommonModel.path = Server.MapPath(tempPath);
            CommonModel.stream = Request.InputStream;
            using (CommonModel.reader = new StreamReader(CommonModel.stream))
                CommonModel.dump = CommonModel.reader.ReadToEnd();
            System.IO.File.WriteAllBytes(CommonModel.path, ConvertStringToByte(CommonModel.dump));
return sessionIdValue;
        }

So Here my requirements is, I want to get this returned value in Javascript, I have to pass this variable to another function. I tried doing alert(webcam.save("@Url.Content("~/Image/ImageCapture")/")), but the message alerted here is true, but I want the sessionIdValue in return. I'm confused, if I have to use Ajax, if so, I'm not sure of how to call this URL. As this is webcam save method.

Please let me know on how can I do this.

Thanks

Upvotes: 0

Views: 182

Answers (1)

Eisa
Eisa

Reputation: 364

This @Url.Content("~/Image/ImageCapture") is your return value. just use this instead of using webcam.save("@Url.Content("~/Image/ImageCapture")/" in alert function.

Upvotes: 1

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