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but this dictionary requires a model item of type

Posted on 2015-01-02
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183 Views
Last Modified: 2015-01-04
The model item passed into the dictionary is of type 'System.Collections.Generic.List`1[MobileWeb.Models.ProductDetailsModel]', but this dictionary requires a model item of type 'MobileWeb.Models.ProductDetailsModel'.

Keep receiving this error, I see the data when it reaches the return view but I get this error.

 public class ProductDetails
    {

        public List<ProductDetailsModel> data { get; set; }
    }
    public class ProductDetailsModel
    {
               
        public string datanumber { get; set; }
        public string name { get; set; }
        public string description { get; set; }
        public string number { get; set; }
        public float avg_price { get; set; }

    }

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  Stream objStream;
                objStream = wrGETURL.GetResponse().GetResponseStream();
              
                StreamReader objReader = new StreamReader(objStream);

                string sLine = "";
                int i = 0;
                WebResponse response = wrGETURL.GetResponse();
                string contentType = response.ContentType;
                //while (sLine != null)
                //{
                //    i++;
                sLine = objReader.ReadLine();
                 JObject o = JObject.Parse(sLine);
             //    Console.WriteLine("description: " + o["description"]);
                 List<ProductDetailsModel> mydata = new List<ProductDetailsModel>()
                 {
                   new ProductDetailsModel { description = (o["description"]).ToString() }
                 };

                 return View(mydata.AsEnumerable());

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0
Comment
Question by:Seven price
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6 Comments
 
LVL 7

Assisted Solution

by:Camillia
Camillia earned 167 total points
ID: 40528752
I think you need to change your View to have this on top

@model IEnumerable<WebApplication1.Models.ProductDetailsModel>

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It's probably like this now

@model WebApplication1.Models.ProductDetailsModel

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0
 
LVL 36

Assisted Solution

by:Miguel Oz
Miguel Oz earned 333 total points
ID: 40529898
Actually there is no need to convert to an enumerable if your code is changed as follows:
Controller: Returns the list, in fact you can use any object as parameter.
return View(mydata);

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View: Define model using mydata type (List<ProductDetailsModel>) and access each item by iterating directly from model as shown below:
@model List<'MobileWeb.Models.ProductDetailsModel>
@* and property Model is type of List<ProductDetailsModel> *@
@foreach(var item in Model)
{
    <span>@item.name</span>
}

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where "MobileWeb.Models" is current namespace where ProductDetailsModel is defined.  (Change it if this is not correct)

Note: Another alternative could be:
Controller Create a ProductDetails instance and return View(instance).
View: Define model as ProductDetails and when you need to access an item iterate  Model.data
0
 
LVL 9

Author Comment

by:Seven price
ID: 40530651
List<ProductDetailsModel> mydata = new List<ProductDetailsModel>()
Line 112:                 {

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Object reference not set to an instance of an object.

Recieving this error now
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LVL 36

Accepted Solution

by:
Miguel Oz earned 333 total points
ID: 40530696
Please be more specific, the posted line seems correct unless the following line set o as null. (JObject o = JObject.Parse(sLine);), thus this construct will fail: o["description"]).
IF that is the case you need to add conditional logic to avoid adding a new item to list.
0
 
LVL 9

Author Comment

by:Seven price
ID: 40530821
ok I need a little help with that, I can do that from a database but I have the data with sLine

how can I return the data description with this list
 List<LookupTable> mydata = new List<LookupTable>()
                 {
                   new LookupTable { description = (o["description"]).ToString() }
                 };

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0
 
LVL 9

Author Closing Comment

by:Seven price
ID: 40530921
Ok it was simple I just had to

var model = new context
                 {
                   description =(o["description"]).ToString()
                 };
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