[RUBY] Development memo ~ I want to display only the first image posted multiple times ~


When multiple images were posted at the same time, I had to devise a description to display the images. I will post the article as a memo so that I will not forget the contents.

Thing you want to do

** I want to display the one with the lowest index number of multiple posted photos on the index screen **


File name.html.haml

  %h2.title Pickup category
      %h3.heading Newly posted products
        - @newProducts.each do |product|
          =link_to "#" do
                = link_to product_path(product), class: "listBtn" do
                    = image_tag product.images[0].name.url, alt: "image_url",size: "200x150", class:"new-img"
                  = product.name
                      = product.price
                      %i.fa.fa-star.likeIcon 0
                  %p (tax included)


def index
    @newProducts = Product.includes(:images).where(status: 0).order("RAND()")


If you have posted multiple images and want to display only the first one, you can get the index number [0] by writing as follows.

= image_tag product.images[0].name.url, alt: "image_url",size: "200x150", class:"new-img"

Also, in this case, you have to define the file to be called by the index action. Therefore, **. Where (status: 0) ** is defined and assigned to the instance variable.

@newProducts = Product.includes(:images).where(status: 0).order("RAND()")

I think this will work!

At the end

This time, "I can write like this! I thought, and wrote an article! I'm still studying, so if you find something wrong, please comment!

Thank you very much!

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