I have made myself a small user control for consuming a feed, the code for said user control looks like this
<%# Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<SyndicationFeed >" %>
<%foreach (var rss in ViewData.Model.Items)
{
Response.Write("<div id={0}><a href={1} target=\"_blank\" /> <strong>{2}</strong></div>",
rss.Links[0].Uri.OriginalString, rss.Title.Text, rss.Title.Text);
Response.Write("<div>" + rss.Summary.Text.Truncate(100) + "</div>")
}%>
The code for the RssController looks like this
public virtual ActionResult Index()
{
string feedUrl = #"http://wdfw.wa.gov/news/newsrss.php";
using (XmlReader reader = XmlReader.Create(feedUrl))
{
SyndicationFeed rss = SyndicationFeed.Load(reader);
return View(rss);
}
}
And I call it in Site.Master like so
<%Html.RenderPartial("Index", Model);%>
I've also tried
<%Html.RenderPartial("Index", ViewData.Model);%>
All resulting in this this error:
Multiple controls with the same ID
'ctl00' were found. Trace requires
that controls have unique IDs.
Can someone help me figure out where I'm going wrong please :)
Can you try to use Html.RenderAction("Index","Rss") instead Html.RederPartial? When Html.RenderAction is called by ViewEngine, all code in your Index action is executed and Model with rss data goes to Index view to be generated to markup. And why do you use Response.Write, in view you can put html tags and this markup will be inserted where you are calling Html.RenderAction. Hope it will be helpful for you, if not, sorry, at least I've tried ))
Best regards,
Dima.
Related
I have one controller (RegisteredController.java) , and I want that the output of the controller is displayed in the JSP (its name is commment_form.jsp). So I use a forEach tag in the jsp to display a list of comments (the comments which the user has inserted about a given resource). For "resource" I usually mean an image. So there are a list of comments about an "image" and I want that all the comments are all displayed in the bottom page, when a comment is going to be inserted into the comment form. My question is how must be written the code into the controller in order to set the output for the jsp ? Should I use a #ModelAttribute , a put-attribute or something else ? Here is the code of the controller and of the jsp :
The comment_form.jsp is:
<form:form modelAttribute="comments">
<table class="commento">
<tr>
<th/>
<th>ID</th>
<th>Contenuto</th>
</tr>
<c:forEach items = "${comments}" var="comment">
<tr>
<td><c:out value="${comments.content}"></c:out></td>
<td><c:out value="${comments.content}</c:out></td>
</c:forEach>
</table>
</form:form>
The RegisteredController.java is:
#RequestMapping("/comment.do")
public String comment(#ModelAttribute Comment comment, BindingResult
bindingResult, Model model, Locale locale) {
User user=userService.getUserCurrent();
comment.setDatePubblication(SQLUtility.getCurrentDate());
comment.setIdUser(user.getId());
commentService.create(comment);
Object[] args = { comment.getId() };
String message = messageSource.getMessage("message.update", args,locale);
List<Comment> comments =
commentService.findAllCommentByResource(comment.getIdResource());
model.addAttribute("comments", comments);
model.addAttribute("id",comment.getIdResource());
model.addAttribute("message", message);
model.addAttribute("comment", comment);
return "redirect:/registered/comment_start.do";
}
Please any help ? I will appreciate . Thanks you.
In case of redirect pass additional data as redirect attributes.
To carry data across a redirect use RedirectAttributes#addFlashAttribute(key, value).
What Java doc says:
A RedirectAttributes model is empty when the method is called and is never used unless the method returns a redirect view name or a RedirectView.
After the redirect, flash attributes are automatically added to the model of the controller that serves the target URL.
Read more...
One extra note :
In JSP it should be ${comment.content} instead of ${comments.content}
If there's a form, and has a textbox and a button, how do you erase the content of the textbox after you submit the form?
