Creating new HTML images using a For loop through an array of image URLs









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I have an array of 40 different image URLs being returned from an AJAX request. I'm trying to create a new HTML image element for each URL in the array using a For loop, as seen in the below code. For some reason, it's only displaying the image at the first URL and that's it. Any idea why the other 39 aren't showing up?



$(document).ready(function() 
$.ajax(
type: 'GET',
dataType: 'json',
url: 'https://****/images',
success: function(data)
console.log(data);
let container = document.getElementById('feed');
let image = document.createElement("img");
for (let i = 0; i < data.length; i++)
image.setAttribute('src', data[i]);
container.appendChild(image);


);
);


<body>

<div id="feed">
</div>

</body>









share|improve this question

























    up vote
    0
    down vote

    favorite












    I have an array of 40 different image URLs being returned from an AJAX request. I'm trying to create a new HTML image element for each URL in the array using a For loop, as seen in the below code. For some reason, it's only displaying the image at the first URL and that's it. Any idea why the other 39 aren't showing up?



    $(document).ready(function() 
    $.ajax(
    type: 'GET',
    dataType: 'json',
    url: 'https://****/images',
    success: function(data)
    console.log(data);
    let container = document.getElementById('feed');
    let image = document.createElement("img");
    for (let i = 0; i < data.length; i++)
    image.setAttribute('src', data[i]);
    container.appendChild(image);


    );
    );


    <body>

    <div id="feed">
    </div>

    </body>









    share|improve this question























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have an array of 40 different image URLs being returned from an AJAX request. I'm trying to create a new HTML image element for each URL in the array using a For loop, as seen in the below code. For some reason, it's only displaying the image at the first URL and that's it. Any idea why the other 39 aren't showing up?



      $(document).ready(function() 
      $.ajax(
      type: 'GET',
      dataType: 'json',
      url: 'https://****/images',
      success: function(data)
      console.log(data);
      let container = document.getElementById('feed');
      let image = document.createElement("img");
      for (let i = 0; i < data.length; i++)
      image.setAttribute('src', data[i]);
      container.appendChild(image);


      );
      );


      <body>

      <div id="feed">
      </div>

      </body>









      share|improve this question













      I have an array of 40 different image URLs being returned from an AJAX request. I'm trying to create a new HTML image element for each URL in the array using a For loop, as seen in the below code. For some reason, it's only displaying the image at the first URL and that's it. Any idea why the other 39 aren't showing up?



      $(document).ready(function() 
      $.ajax(
      type: 'GET',
      dataType: 'json',
      url: 'https://****/images',
      success: function(data)
      console.log(data);
      let container = document.getElementById('feed');
      let image = document.createElement("img");
      for (let i = 0; i < data.length; i++)
      image.setAttribute('src', data[i]);
      container.appendChild(image);


      );
      );


      <body>

      <div id="feed">
      </div>

      </body>






      arrays ajax loops dom






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 10 at 0:40









      Aly Swerdlova

      203




      203






















          2 Answers
          2






          active

          oldest

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          up vote
          1
          down vote



          accepted










          Try to create the element inside the loop.



           for (let i = 0; i < data.length; i++) 
          let image = document.createElement("img");
          image.setAttribute('src', data[i]);
          container.appendChild(image);






          share|improve this answer



























            up vote
            1
            down vote













            The way to create images is with new Image() and when appending multiple nodes to the DOM at once, it's better to first append the image nodes into a document fragment, and only when all the images have been appended to the fragment, then append the fragment itself into the Document (prevents redundant repaints)






            // dummy data
            const data = ['http://placekitten.com/100/100',
            'http://placekitten.com/100/150',
            'http://placekitten.com/100/180',
            'http://placekitten.com/100/200']

            // create a dumpster-node for the images to reside in
            const fragment = document.createDocumentFragment();

            // iterate the data and create <img> elements
            data.forEach(url =>
            let image = new Image()
            image.src = url;
            fragment.appendChild(image);
            )

            // dump the fragment into the DOM all the once (FTW)
            document.body.appendChild(fragment);





            I've used Array forEach iterator in my example, because it's easier in my opinion, but you can use a for loop (or for..of loop)






            share|improve this answer




















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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              Try to create the element inside the loop.



               for (let i = 0; i < data.length; i++) 
              let image = document.createElement("img");
              image.setAttribute('src', data[i]);
              container.appendChild(image);






              share|improve this answer
























                up vote
                1
                down vote



                accepted










                Try to create the element inside the loop.



                 for (let i = 0; i < data.length; i++) 
                let image = document.createElement("img");
                image.setAttribute('src', data[i]);
                container.appendChild(image);






                share|improve this answer






















                  up vote
                  1
                  down vote



                  accepted







                  up vote
                  1
                  down vote



                  accepted






                  Try to create the element inside the loop.



                   for (let i = 0; i < data.length; i++) 
                  let image = document.createElement("img");
                  image.setAttribute('src', data[i]);
                  container.appendChild(image);






                  share|improve this answer












                  Try to create the element inside the loop.



                   for (let i = 0; i < data.length; i++) 
                  let image = document.createElement("img");
                  image.setAttribute('src', data[i]);
                  container.appendChild(image);







