I want data from Iformcollection in nopCommerce 4.0 it getting null value.
This is my post method in controller
[HttpPost]
public ActionResult UpdateCart(int Id, IFormCollection form)
{
var setting = settingService.LoadSetting<DemoSetting>(_storeContext.CurrentStore.Id);
//Check plugin is enabled or not
if (!_setting.DemoSettingEnabled)
return Content("");
//Check null value
if Id,<= 0)
throw new ArgumentNullException("Id,");
This is my view page from which i can post data
<form asp-controller="DemoDiscounts" asp-action="UpdateCart" asp-antiforgery="true"
asp-route-Id="#Model.Id" asp-route-id="product-attributes-form" >
Can you please suggest if any one have solution?
Hey You can try this hope so it will helpful to you
<form method="post" asp-controller="YourControllerName" asp-action="YourActionName" asp-route-Id="#Model.Id" id="product-attributes-form" role="form">
Main thing you forgot Method="Post" and give id="product-attributes-form" role="form" hope so it can helpful to you
You need to add role parameter in tag
i.e. role="form"
<form method="post" role="form">
...
...
</form>
Related
how to get data from a form("multipart/form-data") with springMvc?
i want to upload a file(a photo),so the type of form is 'multipart/form-data'
but i found that springMVC cannot map data(in the form) into the property of User,I know i should use multipartFile to accept the picture,but how can other data in the form be mapped into object User automatically ?
<form enctype="multipart/form-data" type="post" action="....">
<input type"file" name="photo"/>
<input type"text" name="username"/>
.....
</form>
#RequestMapping("...")
public String editUser(User user,MultipartFile multipartFile){
.....
}
I get a 400 bad request error,so does anyone know how to achieve it? thanks a lot if someone help me out
You can do like this:-
<form enctype="multipart/form-data" type="post" action="....">
<input type"file" name="photo"/>
<input type"text" name="username"/>
.....
</form>
add model attribute on this page and use it in
your controller like this:-
#RequestMapping("...")
public String editUser(#RequestParam("photo") MultipartFile photo,
#ModelAttribute("your model attribute") User user){
.....
}
I want to POST to URL but following making it as GET , so How can I POST the
Object portletResponse = webAppAccess.getHttpServletRequest()
.getAttribute(Constants.PORTLET_RESPONSE);
if (portletResponse instanceof javax.portlet.ActionResponse) {
javax.portlet.ActionResponse actionResponse = (javax.portlet.ActionResponse) portletResponse;
actionResponse.sendRedirect(URL);
}
Have you tried to use RequestDispater instead of response.sendRedirect?
It retains the original request, without changing it.
So, it will remain POST if it was POST.
I have done this by using FORM POST method as follows.
webAppAccess.processPage("importedPage");
Added this imported Page in model :
<HTML>
<HEAD>
<title>New Page</title>
</HEAD>
<Body onload="">
<form id="FormID" method="POST" action="actionURL">
<input type="hidden" name="id" id="ID" value="<%=webAppAccess.getVariables().getString("ID")%>"/>
<noscript>
<p>Your browser does not support JavaScript or it is disabled.
Please click the button below to process the request.
</p>
<input type="submit" value="Proceed " name ="submit"></input>
</noscript>
<script>
document.getElementById('FormID').submit();
</script>
</form>
and then MVC controller mapping as follows:
#RequestMapping(value = {"/Details"}, method = RequestMethod.POST)
public String mthDetails(final Model p_model, #RequestParam(value = "id", required = false) final String p_ID){
//code for further logic using ID
}
In Asp.net Entity Framework I need to forward to another page and pass some data processed by the second page along.
In PHP I could do something like
<!-- page1.php -->
<form action="page2.php" method="POST">
<input type="hidden" name="id" />
<input type="submit" value="Go to page 2" />
</form>
<!-- page2.php -->
<?php
echo $_POST['id'];
?>
How can this be implemented in Asp.net?
Edit: There is a simple solution using Javascript and jQuery.
<!-- on page 1 -->
$('input[type=submit]').on('click', function (e) {
// Forward to browsing page and pass id in URL
e.preventDefault();
var id= $('input[name=id]').val();
if ("" == id)
return;
window.location.href = "#Request.Url.OriginalString/page2?id=" + id;
});
<!-- on page 2 -->
alert("#Request.QueryString["id"]");
There are, at least, two options:
Session state, like this:
Putting data into Session (your first page)
Session["Id"] = HiddenFieldId.Value;
Getting data out of Session (your second page)
// First check to see if value is still in session cache
if(Session["Id"] != null)
{
int id = Convert.ToInt32(Session["Id"]);
}
Query string, like this:
Putting the value into the URL for the second page as a query string
http://YOUR_APP/Page2.aspx?id=7
Reading the query string in the second page
int id = Request.QueryString["id"]; // value will be 7 in this example
There's a lot of ways to do this, take a look at this link for some guidance.
HTML page:
<form method="post" action="Page2.aspx" id="form1" name="form1">
<input id="id" name="id" type="hidden" value='test' />
<input type="submit" value="click" />
</form>
Code in Page2.aspx:
protected void Page_Load(object sender, EventArgs e)
{
string value = Request["id"];
}
MVC would look like...