<h:inputText id="name" value="#{bean.name}" />
<h:commandButton id="submit" value="Add Name" action="#{bean.submit}" />
After I enter a value in the textbox and submit, the value still appears in the textbox. I need to clear the content of the textbox once its been submitted. How can I achieve this?
Introduction
There are several ways to achieve this. The naive way is to simply null out the fields in backing bean. The insane way is to grab JS/jQuery for the job which does that after submit or even during page load. Those ways only introduces unnecessary code and indicates a thinking/design problem. All you want is just starting with a fresh request/page/view/bean. Like as you would get with a GET request.
POST-Redirect-GET
The best way is thus to just send a redirect after submit. You probably already ever heard of it: POST-Redirect-GET. It gives you a fresh new GET request after a POST request (a form submit), exactly as you intended. This has the additional benefit that the previously submitted data isn't re-submitted when the enduser ignorantly presses F5 afterwards and ignores the browser warning.
There are several ways to perform PRG in JSF.
Just return to same view with faces-redirect=true query string. Assuming a /page.xhtml, you could do so in action method:
public String submit() {
// ...
return "/page.xhtml?faces-redirect=true";
}
If you're still fiddling around with navigation cases the JSF 1.x way, then it's a matter of adding <redirect/> to the navigation case in question. See also How to make redirect using navigation-rule.
To make it more reusable, you can obtain the view ID programmatically:
public String submit() {
// ...
UIViewRoot view = FacesContext.getCurrentInstance().getViewRoot();
return view.getViewId() + "?faces-redirect=true";
}
Either way, if you've view parameters which needs to be retained in the request URL as well, then append &includeViewParams=true to the outcome. See also Retaining GET request query string parameters on JSF form submit.
If you're making use of some URL rewriting solution which runs outside JSF context, then you'd best grab the current request URL (with query string) and use ExternalContext#redirect() to redirect to exactly that.
public void submit() throws IOException {
// ...
ExternalContext ec = FacesContext.getCurrentInstance().getExternalContext();
StringBuffer requestURL = ((HttpServletRequest) ec.getRequest()).getRequestURL();
String queryString = ((HttpServletRequest) ec.getRequest()).getQueryString();
ec.redirect((queryString == null) ? requestURL.toString() : requestURL.append('?').append(queryString).toString());
}
It's only a mess which should really be refactored to some utility class.
Request/View scoped bean
Note that this all works only nicely in combination with request or view scoped beans. If you've a session scoped bean tied to the form, then the bean wouldn't be recreated from scratch. You've then another problem which needs to be solved as well. Split it into a smaller session scoped one for the session scoped data and a view scoped one for the view scoped data. See also How to choose the right bean scope?
Faces Messages
If you've a faces message to be shown as result of successful action, then just make it a flash message. See also How to show faces message in the redirected page.
public String submit() {
// ...
FacesContext context = FacesContext.getCurrentInstance();
context.addMessage(clientId, message);
context.getExternalContext().getFlash().setKeepMessages(true);
return "/page.xhtml?faces-redirect=true";
}
Ajax
Only if you happen to have an ajax-only page on which a F5 would always trigger a fresh new GET request, then simply nulling out the model field(s) in action method shouldn't harm that much.
See also:
How to navigate in JSF? How to make URL reflect current page (and not previous one)
Pure Java/JSF implementation for double submit prevention
You can blank out the property of the managed bean that should not be repainted when you render the response. This can be done done using code similar to the snippet posted below:
private String name;
public String getName(){return name;}
public void setName(String name){this.name=name};
public String submit()
{
//do some processing
...
// blank out the value of the name property
name = null;
// send the user back to the same page.
return null;
}
The reason for the current behavior can be found in how the JSF runtime processes requests. All JSF requests to a view are processed in accordance with the JSF standard request-response lifecyle. In accordance with the lifecyle, the managed bean contents are updated with the value from request (i.e. the value of DataForm.Name is set) before the application event (DataForm.submit) is executed. When the page is rendered in the Render Response phase, the current value of the bean is used to render the view back to the user. Unless the value is changed in an application event, the value will always be one that is applied from the request.
