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我想点击其ID的参考按钮后删除spacific行表的ID,删除行使用AJAX
这里是我的PHP代码,
while($row = mysqli_fetch_array($values)) {
echo "<tbody>";
echo "<tr id=\"12\">";
echo "<td>" . $row['pid'] . "</td>";
echo "<td>'<img src='/Login/product_avtar/".$row['pic']."' width=\"80\" height=\"60\">'</td>";
echo "<td>" . $row['pname'] . "</td>";
echo "<td>" . $row['pprice'] . "</td>";
echo "<td>" . $row['pdes'] . "</td>";
echo "<td>" . $row['qnt'] . "</td>";
echo "<td><button class=\"btn btn-sm btn-danger delete_class\" id=\"".$row['pid']."\">Delete</button></td>";
echo "</tr>";
echo "</tbody>";
这是阿贾克斯,
$(function() {
$(".delete_class").click(function(){
var element = $(this);
var del_id = element.attr("pid");
var info = 'pid=' + del_id;
if(confirm("Are you sure you want to delete this Record?")){
$.ajax({
type: "POST",
url: ajax_url,
data: info,
success: function(){
alert('Successfully Deleted');
}
});
}
return false;
});
});
,这是ajax_url,
if(($_POST['pid']))
{
$id=$_POST['pid'];
$id = mysqli_real_escape_string($id);
$option = mysqli_query($conn,"DELETE * FROM product where pid = ".$id."");
}
请帮忙找出解决方案, 感谢
非常感谢你的快速回复,我已经改变ID到PID但它仍然不能正常工作, –
您是否尝试过console.log并检查是否检索了正确的值d点击按钮? –
是的,我已经试过控制台,但它并没有显示任何东西,它显示网络错误 –