感谢强大的 Johan 的帮助。但我无法让我的数据库更新。如果你有时间请查看我的两个文件并告诉我 F 有什么问题 我的 hire_staff.php
<?php
session_start();
include("header.php");
?>
<?php
$chief_aerodynamicist = $staff['chief_aerodynamicist'];
$chief_designer = $staff['chief_designer'];
$commercial_director = $staff['commercial_director'];
$pit_crew = $staff['pit_crew'];
$technical_director = $staff['technical_director'];
?>
<head>
<script>
function bye(){
alert('bye');
}
$(document).ready(function(){
function hire(){
$.ajax({
url: "update.php",
type: "POST",
data: {uid: 12341234}, //this sends the user-id to php as a post variable, in php it can be accessed as $_POST['uid']
success: function(data){
data = JSON.parse(data);
//update some fields with the updated data
//you can access the data like 'data["driver"]'
}
});
}
});
</script>
</head>
<center><h2>You can hire new staff here</h2></center>
<table cellpadding="3" cellspacing="5">
<tr>
<td>Chief Aerodynamicist:</td>
<td><i><?php echo "Level $chief_aerodynamicist"; ?></i></td>
<td><form><input type="button" value="Hire!" onClick="hire();"</form></td>
</tr>
<tr>
<td>Chief Designer:</td>
<td><i><?php echo "Level $chief_designer"; ?></i></td>
</tr>
<tr>
<td>Commercial Director:</td>
<td><i><?php echo "Level $commercial_director"; ?></i></td>
</tr>
<tr>
<td>Pit Crew:</td>
<td><i><?php echo "Level $pit_crew"; ?></i></td>
</tr>
<tr>
<td>Technical Director:</td>
<td><i><?php echo "Level $technical_director"; ?></i></td>
</tr>
</table>
<?php
include("footer.php");
?>
我的test1.php文件
<?php
include("functions.php");
connect();
if(isset($_POST['uid'])){
connect();
mysql_query("UPDATE `staff` SET `driver` = '3' WHERE `id`='".$_SESSION['uid']."'") or die(mysql_error());
$query = mysql_query("SELECT FROM `staff` WHERE `id`='".$_POST['uid']."'")or die(mysql_error());
$results = mysql_fetch_assoc($query);
echo json_encode($results);
}
?>
它实际上是我要更新的 中的人员，但司机也在同一张表中
最佳答案
$(document).ready(function(){
$("#button").click(function(){
$.ajax({
url: "update.php",
type: "POST",
data: {uid: 12341234}, //this sends the user-id to php as a post variable, in php it can be accessed as $_POST['uid']
success: function(data){
data = JSON.parse(data);
//update some fields with the updated data
//you can access the data like 'data["driver"]'
}
});
});
});
//the button
<input type="button" id="button" value="Hire!"/>
//update.php
if(isset($_POST['uid'])){
//connect to the db etc...
mysql_query("UPDATE `staff` SET `driver` = '3' WHERE `id`='".$_POST['uid']."'") or die(mysql_error());
//this'll send the new statistics to the jquery code
$query = mysql_query("SELECT FROM `staff` WHERE `id`='".$_POST['uid']."'")or die(mysql_error());
$results = mysql_fetch_assoc($query);
echo json_encode($results);
}
在ajax调用的success函数中使用“update some fields”，我的意思是你可以在数据库中更新你刚刚更改的页面的统计信息，这样统计网页在数据库和页面本身中自动更改
编辑
把它放在标题中:
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<script type="text/javascript">
function bye(){
alert('bye');
}
$(document).ready(function(){
$("#button").click(function(){
$.ajax({
url: "test1.php",
type: "POST",
data: {uid: 12341234}, //this sends the user-id to php as a post variable, in php it can be accessed as $_POST['uid']
success: function(data){
data = JSON.parse(data);
//update some fields with the updated data
//you can access the data like 'data["driver"]'
}
});
});
});
</script>
然后像这样给你的按钮 id 'button':
<form><input type="button" id="button" value="Hire!"/></form>
确保 ajax 请求的 url 参数 (url: "test1.php",) 指向正确的 php 页面。
希望对您有所帮助!
编辑
如果你想选择一个驱动程序，这是代码:
<form><input type="button" class="button" id="3" value="Hire!"/></form> //the id is different for every guy
JavaScript 代码:
$(document).ready(function(){
$(".button").click(function(e){
$.ajax({
url: "test1.php",
type: "POST",
data: {colID: e.target.id}, //this sends the user-id to php as a post variable, in php it can be accessed as $_POST['uid']
success: function(data){
data = JSON.parse(data);
//update some fields with the updated data
//you can access the data like 'data["driver"]'
}
});
});
});
如您所见，我将 # 更改为 .，我将参数 e 添加到回调函数中，并添加了colID 到数据对象。
这将是新的 PHP 代码:
if(isset($_POST['colID'])){
//connect to the db etc...
$colID = mysql_real_escape($_POST['colID']); //for security
$userID = mysql_real_escape($_SESSION['uid']);
mysql_query("UPDATE `staff` SET `$colID` = 'newValue' WHERE `id`='$userID'") or die(mysql_error());
//this'll send the new statistics to the jquery code
$query = mysql_query("SELECT FROM `staff` WHERE `id`='$userID'")or die(mysql_error());
$results = mysql_fetch_assoc($query);
echo json_encode($results);
}
我希望这就是你的意思:)
关于javascript - 使用 onClick 更新数据库(NewB)，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24753319/