Tell us what you are trying to do.
I am having a hard time passing arguments for conditional display within a view. I have read through (old) support answers. Many are not using Toolset Blocks. I am hoping you can resolve my issue—or point me to the correct article, which I cannot seem to find. I want to display all other CPTs in a view that has the same field group value as the current Post. Each post gets one value of a certain custom field as the first and main differentiator.
For example, I have a CPT of Outdoor Kitchen Ideas. One of many custom fields assigned to each Outdoor Kitchen Idea is "Layout Type." These include "U-shaped," "L-shaped," "Island," etc. There are several of each layout type.
I want to show all other kitchen ideas filtered by "layout Type" within a view of any given "Outdoor Kitchen Idea" post. When viewing one kitchen that is L-shaped, later on the page, I want to show the user other L-shaped layout options (not U-shaped, Compact, etc)
I have tried taxonomies, but they still have the same issue of reading the field for the current post type to reference the output of other posts. Do I need to create many views and filter out each view? Do I put each of the many filtered views in a conditional block for formatting?
Do I set a template conditionally by the layout type field- only changing the view in the template? ( I want all the posts to look the same.)
I hope to add conditional filters within a single View- unless that is unadvisable.
Hello. Thank you for contacting the Toolset support.
I need to check how you configure your view.
I checked the problem URL you shared:
- hidden link
As I understand you want to display the posts having same layout assigned as the current post you are displaying under the section "Check Out More U-Shaped Outdoor Kitchen Ideas" - is that correct? if yes:
- Can you please send me admin access details and let me check what's going wrong with your setup.
*** Please make a FULL BACKUP of your database and website.***
I would also eventually need to request temporary access (WP-Admin and FTP) to your site. Preferably to a test site where the problem has been replicated if possible in order to be of better help and check if some configurations might need to be changed.
I have set the next reply to private which means only you and I have access to it.
No. not exactly. The template IS already the same. I want the VIEW for each idea /outdoor-kitchen-idea/ to read and identify the current "layout type" as defined by the field of that name. Then, populate the VIEW with other /outdoor-kitchen-idea/ that match the pages "layout-type."
I can't filter by the pages current (dynamic) "layout-type. I have given up. But at this URL:
hidden link
I want to show L-shaped posts not U-shaped posts. I can filter by taxonomy or field, but I can't filter by a specific taxonomy or field as determined by THAT page.
For instance, I thought surely you would do this for the faux real estate site. a view on every post that "sees" you are looking at houses with basements (for example) and suggests: Check out these other houses with [basements]:
((View on every post that shows related content based on the current post's field value.))
Can you please share what view you are using and you added that view under what section of the page you shared:
=> hidden link
And send me access details and what custom field you want to target and with what value you want to display the posts under what section.
*** Please make a FULL BACKUP of your database and website.***
I would also eventually need to request temporary access (WP-Admin and FTP) to your site. Preferably to a test site where the problem has been replicated if possible in order to be of better help and check if some configurations might need to be changed.
I have set the next reply to private which means only you and I have access to it.
The topic ‘[Closed] Passing Arguments to Views’ is closed to new replies.
