Using orchard cms 1.6 I have a table in my db 'cars'. I want to display the column 'CarName' from the table, as a list on screen with all the records from the table.
carname1
carname2
carname3
When the user clicks on their link it will bring them to that page.
I know how to do this from the view e.g.
#T("Cars")
but I would like to try and create a content type which shows this list.
Content type seems to be all UI related. Im not sure how to take a table and display a column as a list on screen through the content type...any idea on how to do this?
then I can choose to show the content type as a form and the user can view it from the main menu.
thanks
It looks like you want to create a content type called Car, possibly with a CarPart and a record class CarPartRecord (perhaps refactor your Car class to CarPartRecord to follow Orchard's naming conventions). Make sure CarPartRecord derives from ContentPartRecord.
To render a list of Cars, you could use a Projection that renders a list of cars. A Projection renders content based on a Query, which you configure using the dashboard.
Alternatively, you could create a controller that leverages IContentManager to query all Car content items, and returns a view to render them in a table.
For each Car content item, use Html.ItemDisplayLink to render a link to its details page.
Related
I try to create custom content part with following specification.
I need to add list of items in database with same editor template view page of driver.
I want to include one form with required fields with an add button, by using this add button I want to add more number of rows with given field values in database.
I have checked Orchard documentation in this link, they provide sample with single item add in editor template page. In my case I need to add multiple items in editor page.
In display view page of driver, I want to retrieve collection of data from database and rendered in display view page.
Please share some details for this.
Thanks,
Using orchard 1.6. I've created a content type 'ImageUpload' which has 2 fields 'Image' and 'Date'. So the user can select todays date and upload an image. The uploads can be viewed from the 'Content Items' section of the dashboard but...
I would separately like to access/view the uploads from a different section(as the user wont have access to view the content items page) Iv set up a navigation menu but how can I view the records?
The content items are stored in the 'ContentItemVersionRecord' table...?
I don't fully understand your question. View the content items on the front end? Try using the projections module. You create a query that gets content of ImageUpload type and then you create a projection page that will use this query. Enable Projections feature, it comes with 1.6
I am currently working on an Orchard module. This module contains an MVC application including the views. I would like to make the module as configurable as possible. One of the items that I would like the customer to configure is the way the MVC views from the module look. Part of it will be determined by the theme. But not everything. Consider the following scenario:
The module contains a view for placing an order. The view displays a form in ´normal view way´. That is field labels and input labels. But at the head of the form each customer must be able to define his own set of instructions to display. Or maybe the customer wants to put there a message for pointing the customer to some other actions.
In the most ideal way I would have a content page where the customer can put all kind of content and one specific block that is the result of the view of the module. Kind like a web part. I can´t find out if it is possible and how that is achieved.
Edit for clarification
Module creates a page like this:
TITLE
FORM
So both title and form are outputted by the module controller.
I have managed to create a layer with the condition that the url matches the page with my form.
I have added a HTML widget to this layer in the content zone with position 1 (tried 0 to).
However the pages looks like this:
TITLE
FORM
WIDGET
instead of
WIDGET
TITLE
FORM
Returning a ShapeResult from your controller action will ensure that your view is themed and benefits from widgets, which are your "kind like a web part" thingies in Orchard.
I want to create a page in Drupal 6 where I can show list of restaurants.When a user clicks on any restaurant page, I should be redirected to Restaurant details page.
For this :
1.) I created a new content type called "Restaurant" with some fields.
2.) Created 3-4 content items for Restaurant( Restaurant1, Restaurant2, Restaurant3)
3.) Created view called: RestaurantList, Added Fields for it. Then added Page Display and gave the path for it http://website/Restaurants
Now, when I browse to Restaurants page, I only get labels of my fields but no values. How can I get the values but not the labels? Also, I want to go to the RestaurantDetails page. How can that be achieved?
Thanks,
Rashmi
Well if I were to set up a page view this is how I would set it up:
Filters:
Node type - Restaurants
Node published - Yes
Fields
Node title
check the option Link this field to it's node
leave the Label: field empty
check option Hide if empty
And if your view style is a HTML list, for extra you can go to Row style options and check the option: Hide empty fields
Make sure you click Preview to see if you get any values. If you don't then there's something wrong with the view settings, most probably the filters which are to restrictive. Start with something loose, like Node type - Restaurants.
I have a Sharepoint library that is currently rendering as the usual folder and document items table list view.
I would like to use present the same information as a grid of folder and document icons with some nice jQuery hover animations to show tooltips on the item the icon represents.
I suppose what I am really trying to find out is how to add a new library view that allows me to specify the markup rendered per item. I could write a new webpart to query the list and use an ASP:Repeater but I don't want to have to specify a webpart property each page to tell the webpart where it should open the list from.
You can use List View Web Part with custom XSLT.