How to call function Next on View Create where click Next Button?
0
Problem
How to get next id when click on button next ?
meaning if i have textbox employee id 2 then click next
it must show 3 .
Actually what must i write under BtnNext on view to call Next Function
public class EmployeesController : Controller
{
private readonly IEmployees _context;
public EmployeesController(IEmployees context)
{
_context = context;
}
// GET: Employees/Create
public IActionResult Create()
{
var model = new Employee();
model.EmployeeId = _context.GetAll().Max(Employee => Employee.EmployeeId) + 1;
return View(model);
}
public ActionResult Next(int id)
{
var nextID = _context.GetAll().OrderBy(i => i.EmployeeId)
.SkipWhile(i => i.EmployeeId != id)
.Skip(1)
.Select(i => i.EmployeeId);
ViewBag.NextID = nextID;
return View("Create");
}
[HttpPost]
[ValidateAntiForgeryToken]
public async Task<IActionResult> Create(Employee employee)
{
if (ModelState.IsValid)
{
if (employee.EmployeeId <= 0)
{
await _context.InsertAsync(employee);
}
else
{
await _context.UpdateAsync(employee);
}
return RedirectToAction("Index");
}
return View(employee);
}
}
}
view create in employee controller
<div class="row">
<div class="col-md-4">
<button id="BtnNext" style="display:inline"><b>Next</b></button>
<form asp-action="Create">
<div asp-validation-summary="ModelOnly" class="text-danger">
</div>
<div class="form-group">
<label asp-for="EmployeeId" class="control-label"></label>
<input asp-for="EmployeeId" class="form-control" />
<span asp-validation-for="EmployeeId" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="BranchCode" class="control-label">
</label>
<input asp-for="BranchCode" class="form-control" />
<span asp-validation-for="BranchCode" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="EmployeeName" class="control-label"></label>
<input asp-for="EmployeeName" class="form-control" />
<span asp-validation-for="EmployeeName" class="text-danger"></span>
</div>
<div class="form-group">
<input type="submit" value="Create" class="btn btn-default" />
</div>
</form>
</div>
</div>
public class Employee
{
[Key]
public int EmployeeId { get; set; }
public int BranchCode { get; set; }
public string EmployeeName { get; set; }
}
mvc asp.net-core entity-framework-core
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
0
Problem
How to get next id when click on button next ?
meaning if i have textbox employee id 2 then click next
it must show 3 .
Actually what must i write under BtnNext on view to call Next Function
public class EmployeesController : Controller
{
private readonly IEmployees _context;
public EmployeesController(IEmployees context)
{
_context = context;
}
// GET: Employees/Create
public IActionResult Create()
{
var model = new Employee();
model.EmployeeId = _context.GetAll().Max(Employee => Employee.EmployeeId) + 1;
return View(model);
}
public ActionResult Next(int id)
{
var nextID = _context.GetAll().OrderBy(i => i.EmployeeId)
.SkipWhile(i => i.EmployeeId != id)
.Skip(1)
.Select(i => i.EmployeeId);
ViewBag.NextID = nextID;
return View("Create");
}
[HttpPost]
[ValidateAntiForgeryToken]
public async Task<IActionResult> Create(Employee employee)
{
if (ModelState.IsValid)
{
if (employee.EmployeeId <= 0)
{
await _context.InsertAsync(employee);
}
else
{
await _context.UpdateAsync(employee);
}
return RedirectToAction("Index");
}
return View(employee);
}
}
}
view create in employee controller
<div class="row">
<div class="col-md-4">
<button id="BtnNext" style="display:inline"><b>Next</b></button>
<form asp-action="Create">
<div asp-validation-summary="ModelOnly" class="text-danger">
</div>
<div class="form-group">
<label asp-for="EmployeeId" class="control-label"></label>
<input asp-for="EmployeeId" class="form-control" />
<span asp-validation-for="EmployeeId" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="BranchCode" class="control-label">
</label>
<input asp-for="BranchCode" class="form-control" />
<span asp-validation-for="BranchCode" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="EmployeeName" class="control-label"></label>
<input asp-for="EmployeeName" class="form-control" />
<span asp-validation-for="EmployeeName" class="text-danger"></span>
</div>
<div class="form-group">
<input type="submit" value="Create" class="btn btn-default" />
</div>
</form>
</div>
</div>
public class Employee
{
[Key]
public int EmployeeId { get; set; }
public int BranchCode { get; set; }
public string EmployeeName { get; set; }
}
mvc asp.net-core entity-framework-core
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm sorry, [cr] is for improving working code, not to help you implement new features. This simply isn't the place to ask. Once you do have it working bring it back and we can tell you ways to improve it.
– bruglesco
12 mins ago
add a comment |
0
0
0
Problem
How to get next id when click on button next ?
meaning if i have textbox employee id 2 then click next
it must show 3 .
