Javascript HTML拖放OnDrop Ajax事件
我有一个简单的页面,它使用由MySQLi函数创建的HTML拖放缩略图。像这样Javascript HTML拖放OnDrop Ajax事件,javascript,php,jquery,html,ajax,Javascript,Php,Jquery,Html,Ajax,我有一个简单的页面,它使用由MySQLi函数创建的HTML拖放缩略图。像这样 <?php include "database_connection.php"; $query = "SELECT * FROM table where currentZone = 0"; if ($result = mysqli_query($link, $query)) { /* fetch associative array */ while
<?php
include "database_connection.php";
$query = "SELECT * FROM table where currentZone = 0";
if ($result = mysqli_query($link, $query)) {
/* fetch associative array */
while ($row = mysqli_fetch_assoc($result))
{
echo "<img id='{$row["ID"]}' src='{$row["photoLink"]}.jpg' draggable='true' ondragstart='drag(event)' width='75' height='75'>" ;
}
/* free result set */
mysqli_free_result($result);
}
mysqli_close($link);
?>
mainpage.php
功能allowDrop(ev){
ev.preventDefault();
}
功能阻力(ev){
ev.dataTransfer.setData(“文本”,ev.target.id);
}
功能下降(ev){
ev.preventDefault();
var data=ev.dataTransfer.getData(“文本”);
ev.target.appendChild(document.getElementById(数据));
}
?>
而test.php仅仅连接到数据库并插入一条垃圾记录
function drop(id, event) {
$.ajax({
url: "test.php",
type: "POST",
data: {
id: id,
event: event
},
success: function () {
console.log('great success');
return true
}
});
return false;
}
然后在do_sql_stuff.php中,您可以得到
<!DOCTYPE HTML>
<html>
<head>
<script type="text/javascript" src="/js/jquery.js"></script>
<script type="text/javascript" src="/js/script.js"></script>
<header class="main-header" role="banner"><center>
<img src="logo.jpg" height="90" width="400"alt="Banner Image"/></center>
</header>
<script>
function allowDrop(ev) {
ev.preventDefault();
}
function drag(ev) {
ev.dataTransfer.setData("text", ev.target.id);
}
function drop(ev) {
ev.preventDefault();
var data = ev.dataTransfer.getData("text");
ev.target.appendChild(document.getElementById(data));
}
</script>
</head>
<body>
<div id="1" ondrop="drop(event)" ondragover="allowDrop(event)">
<?php
include "database_connection.php";
///////////////////////////////////////////////////////////////////////////////////////////////
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
else
{
}
$query = "SELECT * FROM table where currentZone = 1";
if ($result = mysqli_query($link, $query)) {
/* fetch associative array */
while ($row = mysqli_fetch_assoc($result)) {
echo "<img id='{$row["ID"]}' src='{$row["photoLink"]}.jpg' draggable='true' ondragstart='drag(event)' width='75' height='75'>" ;
}
/* free result set */
mysqli_free_result($result);
}
mysqli_close($link);
/////////////////////////////////////////////////////////////////////////////////
?>
</div>
<div id="2" ondrop="drop(event)" ondragover="allowDrop(event)">
<?php
include "database_connection.php";
$query = "SELECT * FROM table where currentZone = 2";
if ($result = mysqli_query($link, $query)) {
/* fetch associative array */
while ($row = mysqli_fetch_assoc($result)) {
echo "<img id='{$row["ID"]}' src='{$row["photoLink"]}.jpg' draggable='true' ondragstart='drag(event)' width='75' height='75'>" ;
}
/* free result set */
mysqli_free_result($result);
}
mysqli_close($link);
?>
</div>
?>
我试试看。感谢您提供的信息。显然,您需要将id传递到函数中。。。。我也会用php加载它,这样你就可以传入一个变量了,我正要问这个问题。谢谢我可以手动(通过访问url)使php页面正常运行,但在更改div4以执行发布的操作并将函数放入脚本后,它似乎无法发布。关于我遗漏的步骤有什么想法吗?为问题添加了代码-
<!DOCTYPE HTML>
<html>
<head>
<script type="text/javascript" src="/js/jquery.js"></script>
<script type="text/javascript" src="/js/script.js"></script>
<header class="main-header" role="banner"><center>
<img src="logo.jpg" height="90" width="400"alt="Banner Image"/></center>
</header>
<script>
function allowDrop(ev) {
ev.preventDefault();
}
function drag(ev) {
ev.dataTransfer.setData("text", ev.target.id);
}
function drop(ev) {
ev.preventDefault();
var data = ev.dataTransfer.getData("text");
ev.target.appendChild(document.getElementById(data));
}
</script>
</head>
<body>
<div id="1" ondrop="drop(event)" ondragover="allowDrop(event)">
<?php
include "database_connection.php";
///////////////////////////////////////////////////////////////////////////////////////////////
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
else
{
}
$query = "SELECT * FROM table where currentZone = 1";
if ($result = mysqli_query($link, $query)) {
/* fetch associative array */
while ($row = mysqli_fetch_assoc($result)) {
echo "<img id='{$row["ID"]}' src='{$row["photoLink"]}.jpg' draggable='true' ondragstart='drag(event)' width='75' height='75'>" ;
}
/* free result set */
mysqli_free_result($result);
}
mysqli_close($link);
/////////////////////////////////////////////////////////////////////////////////
?>
</div>
<div id="2" ondrop="drop(event)" ondragover="allowDrop(event)">
<?php
include "database_connection.php";
$query = "SELECT * FROM table where currentZone = 2";
if ($result = mysqli_query($link, $query)) {
/* fetch associative array */
while ($row = mysqli_fetch_assoc($result)) {
echo "<img id='{$row["ID"]}' src='{$row["photoLink"]}.jpg' draggable='true' ondragstart='drag(event)' width='75' height='75'>" ;
}
/* free result set */
mysqli_free_result($result);
}
mysqli_close($link);
?>
</div>
?>
function drop(id, event) {
$.ajax({
url: "do_sql_stuff.php",
type: "POST",
data: {
id: id,
event: event
},
success: function () {
console.log('great success');
return true
}
});
return false;
}
$event = $_POST['event'];
$id = $_POST['id'];