#using (Html.BeginForm("page2", "controllername", FormMethod.Post))
{
#Html.Hidden(f => f.id)
<input type="submit" value="click" />
}
also, read through these MVC tutorials, you shouldn't blindly translate what you know in PHP to ASP.NET MVC, since you need to learn the MVC pattern too.
You can also use <form> with method="POST" in ASP.NET. And get value in code:
int id = int.Parse(Request.Form["id"]);
I have a form in ASP.Net and razor.
I need to have two ways of submitting said form: one that goes through the Edit action, and another that goes through the Validate action.
How should I go about doing this?
I don't mind using JavaScript for this.
EDIT:
Using the custom attribute I get this error.
The current request for action 'Resultados' on controller type 'InspecoesController' is ambiguous between the following action methods:
System.Web.Mvc.ActionResult Validar(System.Collections.Generic.ICollection1[Waveform.IEP.Intus.Server.Web.ViewModels.ResultadoViewModel]) on type Waveform.IEP.Intus.Server.Web.Controllers.InspecoesController
System.Web.Mvc.ActionResult Resultados(System.Collections.Generic.ICollection1[Waveform.IEP.Intus.Server.Web.ViewModels.ResultadoViewModel]) on type Waveform.IEP.Intus.Server.Web.Controllers.InspecoesController
That's what we have in our applications:
Attribute
public class HttpParamActionAttribute : ActionNameSelectorAttribute
{
public override bool IsValidName(ControllerContext controllerContext, string actionName, MethodInfo methodInfo)
{
if (actionName.Equals(methodInfo.Name, StringComparison.InvariantCultureIgnoreCase))
return true;
var request = controllerContext.RequestContext.HttpContext.Request;
return request[methodInfo.Name] != null;
}
}
Actions decorated with it:
[HttpParamAction]
public ActionResult Save(MyModel model)
{
// ...
}
[HttpParamAction]
public ActionResult Publish(MyModel model)
{
// ...
}
HTML/Razor
#using (#Html.BeginForm())
{
<!-- form content here -->
<input type="submit" name="Save" value="Save" />
<input type="submit" name="Publish" value="Publish" />
}
name attribute of submit button should match action/method name
This way you do not have to hard-code urls in javascript
You can do it with jquery, just put two methods to submit for to diffrent urls, for example with this form:
<form id="myForm">
<%-- form data inputs here ---%>
<button id="edit">Edit</button>
<button id="validate">Validate</button>
</form>
you can use this script (make sure it is located in the View, in order to use the Url.Action attribute):
<script type="text/javascript">
$("#edit").click(function() {
var form = $("form#myForm");
form.attr("action", "#Url.Action("Edit","MyController")");
form.submit();
});
$("#validate").click(function() {
var form = $("form#myForm");
form.attr("action", "#Url.Action("Validate","MyController")");
form.submit();
});
</script>
If you are working in asp.net with razor, and you want to control multiple submit button event.then this answer will guide you. Lets for example we have two button, one button will redirect us to "PageA.cshtml" and other will redirect us to "PageB.cshtml".
#{
if (IsPost)
{
if(Request["btn"].Equals("button_A"))
{
Response.Redirect("PageA.cshtml");
}
if(Request["btn"].Equals("button_B"))
{
Response.Redirect("PageB.cshtml");
}
}
}
<form method="post">
<input type="submit" value="button_A" name="btn"/>;
<input type="submit" value="button_B" name="btn"/>;
</form>
Here is a good eplanation:
ASP.NET MVC – Multiple buttons in the same form
In 2 words:
you may analize value of submitted button in yout action
or
make separate actions with your version of ActionMethodSelectorAttribute (which I personaly prefer and suggest).
With HTML5 you can use button[formaction]:
<form action="Edit">
<button type="submit">Submit</button> <!-- Will post to default action "Edit" -->
<button type="submit" formaction="Validate">Validate</button> <!-- Will override default action and post to "Validate -->
</form>
<input type="submit" value="Create" name="button"/>
<input type="submit" value="Reset" name="button" />
write the following code in Controler.
[HttpPost]
public ActionResult Login(string button)
{
switch (button)
{
case "Create":
return RedirectToAction("Deshboard", "Home");
break;
case "Reset":
return RedirectToAction("Login", "Home");
break;
}
return View();
}
We can have this in 2 ways,
Either have 2 form submissions within the same View and having 2 Action methods at the controller but you will need to have the required fields to be submitted with the form to be placed within
ex is given here with code Multiple forms in view asp.net mvc with multiple submit buttons
Or
Have 2 or multiple submit buttons say btnSubmit1 and btnSubmit2 and check on the Action method which button was clicked using the code
if (Request.Form["btnSubmit1"] != null)
{
//
}
if (Request.Form["btnSubmit2"] != null)
{
//
}
I have:
<form method="POST" name="f">
<input type="file" name="FileUpload" id="FileUpload" />
</form>
I'm sent the file using ajax,using the uplaod.file method:
document.getElementById('FileUpload').onchange = function() {
file = this.files[0];
ajax = new XMLHttpRequest;
ajax.file = file;
//etc..
ajax.open('post','Default.aspx', true);
ajax.setRequestHeader('foo','baa');
ajax.send(file);
}
Request.Forms["FileUplaod"] // don't works
How I get the value (the file) sent by ajax in my asp.net application?
I tried sent by http headers(is not pratice good,I know) but there problem with long length.
Thanks in advance.
try this form attribute :
form action="some/action" method="POST" enctype="multipart/form-data"