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How to parse JSON string in ASP.NET MVC view

Compile Views in ASP.NET MVCHow do you create a dropdownlist from an enum in ASP.NET MVC?How do you handle multiple submit buttons in ASP.NET MVC Framework?How do I get a consistent byte representation of strings in C# without manually specifying an encoding?How to render an ASP.NET MVC view as a string?In MVC, how do I return a string result?SqlException from Entity Framework - New transaction is not allowed because there are other threads running in the session.NET - JSON serialization of enum as stringFile Upload ASP.NET MVC 3.0Razor View throwing “The name 'model' does not exist in the current context”

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0

I am inserting JSON string into table, than on listing page in View inside foreach loop I want to parse that JSON string using Razor

@foreach (var item in Model) 
 var pr = JsonConvert.DeserializeObject<dynamic>(item.profile);
 //getting error 
 //the 'JsonConvert' does not exist in corrent context
 //and also the return type Profile was showing error so I changed it to `dynamic`
 <tr>
 <td>@pr.Name</td>

c# asp.net-mvc view json-deserialization

share|improve this question

edited Mar 24 at 7:26

marc_s

591k13311301278

asked Mar 24 at 7:13

user10182659

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why don't you deserialize your object of type Profile in your controller and the send it to your razor view as a List?
  
  – Rahul Sharma
  Mar 24 at 7:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  because then I have to use 2 loops, one process it in controller than at view to show
  
  – user10182659
  Mar 24 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You do not need to loop to deserialize your object in your controller. Just usevar pr=JsonConvert.DeserializeObject<Profile>(item.profile.ToString()); in your controller and then loop over pr which I assume is a List to get your properties on your razor view.
  
  – Rahul Sharma
  Mar 24 at 7:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Or if you really want to parse your JSON string in your View, try: var pr = JSON.parse(@Html.Raw(Json.Encode(item.profile))); // returns Object
  
  – Rahul Sharma
  Mar 24 at 7:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Then add the following line in your view: @using Newtonsoft.Json; .This will get your JsonConvert method to run in your view.
  
  – Rahul Sharma
  Mar 24 at 19:21
  
  

  
  
  
  

 | 
show 3 more comments

0

I am inserting JSON string into table, than on listing page in View inside foreach loop I want to parse that JSON string using Razor

@foreach (var item in Model) 
 var pr = JsonConvert.DeserializeObject<dynamic>(item.profile);
 //getting error 
 //the 'JsonConvert' does not exist in corrent context
 //and also the return type Profile was showing error so I changed it to `dynamic`
 <tr>
 <td>@pr.Name</td>

c# asp.net-mvc view json-deserialization

share|improve this question

edited Mar 24 at 7:26

marc_s

591k13311301278

asked Mar 24 at 7:13

user10182659

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Why don't you deserialize your object of type Profile in your controller and the send it to your razor view as a List?
  
  – Rahul Sharma
  Mar 24 at 7:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  because then I have to use 2 loops, one process it in controller than at view to show
  
  – user10182659
  Mar 24 at 7:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You do not need to loop to deserialize your object in your controller. Just usevar pr=JsonConvert.DeserializeObject<Profile>(item.profile.ToString()); in your controller and then loop over pr which I assume is a List to get your properties on your razor view.
  
  – Rahul Sharma
  Mar 24 at 7:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Or if you really want to parse your JSON string in your View, try: var pr = JSON.parse(@Html.Raw(Json.Encode(item.profile))); // returns Object
  
  – Rahul Sharma
  Mar 24 at 7:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Then add the following line in your view: @using Newtonsoft.Json; .This will get your JsonConvert method to run in your view.
  
