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Home/ Questions/Q 8457363
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T12:47:25+00:00 2026-06-10T12:47:25+00:00

So I originally had this function- $(.carImageClass).each( (elem,value)-> imgSrc = $(value).attr(src) $(value).qtip({ content: {

  • 0

So I originally had this function-

$(".carImageClass").each( (elem,value)->
  imgSrc = $(value).attr("src")
$(value).qtip({
  content: {
    text: "<img src=" + imgSrc + ">",
  },
  position:{
    corner:{
      target:'bottomMiddle',
      tooltip:'topLeft'
    }
  }
})
)

Which worked, as it displayed the img that I had. However, when I made the function like this, the qtip no longer appears.

$(".result").each( (elem,value)->
  imgResultDiv = $(value).find(".result-img")
  img = $(imgResultDiv).find("img")
  imgSrc = $(img).attr("src")
  console.log(value)
  console.log(imgSrc)
  $(value).qtip({
    content: {
      text: "<img src=" + imgSrc + ">",
    },
    position:{
        corner:{
          target:'bottomMiddle',
          tooltip:'topLeft'
        }

    }

I can prove that imgSrc is correct, and that value is what I think it is (an element of class “result”), so I’m not sure why the qtip does not appear.

Is there some sort of syntax issue? What’s going on?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T12:47:26+00:00Added an answer on June 10, 2026 at 12:47 pm

    did you paste the entire code?

    $(".result").each( (elem,value) ->
      imgResultDiv = $(value).find(".result-img")
      img = $(imgResultDiv).find("img")
      imgSrc = $(img).attr("src")
      console.log(value)
      console.log(imgSrc)
      $(value).qtip({
        content: {
          text: "<img src=" + imgSrc + ">",
        },
        position:{
            corner:{
              target:'bottomMiddle',
              tooltip:'topLeft'
            }
        }
      })
    )
    

    your paste was missing a few parentheses and a brace

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