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Can't show image depending on item selected from dropdown list

Trying to display an image depending on what user selects from dropdown list. Right now, it only displays image 2 when I select image 1 and I can not figure out what the problem is.

<HTML>
<BODY>

Chose an image:
<select id = "choice">
    <option>Image 1</option>
    <option>Image 2</option>
</select>

<button onclick = "showImage()">Show image</button>
<img src = " " id = "chosenImage" width = 250 height = 250>
    <script>
    function showImage(){
        var dropDownName = document.getElementById("choice");
        var imageChosen = dropDownName.options[dropDownName.selectedIndex].text;
        if (imageChosen == "image 1"){
            document.getElementById("chosenImage").src = "land.jpg";
        }else{
            document.getElementById("chosenImage").src = "code.png";
        }
    }
    </script>

</BODY>


</HTML>
about 4 years ago · Juan Pablo Isaza
2 Respostas
Responde à pergunta

0

You can just add a value attribute to the option elements then use that as your basis for identifying what to show.

1st Example:

<html>
  <body>
    Chose an image:
    <select id = "choice">
      <option value="default">Default</option>
      <option value="image_1">Image 1</option>
      <option value="image_2">Image 2</option>
    </select>

    <button onclick = "showImage()">Show image</button>
    <img src="https://4197r62cmrjs32n9dndpi2o1-wpengine.netdna-ssl.com/wp-content/uploads/2020/07/square-placeholder.jpg" id="chosenImage" width='250' height ='250'>
    
    <script>
      function showImage() {
        var dropDownName = document.getElementById("choice");
        var imageChosen = dropDownName.options[dropDownName.selectedIndex].value;

        if (imageChosen == "image_1"){
          imageChosen = "https://pbs.twimg.com/profile_images/1220067947798024192/30eZhfxx_400x400.png";
        } else if (imageChosen == 'image_2') {
          imageChosen = 'https://upload.wikimedia.org/wikipedia/commons/thumb/9/99/Unofficial_JavaScript_logo_2.svg/2048px-Unofficial_JavaScript_logo_2.svg.png';
        } else {
           imageChosen = "https://4197r62cmrjs32n9dndpi2o1-wpengine.netdna-ssl.com/wp-content/uploads/2020/07/square-placeholder.jpg";
        }
        document.getElementById("chosenImage").src = imageChosen;
      }
    </script>
  </body>
</html>

Alternatively, you can just shorten it by placing the path to your image in the value attribute as shown below.

2nd Example:

<html>

<body>
  Chose an image:
  <select id="choice">
    <option value="https://4197r62cmrjs32n9dndpi2o1-wpengine.netdna-ssl.com/wp-content/uploads/2020/07/square-placeholder.jpg">Default</option>
    <option value="https://pbs.twimg.com/profile_images/1220067947798024192/30eZhfxx_400x400.png">Image 1</option>
    <option value="https://upload.wikimedia.org/wikipedia/commons/thumb/9/99/Unofficial_JavaScript_logo_2.svg/2048px-Unofficial_JavaScript_logo_2.svg.png">Image 2</option>
  </select>

  <button onclick="showImage()">Show image</button>
  <img src="https://4197r62cmrjs32n9dndpi2o1-wpengine.netdna-ssl.com/wp-content/uploads/2020/07/square-placeholder.jpg" id="chosenImage" width='250' height='250'>

  <script>
    function showImage() {
      var dropDownName = document.getElementById("choice");
      var imageChosen = dropDownName.options[dropDownName.selectedIndex].value;

      document.getElementById("chosenImage").src = imageChosen;
    }
  </script>
</body>

</html>

about 4 years ago · Juan Pablo Isaza Relatório

0

the problem is

if (imageChosen == "image 1"){

it should be if (imageChosen == "Image 1"){ with I capitalize

about 4 years ago · Juan Pablo Isaza Relatório
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