How can i get multile record in web api

selvi jp

here upload multiple upload file with same id .i want to show multiple record imagename field.but only one record is display.if i pass id 3.then 4 record will disply bcoz 4 record is storing in the upload id 4 .

public ImageUpload Get(int id)
{
using (WebApiDatabaseEntities dbContext = new WebApiDatabaseEntities())
{
return dbContext.ImageUploads.FirstOrDefault(e => e.UploadId == id);
}
}

3 Replies

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Sachin Singh5y ago
Accepted answer
you are using FirstOrDefault that causing to return a single entity.
instead, use where() method.
public List Get(int id)
{
    using (WebApiDatabaseEntities dbContext = new WebApiDatabaseEntities())
    {
        return dbContext.ImageUploads.Where(e => e.UploadId == id).ToList();
    }
}
Arnab Mukherjee5y ago
Hi Selvi,
You need to return a list from the method as you want to return multiple records.Please try the steps given below. Hope it will help to solve your issue.

public class ImageUpload
{
        public string ImagePath { get; set; }
        public string ImageName { get; set; }
        public string ImageType { get; set; }
        public int UploadId { get; set; }
}

public List Get(int id)
{
    using (WebApiDatabaseEntities dbContext = new WebApiDatabaseEntities())
    {
         List lobj_List = new List();
         lobj_List = dbContext.ImageUploads.Where(e => e.UploadId == id).ToList();
         return lobj_List;
    }
}
Pranam Bhat5y ago
Try this one: [Route("api/test")] public HttpResponseMessage Get() { HttpResponseMessage response = null; List obj = new List(); try { obj = db.table_name.AsNoTracking().ToList(); response = Request.CreateResponse(HttpStatusCode.OK, obj); } catch (Exception ex) { response = Request.CreateResponse(HttpStatusCode.NotFound); } return response; } then access this api with http://192.168.0.1/api/test