Actually what must i write under BtnNext on view to call Next Function
public class EmployeesController : Controller
{
private readonly IEmployees _context;
public EmployeesController(IEmployees context)
{
_context = context;
}
// GET: Employees/Create
public IActionResult Create()
{
var model = new Employee();
model.EmployeeId = _context.GetAll().Max(Employee => Employee.EmployeeId) + 1;
return View(model);
}
public ActionResult Next(int id)
{
var nextID = _context.GetAll().OrderBy(i => i.EmployeeId)
.SkipWhile(i => i.EmployeeId != id)
.Skip(1)
.Select(i => i.EmployeeId);
ViewBag.NextID = nextID;
return View("Create");
}
[HttpPost]
[ValidateAntiForgeryToken]
public async Task<IActionResult> Create(Employee employee)
{
if (ModelState.IsValid)
{
if (employee.EmployeeId <= 0)
{
await _context.InsertAsync(employee);
}
else
{
await _context.UpdateAsync(employee);
}
return RedirectToAction("Index");
}
return View(employee);
}
}
}
view create in employee controller
<div class="row">
<div class="col-md-4">
<button id="BtnNext" style="display:inline"><b>Next</b></button>
<form asp-action="Create">
<div asp-validation-summary="ModelOnly" class="text-danger">
</div>
<div class="form-group">
<label asp-for="EmployeeId" class="control-label"></label>
<input asp-for="EmployeeId" class="form-control" />
<span asp-validation-for="EmployeeId" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="BranchCode" class="control-label">
</label>
<input asp-for="BranchCode" class="form-control" />
<span asp-validation-for="BranchCode" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="EmployeeName" class="control-label"></label>
<input asp-for="EmployeeName" class="form-control" />
<span asp-validation-for="EmployeeName" class="text-danger"></span>
</div>
<div class="form-group">
<input type="submit" value="Create" class="btn btn-default" />
</div>
</form>
</div>
</div>
public class Employee
{
[Key]
public int EmployeeId { get; set; }
public int BranchCode { get; set; }
public string EmployeeName { get; set; }
}
mvc asp.net-core entity-framework-core
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Problem
How to get next id when click on button next ?
meaning if i have textbox employee id 2 then click next
it must show 3 .
Actually what must i write under BtnNext on view to call Next Function
public class EmployeesController : Controller
{
private readonly IEmployees _context;
public EmployeesController(IEmployees context)
{
_context = context;
}
// GET: Employees/Create
public IActionResult Create()
{
var model = new Employee();
model.EmployeeId = _context.GetAll().Max(Employee => Employee.EmployeeId) + 1;
return View(model);
}
public ActionResult Next(int id)
{
var nextID = _context.GetAll().OrderBy(i => i.EmployeeId)
.SkipWhile(i => i.EmployeeId != id)
.Skip(1)
.Select(i => i.EmployeeId);
ViewBag.NextID = nextID;
return View("Create");
}
[HttpPost]
[ValidateAntiForgeryToken]
public async Task<IActionResult> Create(Employee employee)
{
if (ModelState.IsValid)
{
if (employee.EmployeeId <= 0)
{
await _context.InsertAsync(employee);
}
else
{
await _context.UpdateAsync(employee);
}
return RedirectToAction("Index");
}
return View(employee);
}
}
}
view create in employee controller
<div class="row">
<div class="col-md-4">
<button id="BtnNext" style="display:inline"><b>Next</b></button>
<form asp-action="Create">
<div asp-validation-summary="ModelOnly" class="text-danger">
</div>
<div class="form-group">
<label asp-for="EmployeeId" class="control-label"></label>
<input asp-for="EmployeeId" class="form-control" />
<span asp-validation-for="EmployeeId" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="BranchCode" class="control-label">
</label>
<input asp-for="BranchCode" class="form-control" />
<span asp-validation-for="BranchCode" class="text-danger"></span>
</div>
<div class="form-group">
<label asp-for="EmployeeName" class="control-label"></label>
<input asp-for="EmployeeName" class="form-control" />
<span asp-validation-for="EmployeeName" class="text-danger"></span>
</div>
<div class="form-group">
<input type="submit" value="Create" class="btn btn-default" />
</div>
</form>
</div>
</div>
public class Employee
{
[Key]
public int EmployeeId { get; set; }
public int BranchCode { get; set; }
public string EmployeeName { get; set; }
}
mvc asp.net-core entity-framework-core
mvc asp.net-core entity-framework-core
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
asked 31 mins ago
ahmed abedelazizahmed abedelaziz
1
1
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
ahmed abedelaziz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm sorry, [cr] is for improving working code, not to help you implement new features. This simply isn't the place to ask. Once you do have it working bring it back and we can tell you ways to improve it.
– bruglesco
12 mins ago
add a comment |
I'm sorry, [cr] is for improving working code, not to help you implement new features. This simply isn't the place to ask. Once you do have it working bring it back and we can tell you ways to improve it.
– bruglesco
12 mins ago
I'm sorry, [cr] is for improving working code, not to help you implement new features. This simply isn't the place to ask. Once you do have it working bring it back and we can tell you ways to improve it.
– bruglesco
12 mins ago
I'm sorry, [cr] is for improving working code, not to help you implement new features. This simply isn't the place to ask. Once you do have it working bring it back and we can tell you ways to improve it.
– bruglesco
12 mins ago
add a comment |
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I'm sorry, [cr] is for improving working code, not to help you implement new features. This simply isn't the place to ask. Once you do have it working bring it back and we can tell you ways to improve it.
– bruglesco
12 mins ago