  – Rahul Sharma
  Mar 24 at 19:21
  
  

  
  
  
  

 | 
show 3 more comments

I am inserting JSON string into table, than on listing page in View inside foreach loop I want to parse that JSON string using Razor

@foreach (var item in Model) 
 var pr = JsonConvert.DeserializeObject<dynamic>(item.profile);
 //getting error 
 //the 'JsonConvert' does not exist in corrent context
 //and also the return type Profile was showing error so I changed it to `dynamic`
 <tr>
 <td>@pr.Name</td>

c# asp.net-mvc view json-deserialization

share|improve this question

edited Mar 24 at 7:26

marc_s

591k13311301278

asked Mar 24 at 7:13

user10182659

@foreach (var item in Model) 
 var pr = JsonConvert.DeserializeObject<dynamic>(item.profile);
 //getting error 
 //the 'JsonConvert' does not exist in corrent context
 //and also the return type Profile was showing error so I changed it to `dynamic`
 <tr>
 <td>@pr.Name</td>

share|improve this question

edited Mar 24 at 7:26

marc_s

591k13311301278

asked Mar 24 at 7:13

user10182659

share|improve this question

edited Mar 24 at 7:26

marc_s

591k13311301278

asked Mar 24 at 7:13

user10182659

edited Mar 24 at 7:26

marc_s

591k13311301278

edited Mar 24 at 7:26

marc_s

591k13311301278

1 Answer
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0

When your application grows large you will feel it's better to use the standard approach of Model-View-Controller. It may become difficult to recode your application at that time. You can use this approach instead:

1. Create a class for JSON response.


2. Create a Model class that includes JSON class as a property, and all the other things required by your View.


3. Fetch the JSON from Controller, feed it to the Model, and return the Model to the View.


4. Traverse the Model (not JSON) in your View.

If you're just testing something temporarily then you can add @using Newtonsoft.Json on top of your view to make it recognize the JsonConvert.DeserializeObject method.

share|improve this answer

answered Mar 25 at 2:03

joym8joym8

1,81122248

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, this is pretty much what I was trying to explain to the OP. Using the traditional MVC approach is the best and most efficient way.
  
  – Rahul Sharma
  Mar 25 at 4:38
  

  
  
  
  

add a comment | 

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0

When your application grows large you will feel it's better to use the standard approach of Model-View-Controller. It may become difficult to recode your application at that time. You can use this approach instead:

1. Create a class for JSON response.


2. Create a Model class that includes JSON class as a property, and all the other things required by your View.


3. Fetch the JSON from Controller, feed it to the Model, and return the Model to the View.


4. Traverse the Model (not JSON) in your View.

If you're just testing something temporarily then you can add @using Newtonsoft.Json on top of your view to make it recognize the JsonConvert.DeserializeObject method.

share|improve this answer

answered Mar 25 at 2:03

joym8joym8

1,81122248

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, this is pretty much what I was trying to explain to the OP. Using the traditional MVC approach is the best and most efficient way.
  
  – Rahul Sharma
  Mar 25 at 4:38
  

  
  
  
  

add a comment | 

0

When your application grows large you will feel it's better to use the standard approach of Model-View-Controller. It may become difficult to recode your application at that time. You can use this approach instead:

1. Create a class for JSON response.


2. Create a Model class that includes JSON class as a property, and all the other things required by your View.


3. Fetch the JSON from Controller, feed it to the Model, and return the Model to the View.


4. Traverse the Model (not JSON) in your View.

If you're just testing something temporarily then you can add @using Newtonsoft.Json on top of your view to make it recognize the JsonConvert.DeserializeObject method.

share|improve this answer

answered Mar 25 at 2:03

joym8joym8

1,81122248

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, this is pretty much what I was trying to explain to the OP. Using the traditional MVC approach is the best and most efficient way.
  
  – Rahul Sharma
  Mar 25 at 4:38
  

  
  
  
  

add a comment | 

When your application grows large you will feel it's better to use the standard approach of Model-View-Controller. It may become difficult to recode your application at that time. You can use this approach instead:

1. Create a class for JSON response.


2. Create a Model class that includes JSON class as a property, and all the other things required by your View.


3. Fetch the JSON from Controller, feed it to the Model, and return the Model to the View.


4. Traverse the Model (not JSON) in your View.

If you're just testing something temporarily then you can add @using Newtonsoft.Json on top of your view to make it recognize the JsonConvert.DeserializeObject method.

share|improve this answer

answered Mar 25 at 2:03

joym8joym8

1,81122248

share|improve this answer

answered Mar 25 at 2:03

joym8joym8

1,81122248

answered Mar 25 at 2:03

joym8joym8

1,81122248

answered Mar 25 at 2:03

joym8joym8

1,81122248