You can clear the form from the Bean method that gets called when the form is submitted;`
private String name;
private String description;
private BigDecimal price;
/*----------Properties ------------*/
/*-----Getter and Setter Methods---*/
public void save()throws SQLException{
String sql = "INSERT INTO tableName(name,description,price) VALUES (?,?,?)";
Connection conn = ds.getConnection();
try {
PreparedStatement pstmt = conn.prepareStatement(sql);
pstmt.setString(1, getName());
pstmt.setString(2, getDescription());
pstmt.setBigDecimal(3, getPrice());
pstmt.executeUpdate();
} catch (SQLException e) {
e.getMessage();
e.toString();
}finally{
conn.close();
clear();
}
}//End Save Method
public void clear(){
setName(null);
setDescription(null);
setPrice(null);
}//end clear`
Notice that the clear() method is called from the save method after all the operations of the save method is complete. As an option you could perform the clearing only if the methods operation was successful...The method below is placed in the ProductController Class...
public String saveProduct(){
try {
product.save();
} catch (SQLException e) {
e.printStackTrace();
}
return null;
}
The method call from the view/jsp would look like the Following:
<h:commandButton value="Save" action="#{productController.saveProduct}"/>
You can do it with jQuery.
I had the similar problem. I needed to clear popup window form.
<rich:popupPanel id="newProjectDialog" autosized="true"
header="Create new project">
<h:form id="newProjectForm">
<h:panelGrid columns="2">
<h:outputText value="Project name:" />
<h:inputText id="newProjectDialogProjectName"
value="#{userMain.newProject.projectName}" required="true" />
<h:outputText value="Project description:" />
<h:inputText id="newProjectDialogProjectDescription"
value="#{userMain.newProject.projectDescription}" required="true" />
</h:panelGrid>
<a4j:commandButton id="newProjectDialogSubmit" value="Submit"
oncomplete="#{rich:component('newProjectDialog')}.hide(); return false;"
render="projects" action="#{userMain.addNewProject}" />
<a4j:commandButton id="newProjectDialogCancel" value="Cancel"
onclick="#{rich:component('newProjectDialog')}.hide(); return false;" />
</h:form>
</rich:popupPanel>
jQuery code:
$('#newProjectForm').children('input').on('click', function(){$('#newProjectForm').find('table').find('input').val('');});
I added a code snippet how to reset all values for the current ViewRoot recursively for JSF 2 here:
Reset all fields in form
This works for submitted forms showing validation errors as well as for newly entered values in a form.
I'm trying to create something like *.ascxs' factory.
Scenario:
I would like to render controls which depends on model, which i've passed to partialView.
I'd like to achieve something like this:
<%# Control Language="C#" Inherits="System.Web.Mvc.ViewUserControl<MyAbstractModel>" %>
<%= Model.Property1 %>
<!-- other more sophisticated displays on model -->
<% Html.RenderAction("RenderControl", "Factory", new { model = Model}); %>
FactoryController:
public ActionResult RenderControl(object model) {
if (model.GetType() == typeof(Model1) {
return RenderPartial("Partial2", model);
} else {
return RenderPartial("Partial1", model);
}
}
I'd like to know is there any better way to cope with such situation. I suppose It's not the most efficient method to build web page in ASP.MVC 2.
If this method is acceptable, how can i restrict access to such controller? I would like to use this class only on server side and only by ascxs' pages
Use the ChildActionOnly() attribute to restrict access to your actions.
What you are trying to do is already builtin to MVC: Html.DisplayFor()
See: http://bradwilson.typepad.com/blog/2009/10/aspnet-mvc-2-templates-part-1-introduction.html
I'm a little new to asp.net mvc and I have a question (very basic). I have hacked around but I am not totally sure about this and I could'nt find anything particularly helpful.
