I have an Index method that does double duty as showing a list of posts and a queried list of posts and can also have pages so you get urls like /News/Page/1 or /News?query=test
When a user clicks through to a post at say News/Details/1 they get a simple ActionLink that takes them back to the list. BUT I want this link to take them back to the actual page they were on in terms of the paging or the query. How could I do this? I don't want to use the JavaScript history method. Here is my current ActionLink: <%=Html.ActionLink("<< Back to News List", "Index")%> and this is an example of the paging links: <%= Html.RouteLink("<< First Page", "NewsPaging", new { query = ViewData["query"], page = 0 })%>
Thanks
It would probably be easier to just use javascript to send them back to the last page in their history (without creating a link to the specific page).
history.go(-1)
I will go for a ActionLink like
<%=Html.ActionLink("<< Back to News List", "Index")%>
because user can access News/Details/1 directly and if then you have a link that uses history.go(-1) or history.back() function it will not redirect it to the Index action
Another way to do this is to create the link in the controller and assign it to a property of the model or a simple ViewBag variable (up to you). For this example, I'm just going to use ViewBag.
So in your example, going from /News/Details/1 back to /News/Page/1?some-querystring, you can do the following:
In your controller (assuming '1' is the 'id'):
ViewBag.BackButton = String.Format("/News/Page/{0}{1}", id.ToString(), Request.Url.Query);
In your view:
< Back
One draw back to this is that you must know what the previous page was/could be. In your case, you want your back button to go back to your list, so this technique should be fine.
Related
I have made an app with jQuery mobile and laravel.
On the first page "mobilepage1" if have a form with a next button.
Also some of the form elements are dropdown menu's. The dropdown select options I get from a table in the database.
In the form on mobilepage1 I have a dropdown list with all department names:
I get this list from my controller like so:
$departmentlist = Company::find($usercompanyid)->departments;
This is how I display the list in the mobilepage1 form:
{{ Form::label('department', 'Department:')}}
<?php
foreach($afdelingslijst as $item) {
$afdelingsarray[$item->afdelingen] = $item->afdelingen;
}
?>
{{Form::select('size',$afdelingsarray)}}
On the second page I want to make a back button.
I tried the jQuery mobile way with a link with: data-rel="back"
But this does not work. I get a page error, it's missing the departmentlist used in mobilepage1. I can make a link to a route to mobilepage1, but I think all the information the user entered in mobilepage1 would be gone.
I see this url in the address bar when I am on mobilepage2 (this is after I filled in the form on mobilepage1 and clicked next).
/public/mobilepage2?size=Personeelszaken&size=32&directechef=Tineke&size=1&size=1&size=1&date-2=&size=1&time=&omschrijving=
Is there a way I can link to the the mobilepage1 route and still keep the information the user entered?
It looks like it would work just by hitting the back button if you passed in the department list with a view composer rather than through a route. Remove it from your route and place this in a file that's being autoloaded...
View::composer('mobilepage2', function($view)
{
$departmentlist = Company::find($usercompanyid)->departments()->lists('afdelingen','afdelingen');
$view->with('departmentlist', $departmentlist);
});
I'm not sure how you are finding $usercompanyid but you should be able to find it within the closure or pass it in by adding use ($usercompanyid) right after the function($view) part.
Also using the lists() function like that will build your array for you so you don't need to worry about the foreach loop in your view.
Since the values are getting passed back and forth as get variables, I think if you use Form::model() instead of Form::open(), it should place all those values back into the input fields for you.
I am trying to implement two pages. The first page being a selection of the items I want to show on the second page.
On my first page, I have done a selection of the items which I want to render.
On click of a button, 1. the ids of these items will be placed in an array and 2. the user will be redirected to a new page through:
window.location.replace("/schedule");
Right now, I want to achieve the following:
Do an ajax call through rails to get each of the array items. i.e. make an ajax calls to the urls of myclass/[id]. The id belonging to the items in the array.