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 10 at 0:49









                  eag845

                  1698




                  1698






















                      up vote
                      1
                      down vote













                      The way to create images is with new Image() and when appending multiple nodes to the DOM at once, it's better to first append the image nodes into a document fragment, and only when all the images have been appended to the fragment, then append the fragment itself into the Document (prevents redundant repaints)






                      // dummy data
                      const data = ['http://placekitten.com/100/100',
                      'http://placekitten.com/100/150',
                      'http://placekitten.com/100/180',
                      'http://placekitten.com/100/200']

                      // create a dumpster-node for the images to reside in
                      const fragment = document.createDocumentFragment();

                      // iterate the data and create <img> elements
                      data.forEach(url =>
                      let image = new Image()
                      image.src = url;
                      fragment.appendChild(image);
                      )

                      // dump the fragment into the DOM all the once (FTW)
                      document.body.appendChild(fragment);





                      I've used Array forEach iterator in my example, because it's easier in my opinion, but you can use a for loop (or for..of loop)






                      share|improve this answer
























                        up vote
                        1
                        down vote













                        The way to create images is with new Image() and when appending multiple nodes to the DOM at once, it's better to first append the image nodes into a document fragment, and only when all the images have been appended to the fragment, then append the fragment itself into the Document (prevents redundant repaints)






                        // dummy data
                        const data = ['http://placekitten.com/100/100',
                        'http://placekitten.com/100/150',
                        'http://placekitten.com/100/180',
                        'http://placekitten.com/100/200']

                        // create a dumpster-node for the images to reside in
                        const fragment = document.createDocumentFragment();

                        // iterate the data and create <img> elements
                        data.forEach(url =>
                        let image = new Image()
                        image.src = url;
                        fragment.appendChild(image);
                        )

                        // dump the fragment into the DOM all the once (FTW)
                        document.body.appendChild(fragment);





                        I've used Array forEach iterator in my example, because it's easier in my opinion, but you can use a for loop (or for..of loop)






                        share|improve this answer






















                          up vote
                          1
                          down vote










                          up vote
                          1
                          down vote









                          The way to create images is with new Image() and when appending multiple nodes to the DOM at once, it's better to first append the image nodes into a document fragment, and only when all the images have been appended to the fragment, then append the fragment itself into the Document (prevents redundant repaints)






                          // dummy data
                          const data = ['http://placekitten.com/100/100',
                          'http://placekitten.com/100/150',
                          'http://placekitten.com/100/180',
                          'http://placekitten.com/100/200']

                          // create a dumpster-node for the images to reside in
                          const fragment = document.createDocumentFragment();

                          // iterate the data and create <img> elements
                          data.forEach(url =>
                          let image = new Image()
                          image.src = url;
                          fragment.appendChild(image);
                          )

                          // dump the fragment into the DOM all the once (FTW)
                          document.body.appendChild(fragment);





                          I've used Array forEach iterator in my example, because it's easier in my opinion, but you can use a for loop (or for..of loop)






                          share|improve this answer












                          The way to create images is with new Image() and when appending multiple nodes to the DOM at once, it's better to first append the image nodes into a document fragment, and only when all the images have been appended to the fragment, then append the fragment itself into the Document (prevents redundant repaints)






                          // dummy data
                          const data = ['http://placekitten.com/100/100',
                          'http://placekitten.com/100/150',
                          'http://placekitten.com/100/180',
                          'http://placekitten.com/100/200']

                          // create a dumpster-node for the images to reside in
                          const fragment = document.createDocumentFragment();

                          // iterate the data and create <img> elements
                          data.forEach(url =>
                          let image = new Image()
                          image.src = url;
                          fragment.appendChild(image);
                          )

                          // dump the fragment into the DOM all the once (FTW)
                          document.body.appendChild(fragment);





                          I've used Array forEach iterator in my example, because it's easier in my opinion, but you can use a for loop (or for..of loop)






                          // dummy data
                          const data = ['http://placekitten.com/100/100',
                          'http://placekitten.com/100/150',
                          'http://placekitten.com/100/180',
                          'http://placekitten.com/100/200']

                          // create a dumpster-node for the images to reside in
                          const fragment = document.createDocumentFragment();

                          // iterate the data and create <img> elements
                          data.forEach(url =>
                          let image = new Image()
                          image.src = url;
                          fragment.appendChild(image);
                          )

                          // dump the fragment into the DOM all the once (FTW)
                          document.body.appendChild(fragment);





                          // dummy data
                          const data = ['http://placekitten.com/100/100',
                          'http://placekitten.com/100/150',
                          'http://placekitten.com/100/180',
                          'http://placekitten.com/100/200']

                          // create a dumpster-node for the images to reside in
                          const fragment = document.createDocumentFragment();

                          // iterate the data and create <img> elements
                          data.forEach(url =>
                          let image = new Image()
                          image.src = url;
                          fragment.appendChild(image);
                          )

                          // dump the fragment into the DOM all the once (FTW)
                          document.body.appendChild(fragment);






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 13 at 9:12









                          vsync

                          44.6k35154217




                          44.6k35154217



























                               

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