使用PHP Ajax在MySQL数据库中存储开/关按钮的值
我找到了一个单一ID的教程 我已经根据我的知识修改了多个id,我无法使用ajax获取id。我不擅长ajax。我已经附上了两个代码。谁能告诉我如何修复多个id 我还附上了一个屏幕截图 index.php使用PHP Ajax在MySQL数据库中存储开/关按钮的值,php,jquery,mysql,ajax,Php,Jquery,Mysql,Ajax,我找到了一个单一ID的教程 我已经根据我的知识修改了多个id,我无法使用ajax获取id。我不擅长ajax。我已经附上了两个代码。谁能告诉我如何修复多个id 我还附上了一个屏幕截图 index.php <?php $query=mysql_connect("localhost","pma","pmapass"); mysql_select_db("testdb",$query); include("connection1.php"); $result = mysql_query("
<?php
$query=mysql_connect("localhost","pma","pmapass");
mysql_select_db("testdb",$query);
include("connection1.php");
$result = mysql_query("SELECT * FROM mytable ORDER BY id");
?>
<script type="text/javascript"src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script>
<div>
<table id="datatables" class="display">
<thead>
<tr>
<th>ID</th>
<th>Name</th>
<th>Text</th>
</tr>
</thead>
<tbody>
<?php
while ($row = mysql_fetch_array($result)) {
?>
<tr>
<td><?php echo $row["id"]; ?></td>
<td><?php echo $row["name"]; ?></td>
<td><?php echo $row["text"]; ?></td>
<td><script type="text/javascript">
$(document).ready(function(){
$('#myonoffswitch<?php echo $row["id"]; ?>').click(function(){
var myonoffswitch<?php echo $row["id"]; ?>=$('#myonoffswitch<?php echo $row["id"]; ?>').val();
if ($("#myonoffswitch<?php echo $row["id"]; ?>:checked").length == 0)
{
var a<?php echo $row["id"]; ?>=myonoffswitch<?php echo $row["id"]; ?>;
}
else
{
var a<?php echo $row["id"]; ?>="off";
}
$.ajax({
type: "POST",
url: "ajax.php",
data: "value="+a<?php echo $row["id"]; ?>,
success: function(html){
$("#display").html(html).show();
}
});
});
});
</script><div class="onoffswitch">
<input type="hidden" value="<?php echo $row["id"]; ?>" name="ids" />
<input type="checkbox" name="onoffswitch<?php echo $row["id"]; ?>" class="onoffswitch-checkbox" id="myonoffswitch<?php echo $row["id"]; ?>"
<?php
$query3=mysql_query("select * from mytable where id=$row[id]");
$query4=mysql_fetch_array($query3);
if($query4['text']=="off")
{
echo "checked";
}
?>>
<label class="onoffswitch-label" for="myonoffswitch<?php echo $row["id"]; ?>">
<div class="onoffswitch-inner"></div>
<div class="onoffswitch-switch"></div>
</label>
</div>
<div id="display"></div>
</td>
</tr>
<?php
}
?>
</tbody>
</table>
</div>
<?php
$query=mysql_connect("localhost","pma","pmapass");
mysql_select_db("testdb",$query);
include("connection1.php");
$result = mysql_query("SELECT * FROM mytable ORDER BY id");
?>
<script type="text/javascript"src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script>
<div>
<table id="datatables" class="display">
<thead>
<tr>
<th>ID</th>
<th>Name</th>
<th>Text</th>
</tr>
</thead>
<tbody>
<?php while ($row = mysql_fetch_array($result)) { ?>
<tr>
<td><?php echo $row["id"]; ?></td>
<td><?php echo $row["name"]; ?></td>
<td><?php echo $row["text"]; ?></td>
<td>
<div class="onoffswitch">
<input type="hidden" value="<?php echo $row["id"]; ?>" />
<input type="checkbox" class="onoffswitch-checkbox"
<?php
if($row['text']=="off")
{
echo "checked";
}
?>>
<label class="onoffswitch-label" for="myonoffswitch<?php echo $row["id"]; ?>">
<div class="onoffswitch-inner"></div>
<div class="onoffswitch-switch"></div>
</label>
</div>
<div id="display">
</div>
</td>
</tr>
<?php
}
?>
</tbody>
</table>
<script type="text/javascript">
$(document).ready(function(){
$('.onoffswitch').click(function(){
var hiddenValueID = $(this).children(':hidden').val();
if ($(this).children(':checked').length == 0)
{
var valueData = 'on';
}
else
{
var valueData = 'off';
}
$.ajax({
type: "POST",
url: "ajax.php",
data: {value: valueData, id: hiddenValueID} ,
done: function(html){
$("#display").html(html).show();
}
});
});
});
</script>
</div>
Ajax.php
为了你自己;:
请整理你的代码
如果无法从数据库加载某些内容,请在
为什么是2。查询时,您已使用select*?
