嗨,伙计们,即使我成功删除了消息,除了成功删除消息外,还返回错误的删除消息(“无法删除”),但查询正在运行 请告诉我我做错了什么这是我的代码\
删除.php
if(isset($_POST['delete_id']) && !empty($_POST['delete_id'])) {
$delete_id = mysql_real_escape_string($_POST['delete_id']);
$query=mysql_query("DELETE FROM color WHERE idColor='$delete_id'");
if($query != false) {
echo 'true';
}
mysql_close();
<script>
$(document).ready(function(){
$('.del_btn').click(function(){
if (confirm("Are you sure you want to delete this Color?"))
var del_id = $(this).attr('rel');
$.post('script/delete_color.php', {delete_id:del_id}, function(data) {
if(data == 'true') {
$('#'+del_id).remove();alert('Color has been deleted!');
}
else {
alert('Color could not delete!');
}
document.messages.submit();
return false; // This line added}
});
});
});
</script>
最佳答案
我不是很明白你的问题,但我会进行有根据的猜测......
您需要将您的逻辑包装在“确认”IF 语句中。此外,您还有一个额外的“}”)。
$(document).ready(function(){
$('.del_btn').click(function() {
if (confirm("Are you sure you want to delete this Color?")) {
var del_id = $(this).attr('rel');
$.post('script/delete_color.php', {delete_id:del_id}, function(data) {
if(data == 'true') {
$('#'+del_id).remove();alert('Color has been deleted!');
} else {
alert('Color could not delete!');
}
document.messages.submit();
return false; // This line added
}
}
});
});
关于php - 检查删除是否成功消息 php ajax jquery mysql,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8611037/