I am not sure how I could get my array items from the current page to the next page or if there is a better way of achieving this. Any advice or suggestion will be greatly appreciated.
Add :remote => true to your form. Then, app will render your action_name.js.erb view instead of action_name.html.erb. In js file you can put code, which will repleace div without redirection, f.e.
$('#div_id').html('<%= escape_javascript render(your_partial_name) %>');
You don't need to do anything special in partial and controller action, its all the same.
I have the following relation in my model:
A user can have several consumers
According to that, I call several actions from different controllers that depend on that consumer_id in the URL. So, I do stuff like:
/:consumer_id/products/all
/:consumer_id/locals/all?params...
I want to be able to given that I am in a particular view, let's say /3/products/all, be able to refresh, redirect or whatever is the best to /4/products/all, with a form that shows the different consumers attached to that user.
I know how to display the form, but I fail to see which action should I put given that I want to load the current controller, current action and the current params, except that I want to change the params[:consumer_id] and the session[:consumer_id] to the one chosen by the user in the form of the view.
What's the best or appropriate way of doing so?
This is a link to the same page with the same parameters and one additional parameter:
<%= link_to "Show more results", url_for(params.merge(:per => 100)) %>
If you want to modify a parameter rather than add a new one, you can modify the params hash just as you would modify any hash, and use url_for again.
Inside the view you can access both the current controller using controller_name and the current action using action_name
Your question is not clear. You want to choose consumer_id from your form, then use that consumer id inside the same page or for links to other pages?
If you want to use the consumer id for the current page, you need to use javascript or JQuery to update all attributes of the current page.
If you want to use that consumer id for links,
you may also use JQuery to update all links on your page
Or, you can submit the form to the server with new consumer_id.
I've got a basic table listing rows from a database, with edit links beside each one. The table may be several pages long, so after the user edits a row, I want the user redirected back to the List view, and I want the page scrolled down to the row that was just edited.
In my controller, I have...
Return RedirectToAction("List")
Which redirects to ...\List
How do I make it redirect to ...\List#100
Using the routing system, it is not possible to do that AFAIK. One way of doing it is to append the hash after the URL like below:
return Redirect(Url.Action("List", "Home") + "#poo");
I'm using ASP.NET MVC 2 for the first time on a project at work and am feeling like a bit of a noob.
I have a page with a customer search control/partial view. The control is a textbox and a button. You enter a customer id into the textbox and hit search. The page then "refreshes" and shows the customer details on the same page. In other words, the customer details appear below the customer search control.
This is so that if the customer isn't the right one, the user can search again without hitting back in the browser. Or, perhaps they mistyped the customer id and need to try again.
I want the URL to look like this:
/Customer/Search/1
Obviously, this follows the default route in the project.
Now, if I type the URL above directly into my browser, it works fine. However, when I then use the search control on that page to search for say customer 2, the page refreshes with the correct customer details but the URL does not change! It stays as
/Customer/Search/1
When I want it to be
/Customer/Search/2
How can I get it to change to the correct URL?
I am only using the default route in Global.asax.
My Search method looks like this:
<AcceptVerbs(HttpVerbs.Get)> _
Function Search(ByVal id As String) As ActionResult
Dim customer As Customer = New CustomerRepository().GetById(id)
Return View("SearchResult", customer)
End Function
A good place to start might be NerdDinner if you havn't already.
In the mean time though The approach I'd use is to have a page that has my search box on it.
Then I'd have a <div> that I name "SearchResults". This will ultimately hold my results to the search.
I then have a PartialView which takes a model that has all the search results and renders them.
So when I click the button I do a call out to a jQuery action that takes the search parameter, performs the search and then returns my PartialView as rendered HTML.
Back in the client side I take that rendered HTML and replace the contents of my div with the HTML.
The keywords to google, or SO, are RenderPartial. This is back end code to render a partial view and give you html.
Also jQuery postbacks so that you can call an action in your controller.
use RedirectToRoute action result
link