Picture not changing when source changes

user_8719734 0
Viewed 56

I hide input type and show only image which is used to choose file and on change of input I call a function :

function showpreview(file) {
  var previewImg = $(file).parent().find('img').eq(0);

  if (file.files && file.files[0]) {
    console.log(window.URL.createObjectURL(file.files[0]));
    previewImg.src = window.URL.createObjectURL(file.files[0]);
  } else {
    previewImg.src = "blankpreview.png";
  }

  console.log(previewImg.src);
  console.log(previewImg);
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="col">
  <label for="exampleFormControlFile2">
    <img style="height:170px;" class="img-thumbnail" src="blankpreview.png" />
  </label>
  <input type="file" onchange="showpreview(this)" name="Screen_2" accept="image/jpeg, image/png" class="form-control-file d-none" id="exampleFormControlFile2" required />
</div>

This show correct source and element in console but image doesn't change

html jquery javascript
2 Answers

You can use .attr('src',yourimage) to set src of your image tag.

Demo Code :

function showpreview(file) {
  var previewImg = $(file).parent().find('img') //no need of eq..
  if (file.files && file.files[0]) {
    console.log(window.URL.createObjectURL(file.files[0]));
    previewImg.attr('src', window.URL.createObjectURL(file.files[0])); //use .attr here 
  } else {
    previewImg.attr('src', "blankpreview.png");
  }
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="col">
  <label for="exampleFormControlFile2">
     <img style="height:170px;" class="img-thumbnail" src="blankpreview.png" />
     </label>
  <input type="file" onchange="showpreview(this)" name="Screen_2" accept="image/jpeg, image/png" class="form-control-file d-none" id="exampleFormControlFile2" required />
</div>

user_106064000

Because you want to work with the raw DOM element, you need to grab that in your selector...

var previewImg = $(file).parent().find('.img-thumbnail').eq(0)[0];

Note the additional [0] at the end. Let's deal with it in steps below...

$(file).parent()
.find('.img-thumbnail') // find the thumbnail image
.eq(0) // "I only want the first one you find"
[0]; // Give me the DOM element that's inside the jQuery object


function showpreview(file) {
  var previewImg = $(file).parent().find('.img-thumbnail').eq(0)[0];

  if (file.files && file.files[0]) {
    previewImg.src = window.URL.createObjectURL(file.files[0]);
  } else {
    previewImg.src = "blankpreview.png";
  }

  console.log(previewImg.src);
  console.log(previewImg);
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="col">
  <label for="exampleFormControlFile2">
    <img style="height:170px;" class="img-thumbnail" src="blankpreview.png" />
  </label>
  <input type="file" onchange="showpreview(this)" name="Screen_2" accept="image/jpeg, image/png" class="form-control-file d-none" id="exampleFormControlFile2" required />
</div>

user_755250
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