I've been experimenting with an Ajax image uploader that I found on this website. Currently, I have managed to create duplicate preview images: one displayed under the input field and the other elsewhere on the page labeled as "this what you chose". However, a problem arises when the user selects a file, changes their selection, and ends up with both old and new images being displayed in the yourCustomPreview
.
Is there a way to only show the most recent preview picture without the previous one lingering? For more clarity, please refer to the source files here
uploaderPreviewer.js- Original function
<script>
function displayImage($previewDiv, imageUrl) {
var imageFilename = imageUrl.substr(imageUrl.lastIndexOf('/') + 1);
$previewDiv
.removeClass('loading')
.addClass('imageLoaded')
.find('img')
.attr('src', imageUrl)
.show();
$previewDiv
.parents('table:first')
.find('input:hidden.currentUploadedFilename')
.val(imageFilename)
.addClass('imageLoaded');
$previewDiv
.parents('table:first')
.find('button.removeImage')
.show();
}
</script>
uploaderPreviewer.js- Modified function
<script>
function displayImage($previewDiv, imageUrl) {
//New
var yourCustomPreview = $('#custompreview');
var imageFilename = imageUrl.substr(imageUrl.lastIndexOf('/') + 1);
$previewDiv
.removeClass('loading')
.addClass('imageLoaded')
.find('img')
.attr('src', imageUrl)
.show();
$previewDiv
.parents('table:first')
.find('input:hidden.currentUploadedFilename')
.val(imageFilename)
.addClass('imageLoaded');
$previewDiv
.parents('table:first')
.find('button.removeImage')
.show();
//New
yourCustomPreview.append('<img src="' + imageUrl + '"/>');
}
</script>