Assume that I have 2 controllers A and B and 2 views FullView and PartView
public class AController:...
{
//Renders FullView
public ActionResult Create
{
....
}
[HttpPost]
public ActionResult Create
{
....
}
}
public class BController:...
{
//Renders an Arbitrary partial View (PartView)
public ActionResult Create
{
....
}
//Saves the data of the partial View
[HttpPost]
public ActionResult Create
{
....
}
}
the 1st view (FullView) has the code
<%Html.RenderAction("Create", "B"); %>
my question is on submit will BController's action ([HttpPost] Create) run?
Thank you
That depends on what action you specify in your <form /> tag. This doesn't have anything to do with asp.net mvc. If you use Html.BeginForm() without parameters it will post to the current url (not the create action on BController).
Well 1st thing you could do is toggle some breakpoints in your actions and hit f5.
Second - what action is called purely depends on what url you hit with what http method.
But for your case, when you post form A and controller A processes post you might get into validation problems and that's when you return View() on a post action and that's why form B is rendered via its post method.
I'm new to MVC!
I am trying to use two DropDownLists (Cities, Categories) in a PartialView that will be used in MasterPage, meaning they will be visble all the time.
I tried to load them in HomeCOntroller, but that didn't work. I got an Exception.
I read something about making a baseController that the other controllers will inherit from, I have tried that, kind of, but I guess i'm doing something wrong.
This is the only code I got today:
Masterpage
<% Html.RenderPartial("SearchForm"); %>
PartialView (SearchForm.ascx)
<% using (Html.BeginForm("Search", "Search")) { %>
<% } %> // dont know why I need two BeginForms, if I dont have this the other form won't trigger at all! Weird!
<% using (Html.BeginForm("Search", "Search", FormMethod.Get)) { %>
<%= Html.DropDownList("SearchForm.Category", new SelectList(ViewData["Categories"] as IEnumerable, "ID", "Name", "--All categories--")) %>
<%= Html.DropDownList("Search.City", Model.Cities, "--All cities--") %>
<input name="search" type="text" size="16" id="search" />
<input type="submit" id="test" title="Search" />
<% } %>
Two question:
Where and how to load the DropDownLists is the problem. I have tried to load it in the HomeController, but when go to another page then it says that the DDLs is empty and I get a Excecption.
Why do I have to use two forms for the ActionMethod to trigger ?
Hope anyone can help me out!
It sounds like you're only setting the property for a single action result. The Model.Cities data will have to be populated for every single view that needs to use it.
One solution would be to move the population of it to an ActionFilter
public class CityListAttribute : ActionFilterAttribute
{
public override void OnActionExecuted(ActionExecutedContext filterContext) {
var result = filterContext.Result as ViewResult;
result.ViewData.Model = //populate model
base.OnActionExecuted(filterContext);
}
}
and then add the filter to your controller
[CityList]
public class HomeController : Controller {
public ActionResult Index() {
return View();
}
}
As for the two forms issue, there should be no reason that i can think of that you need an empty form.
Take a look at the html that's being output and make sure it's ok. Also check the action is being generated correcly
Better way to do this, is to create something like MasterController and have action method on it like this:
[ChildActionOnly]
public ActionResult SearchForm()
{
//Get city data, category data etc., create SearchFormModel
return PartialView(model);
}
I recommend you create strongly typed view (SearchForms.ascx of type ViewUserControl<SearchFormModel>). Also it may be a good idea to have a model like this:
public class SearchViewModel
{
public IList<SelectListItem> Cities { get; set; }
public IList<SelectListItem> Categories { get; set; }
}
and use a helper like this: http://github.com/Necroskillz/NecroNetToolkit/blob/master/Source/NecroNet.Toolkit/Mvc/SelectHelper.cs to convert raw data to DDL friendly format beforehand.
In any case, you now use Html.RenderAction() instead of Html.RenderPartial() and specify you want "SearchForm" action from "MasterController".