然而,这里有一个基于未测试代码的示例:
index.php
<?php
$query=mysql_connect("localhost","pma","pmapass");
mysql_select_db("testdb",$query);
include("connection1.php");
$result = mysql_query("SELECT * FROM mytable ORDER BY id");
?>
<script type="text/javascript"src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script>
<div>
<table id="datatables" class="display">
<thead>
<tr>
<th>ID</th>
<th>Name</th>
<th>Text</th>
</tr>
</thead>
<tbody>
<?php
while ($row = mysql_fetch_array($result)) {
?>
<tr>
<td><?php echo $row["id"]; ?></td>
<td><?php echo $row["name"]; ?></td>
<td><?php echo $row["text"]; ?></td>
<td><script type="text/javascript">
$(document).ready(function(){
$('#myonoffswitch<?php echo $row["id"]; ?>').click(function(){
var myonoffswitch<?php echo $row["id"]; ?>=$('#myonoffswitch<?php echo $row["id"]; ?>').val();
if ($("#myonoffswitch<?php echo $row["id"]; ?>:checked").length == 0)
{
var a<?php echo $row["id"]; ?>=myonoffswitch<?php echo $row["id"]; ?>;
}
else
{
var a<?php echo $row["id"]; ?>="off";
}
$.ajax({
type: "POST",
url: "ajax.php",
data: "value="+a<?php echo $row["id"]; ?>,
success: function(html){
$("#display").html(html).show();
}
});
});
});
</script><div class="onoffswitch">
<input type="hidden" value="<?php echo $row["id"]; ?>" name="ids" />
<input type="checkbox" name="onoffswitch<?php echo $row["id"]; ?>" class="onoffswitch-checkbox" id="myonoffswitch<?php echo $row["id"]; ?>"
<?php
$query3=mysql_query("select * from mytable where id=$row[id]");
$query4=mysql_fetch_array($query3);
if($query4['text']=="off")
{
echo "checked";
}
?>>
<label class="onoffswitch-label" for="myonoffswitch<?php echo $row["id"]; ?>">
<div class="onoffswitch-inner"></div>
<div class="onoffswitch-switch"></div>
</label>
</div>
<div id="display"></div>
</td>
</tr>
<?php
}
?>
</tbody>
</table>
</div>
<?php
$query=mysql_connect("localhost","pma","pmapass");
mysql_select_db("testdb",$query);
include("connection1.php");
$result = mysql_query("SELECT * FROM mytable ORDER BY id");
?>
<script type="text/javascript"src="http://ajax.googleapis.com/ajax/libs/jquery/1.6.2/jquery.min.js">
</script>
<div>
<table id="datatables" class="display">
<thead>
<tr>
<th>ID</th>
<th>Name</th>
<th>Text</th>
</tr>
</thead>
<tbody>
<?php while ($row = mysql_fetch_array($result)) { ?>
<tr>
<td><?php echo $row["id"]; ?></td>
<td><?php echo $row["name"]; ?></td>
<td><?php echo $row["text"]; ?></td>
<td>
<div class="onoffswitch">
<input type="hidden" value="<?php echo $row["id"]; ?>" />
<input type="checkbox" class="onoffswitch-checkbox"
<?php
if($row['text']=="off")
{
echo "checked";
}
?>>
<label class="onoffswitch-label" for="myonoffswitch<?php echo $row["id"]; ?>">
<div class="onoffswitch-inner"></div>
<div class="onoffswitch-switch"></div>
</label>
</div>
<div id="display">
</div>
</td>
</tr>
<?php
}
?>
</tbody>
</table>
<script type="text/javascript">
$(document).ready(function(){
$('.onoffswitch').click(function(){
var hiddenValueID = $(this).children(':hidden').val();
if ($(this).children(':checked').length == 0)
{
var valueData = 'on';
}
else
{
var valueData = 'off';
}
$.ajax({
type: "POST",
url: "ajax.php",
data: {value: valueData, id: hiddenValueID} ,
done: function(html){
$("#display").html(html).show();
}
});
});
});
</script>
</div>
ajax.php
<?php
$query=mysql_connect("localhost","pma","pmapass");
mysql_select_db("testdb",$query);
if(isset($_POST['value']))
{
$value=$_POST['value'];
$id=$_POST['id'];
mysql_query("update mytable set text='$value' where id=$id");
echo "<h2>You have Chosen the button status as:" .$value."</h2>";
}
?>
如果从ajax请求中得到答案,您可以查看浏览器控制台。很难理解您的代码。。。用php生成js变量名可能不是最好的解决方案。。。你到底想干什么?你的按钮在哪里?我想用那个按钮隐藏数据。我有超过10个带有数据的ID。我希望每个ID上的按钮都能正常工作。这是一个不错的按钮,但按钮滑动不起作用。朋友,我已经解决了这个问题,谢谢你的快速帮助。你能告诉我一个更好的学习ajax的方法吗?对于纯ajax,你可以在这里看看,或者在使用jQuery这样的包装器时看看,或者