如何隐藏/删除后的div ajax php

Question

我正在用"id"(动态)抓取图像，我想使用ajax删除图像，图像被成功删除，但无法隐藏(div)动态(我删除)

这是我的html代码

<?php 
$banner=$store->banner;
if(!empty($banner))
{   
$ban = explode(',', $banner);
$i="1";
foreach($ban as $key => $img)
{
$img;
$info=$img;
?>  
<input type="hidden" id="id" name="id" value="<?php echo $store->id; ?>">
<input type="hidden" id="imgname" name="imgname" value="<?php echo $img; ?>">

<div class="form-group" id="<?php echo $key; ?>"> <!-- Creating div with id (dynamic) -->
<label for="status">image <?php  ?></label>
<img src="<?php echo base_url(); ?>/<?php echo $img; ?>" width="80" height="80">
<a class="btn-sm btn-danger text-light" onclick="deleteFun(<?php echo $key; ?>)" href="#"> Delete</a>
</div>
<?php 
$i++;
}
}
?>

这是我的ajax函数，我只是想隐藏div(我选择/删除)，我怎么能做到这一点?

<script type="text/javascript">
function deleteFun(ImgId)
{
if (confirm("Are you sure you want to delete this banner ?")) {
var imgname = $('#imgname').val();
var id = $('#id').val();
$.ajax({
type: "POST",
url: "<?php echo base_url('upload_controller/deleteImage'); ?>", 
data: {'id': id,'imgname':imgname,'ImgId':ImgId},
success: function(data){ 
$("#ImgId").remove();

console.log(data);
} 
});
}
}
</script>

Answer 1

在你的JS中，你必须指定div id作为一个连接值来针对特定的id，因为你正在采取的id将作为参数进入函数。

<script type="text/javascript">
function deleteFun(ImgId)
{
if (confirm("Are you sure you want to delete this banner ?")) {
var imgname = $('#imgname').val();
var id = $('#id').val();
$.ajax({
type: "POST",
url: "<?php echo base_url('upload_controller/deleteImage'); ?>", 
data: {'id': id,'imgname':imgname,'ImgId':ImgId},
success: function(data){ 
$("#" +ImgId).hide();

console.log(data);
} 
});
}
}
</script>