1
我有這個代碼更換並預覽多個圖像上傳
<?php
$query = "SELECT post_gallery_img FROM posts WHERE post_id = $get_post_id";
$select_gallery_imgs = mysqli_query($connection, $query);
while($row = mysqli_fetch_assoc($select_gallery_imgs)) {
$post_gallery_img = $row['post_gallery_img'];
$post_gallery_img = explode(',', $row['post_gallery_img']);
foreach($post_gallery_img as $out) {
echo '<img class="imgFile" id="editPostGalleryImgs" src="../userImages/' . $out . '" alt="Post Image">';
}
}
?>
被動態顯示的圖像,其結果是該
現在我試圖讓用戶上傳多個新圖像,但我希望他們能夠預覽新圖像,而不是動態圖像。
我有此代碼爲多個預覽
<input id="galleryImgs" type="file" multiple="multiple" name="files[]"><button style="display: none;" onclick="reset2($('#galleryImgs'));event.preventDefault()"></button>
<div class="gallery"></div>
<script>
$(function() {
// Multiple images preview in browser
var imagesPreview = function(input, editPostGalleryImgs) {
if (input.files) {
var filesAmount = input.files.length;
for (i = 0; i < filesAmount; i++) {
var reader = new FileReader();
reader.onload = function(event) {
$($.parseHTML('<img>')).attr('src', event.target.result).appendTo(editPostGalleryImgs);
}
reader.readAsDataURL(input.files[i]);
}
}
};
$('#galleryImgs').on('change', function() {
imagesPreview(this, 'div.gallery');
});
});
</script>
的問題是,我想要的預覽圖像來代替動態圖像來代替他們在上面顯示。
定位現有圖像並使用['.replaceWith()'](http://api.jquery.com/replacewith/)? –
只是使用..empty()來擺脫元素中現有的html,然後替換它。 – Difster
@LouysPatriceBessette玩耍,但沒有運氣,你有一個例子嗎? – Brad