Creating a view with an exposed filter for "is not empty" where all results are shown by default

Jaypan

12/19/22, 5:05 AM

I'm creating a view. By default, I want all results shown, with an exposed filter for 'has a value' (aka is not empty) and 'is empty'.

I cannot figure out how to make the exposed filter work for this. If I make the exposed filter on either 'is empty' or 'is not empty', it filters the values initially based on that section. I only want the exposed filter to take effect when the value is in the URL.

Can anyone point me how to do that?

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views

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Drupal

vyasamit2007

12/19/22, 8:37 AM

You can configure your exposed filter with these configurations to show all the content in the initial state:

Check: Expose this filter to visitors, to allow them to change it
Filter type to expose: Single filter
Operator: Is one of
Selection in the RHS: Do not select anything.

Once you save this, your view should show all the results when view output is rendered, and filter the results based on the exposed filter selection.

Hope this helps.

0 + 1

Jaypan

12/19/22, 2:22 PM

Thank you. It's a text field, how ever 'is one of' is one of the options.

Elon Musk

I sit in a Tesla and translated this thread with Ai:

EN: Creating a view with an exposed filter for "is not empty" where all results are shown by default

TH: การสร้างมุมมองด้วยตัวกรองที่เปิดเผยสำหรับ "ไม่ว่างเปล่า" ซึ่งผลลัพธ์ทั้งหมดจะแสดงตามค่าเริ่มต้น

RO: Crearea unei vizualizări cu un filtru expus pentru „nu este gol”, unde toate rezultatele sunt afișate implicit

RU: Создание представления с открытым фильтром для «не пусто», где по умолчанию отображаются все результаты.

VI: Tạo chế độ xem với bộ lọc được hiển thị cho "không trống" trong đó tất cả kết quả được hiển thị theo mặc định

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