我正在使用此Image Picker jQuery插件(http://rvera.github.io/image-picker/)来选择PicDiv中的图片进行删除。
选择图像后,我单击DeleteButton删除DeletePic.php选择的图像,并想刷新PicDiv,但在ajax请求后,javascript不再有效。
我从其他重复的问题中找到答案,告诉我绑定的事件处理程序,但它仍然不起作用
<script>
$(document).ready(function () {
$(".image-picker").imagepicker({
limit_reached: function () {
alert('Limit is 5');
}
});
});
$(document).on('click',"#DeleteButton",function(){
/*get image path*/
var req = $("div[class='thumbnail selected']").children('img');
var imagessource = [];
var srcc = [];
$(req).each(function (datakey, datavalue) {
src = $(datavalue).attr('src');
imagessource.push(src);
});
/*delete and refresh*/
var pp = $.post("DeletePic.php", {
'del[]': imagessource
},
function(){
alert("OK");
}
)
.done(function(){
$('#PicDiv').load(location.href + " #PicDiv");
})
});
</script>
HTML
<div id="PicDiv">
<?php
$arr = array();
$us = $_SESSION['Username'];
$i = 0;
mysql_connect("localhost", "AAAA", "AAAA");
mysql_select_db("test");
$strSQL = "SELECT path FROM tb_pic WHERE Username = '" . $us . "' ";
$objQuery = mysql_query($strSQL);
?>
<select class="image-picker" multiple="multiple" data-limit="5">
<?php
while ($objResult = mysql_fetch_array($objQuery)) {
?>
<option data-img-src="<?php echo $objResult['path']; ?>" value="<?php echo $i; ?>">AAAAAAAA</option>
<?php
$i++;
}
?>
</select>
</div>
<input id="DeleteButton" name="Submit" type="button" value="Delete">