Return data using razor.

Question

Hello everyone I have call this function by ajax call
public JsonResult OnGetUserDetails(string User)
{
var currentDate = DateTime.UtcNow.Date;
var usersDetails = _con.Users
.Join(
_con.Claims.Where(c => c.Type == "Status" && c.Value == "active"),
u => u.Id,
c1 => c1.UserId,
(u, c1) => new { User = u, Claim1 = c1 }
)
.Join(
_con.Claims.Where(c => c.Type == "End"),
uc1 => uc1.User.Id,
c2 => c2.UserId,
(uc1, c2) => new { Claim = uc1, Claim2 = c2 }
)
.AsEnumerable()
.Where(ucc => DateTime.Parse(ucc.Claim2.ClaimValue).Date > currentDate)
.Select(ucc => new UserDetailsViewModel
{
UserEmail = ucc.Claim.User.Email,
Name = ucc.Claim.User.FirstName
}).ToList();
return new JsonResult(usersDetails);
}
It's get successfully data from data base But I have faced a issue , how to return this razor view pages with foreach loop.

This is view code

@model List<App.Pages.Search.Partial.UserListForEmailSendModel>

<table>
<thead>
<tr>
<th>User Email</th>
<th>Name</th>
</tr>
</thead>
<tbody>
@foreach (var userDetails in )
{
<tr>
<td>@userDetails.UserEmail</td>
<td>@userDetails.Name</td>
</tr>
}
</tbody>
</table>

Vishal Joshi · Answer

Hello Sunny You just need to use Model to get the data in razor view. please check replace below view code with your code. @model List User Email Name @foreach (var userDetails in Model) { @userDetails.UserEmail @userDetails.Name } Thanks

Mohamed Azarudeen Z · Answer

Yes it is possible sunny.

Sourabh Dhiman · Answer

Mohamed Azarudeen -- Please let me know It is without PartialView possible or not.

Mohamed Azarudeen Z · Answer

To return the razor view page with the fetched data from the database using an AJAX call, you can make use of the `PartialView` method in your `JsonResult` action. Here's an updated version of your code: 1. Modify the `JsonResult` action method to return a partial view instead of a JSON result: ```csharp public IActionResult OnGetUserDetails(string User) { var currentDate = DateTime.UtcNow.Date; var usersDetails = _con.Users .Join( _con.Claims.Where(c => c.Type == "Status" && c.Value == "active"), u => u.Id, c1 => c1.UserId, (u, c1) => new { User = u, Claim1 = c1 } ) .Join( _con.Claims.Where(c => c.Type == "End"), uc1 => uc1.User.Id, c2 => c2.UserId, (uc1, c2) => new { Claim = uc1, Claim2 = c2 } ) .AsEnumerable() .Where(ucc => DateTime.Parse(ucc.Claim2.ClaimValue).Date > currentDate) .Select(ucc => new UserDetailsViewModel { UserEmail = ucc.Claim.User.Email, Name = ucc.Claim.User.FirstName }).ToList(); return PartialView("_UserListPartial", usersDetails); } ``` 2. Create a partial view named "_UserListPartial.cshtml" (or any name you prefer) and place it in the appropriate folder. This partial view will render the table rows for the user details: ```html @model List User Email Name @foreach (var userDetails in Model) { @userDetails.UserEmail @userDetails.Name } ``` 3. Make sure your main view includes the necessary JavaScript code to perform the AJAX call and update the HTML with the returned partial view. You can use the jQuery library for this purpose. Here's an example: ```html $(document).ready(function() { $.ajax({ url: '/YourController/OnGetUserDetails', // Replace 'YourController' with the actual controller name type: 'GET', success: function(result) { $('#userListContainer').html(result); }, error: function() { alert('Failed to fetch user details.'); } }); }); ``` Make sure to replace `'YourController'` with the actual name of your controller that contains the `OnGetUserDetails` action. With these changes, when the page loads, the AJAX call will be made to the `OnGetUserDetails` action, and the returned partial view will be inserted into the `userListContainer` div on the page. The table rows with user details will be rendered based on the data returned from the server.