I have a site running in Drupal 6 and I'm using Views to display content from several feeds I am importing.
There is a node being created in one of the feeds that I do not want to have shown (or even created at all, if I can help it) & there doesn't appear to be a way to negatively filter out nodes from being displayed by title. Is there a way to do this with any efficiency or might there be some other hack around this?
Thanks in advance for any help you can provide.
-B
Views does allow you to filter out a node with a specific title. On the edit-view page, do the following:
Click the [+] on the Filters box
Check 'Node: Title' and click [Add]
Select 'Is not equal to' as operator and enter the unwanted title as value
Click [Update] and save your view
Why don't you just unpublish the nodes you don't want and add a filter for 'node:published = yes'. Unless the nodes you want to hide always have certain words in the title it will be hard to do it with a title filter.
Related
I am using views 3 and drupal 7. I have a view setup where it will return all related articles based on the keywords which works perfectly.
The problem is that I don't want the article that I'm on to show up in the results. Is there a way to hide specific nodes from displaying? For instance, lets say I have nodes: 43908 and 43909 that I don't want to show up in the view.
I've added a filter criteria (not contextual filter) to remove these nodes, but is it possible to combine node ID's into one filter? So if I wanted to show all but 43908 and 43909 how would I do that? I've tried to add them as 43908+43909 but that didn't seem to work.
You can use the regular expression option for the nid filter. So for the nid you mention you would do:
[^43908|^43909]
This filters out nodes with nid 43908 and 43909.
As for the scenario you describe, I think a better approach would be to add a separate field to your content type, e.g. field_in_progress, a tick box only visible when editing. Then you can add a filter on that field for the view.
This article may help. It involves excluding the current node from a list view.
I have a WordPress blog that let users (only for internal usage) write reports. There are "tasks" that they can select (checkboxes) to include in the report. Those that are selected pop a title and blank text zone on the bottom to include some informations. There is a task that is for feedback.
My question: I wanted to know if it is possible to replace the text zone with a form? Or add the form inside the text zone?
This is what the tasks list looks like, using the twentyten theme and ACF;
One of the tasks (checkboxes) is for feedback. I would like to have a complete form instead, with questions like "Did you see that?" with checkboxes, if they check "yes" then a text zone pop under it so people can write down feedback on that.
I cannot simply do a different type of page with a form and that's it, as the feedback is related to the tasks (to one specific report). It needs to be "inside" the report.
It seems impossible since one checkbox (for the tasks) only open one "thing" like a text zone. But I still hope there is something I can do.
Is there a way or it's impossible?
Thank you very much for all of your help.
Have a nice day.
This is accomplished using conditional logic within the Advanced Custom Fields Plugin. Go to Custom Fields -> Edit the Group you want to change -> Edit the Field you want to be conditional -> Turn Conditional Logic on and add your conditions.
I'm building a custom view ona drupal site and i added all the fields i wanted, but i can't seem to find out how to add the summary that i created for the node. Anyone can help?
I've added Body, put it doesn't include the summary.
Thanks
Ahah - just ran into the same problem - confused me for a moment too.
Go to the View in question, click the + icon to add a field, and select Fields: body. Then when the configuration options for that comes up, go to the very bottom and you'll see a drop down labelled Formatter - select Summary or trimmed. I think this gives you the summary if it exists, if not it will trim it for you.
While the node form refers to this as the 'Summary', elsewhere, such as in the Views UI, it is typically called the 'teaser'. You should be able to find this field listed along with other node fields when editing your view.
I've encountered a Drupal problem: I'm using the Views module for rendering nodes of a kind, based on the user id of it's author (it is a Content Profile actually). I want the view to show the comments for the node, just like in node/%. I could not find any option in views or any relevant module. Am I in the wrong direction and should reorganize stuff for this...?
Any ideas, how can it be done?
Regards,
Laci
Using views is really not the best plan of action. You should instead create a node template in your theme and customize it. If needed you can put some logic in a preprocess function. It requires more coding but will get you where you want.
If you use view node display type
Check in it's settings show comments
if you use view fields display type
Use relationship to comments and select fields you need and theme them
I know this is old request, but I was just struggling with the same issue and came across this post. I thought it'd be helpful to share my solution.
I'm using Drupal 7, with Views 3 and Display Suite.
In your view, choose the display in question.
Under Format, click the first link to the right of 'Show'.
Choose 'Content' (or 'Display Suite' in my case).
Click 'Apply'.
On the next screen, you'll have the option to 'Display comments'. Check this box and save your view.
You should now see the comments displayed under each item in the view.
You could create a second view (with URL e.g. /comments/% where the placeholder will be the node ID, and not the comment ID) that lists comments for a given node, with a contextual filter to only show them based on the NID in the URL.
Then, add that view to the footer (as a 'view area') of the single-node view you've already got.
There's some tweaking required for layout (inline fields etc.) but the basic structure should work.
I've just started using Drupal and I must admit its brilliant. Im still finding my way around it, so this might be a stupdi question, but Id like to know what the best and least perofrmance impacting way to achieve the below.
I have created a view with the "Node: Nid" argument to bascially list certain information about additioanl information about an artist, his events, albums etc..... The path has been set using the %. The idea being that when a particular artist node is in the node URL it will display all of the other info.
What I would like to do is add a link to the artist page called more info. So when clicked it takes you to that particular view and display all the other information.
I have realised that you cannot create menu items, using % argument.
How you would you guys do this ?
thanks :)
Charles
You are correct that you can't use wildcards in menus.
However, you can create a fake menu block outside the normal Drupal menu system that contains the menu items you want: just create a view and use the HTML list style. Each row in the view will correspond to one of your fake menu's items. When you're finished creating the view, create a block display for that view.
Add the block to a region using Site Building -> Blocks, and style it to look like your other menus.