我正试图在带有复选框的表单所在的同一页面上显示mysql中的数据。问题是如何编写js脚本来显示它。代码为：
<form id="myForm" action="pdoakcja.php" method="post">
<!--Instruktor: <input type="text" name="name" /> -->
Permissions:<input type="checkbox" name="M1" value="M1" />M1
<input type="checkbox" name="M2" value="M2" />M2
<input type="submit" value="Szukaj" />
</form>
<div id='name-data'>Instruktorzy o podanych uprawnieniach:</div>
<script src="http://code.jquery.com/jquery-1.8.0.min.js"></script>
<script>
............??????
</script>
您可以使用jquery表单插件来解决您的问题，这将帮助您提交表单，而无需重新加载页面，并在同一页面中显示目标页面的返回信息。只需按照说明操作：
首先下载这个jquery表单插件并保存。
然后
<script type="text/javascript" src="http://code.jquery.com/jquery-1.7.1.min.js"></script>
<!-- This jquery.form.js is for Submitting form data using jquery and Ajax -->
<script type="text/javascript" src="js/jquery.form.js"></script>
<script type="text/javascript">
$(document).ready(function() {
var options = {
success: showResponse
};
// bind form using 'ajaxForm'
$('#myForm').ajaxForm(options);
});
// post-submit callback
function showResponse(responseText, statusText, xhr, $form) {
if(responseText==1){
$("#error").html('No Result Found');
} else{
$("#result").html(responseText);
}
}
</script>
<form id="myForm" enctype="multipart/form-data" action="pdoakcja.php"
method="post" name="myForm">
<!--Instruktor: <input type="text" name="name" /> -->
Permissions:<input type="checkbox" name="M1" value="M1" />M1
<input type="checkbox" name="M2" value="M2" />M2
<input type="submit" value="Szukaj" />
</form>
<span id="error"></span>
<span id="result"></span>
你的pdoakcja.php文件：（我从你的另一篇文章中得到了以下代码，但还没有检查）
<?php
$query = mysql_query("SELECT * FROM permissions WHERE m LIKE '".$_POST['M1']."' OR m LIKE '".$_POST['M2']."' OR mn LIKE '".$_POST['MN1']."' ");
if($query) {
while($permissions = mysql_fetch_assoc($query)){
$query2 = mysql_query("SELECT name_surname FROM instruktorzy WHERE instruktor_id='".$permissions['instruktor_id']."'");
while($Mdwa = mysql_fetch_assoc($query2)){
echo "<p style='"font-size: 14px; font-family: Helvetica; background-color: #FFFFFF'"> ".$Mdwa['name_surname']."<br />" ; "</p>" ;
}
}
} else {echo "1";}
?>
我希望这对你有用。有关详细信息，您可以查看jquery表单插件的网站。
下面是一个伪示例，展示了如何使用jQuery完成此操作，它还会在您单击复选框时更新，以便您可以完全删除提交；你说你已经有一个数据库在做这项工作，所以我不会把它包括在内。只需复制粘贴即可。
<?php
//Some pseudo data kinda as your receive it from a query
$datafromSql = array(
array('id'=>1,'permission'=>'M1','theData'=>'User has M1 permission'),
array('id'=>2,'permission'=>'M2','theData'=>'User has M2 permission'),
array('id'=>3,'permission'=>'M1','theData'=>'User has M1 permission'),
array('id'=>4,'permission'=>'M1','theData'=>'User has M1 permission'),
);
//Access the data
if($_SERVER['REQUEST_METHOD']=='POST'){
$is_ajax = false;
if(isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) === 'xmlhttprequest'){
$is_ajax = true;
}
//pseudo code, really you would put your query here
// SELECT theData FROM your_table WHERE permission=POST_VALUE ... ...
//And then format your output
$result=array();
foreach($datafromSql as $row){
if($is_ajax == true){
foreach($_POST as $key=>$value){
if($_POST[$key] == 'true' && $row['permission']==$key){
$result[]=$row['theData'].'<br />';
}
}
}else{
foreach($_POST as $key=>$value){
if($_POST[$key] == $row['permission']){
$result[]=$row['theData'].'<br />';
}
}
}
}
$result = implode('<hr />',$result);
//AJAX Response, echo and then die.
if($is_ajax === true){
header('Content-Type: text/html');
//example output sent back to the jQuery callback
echo $result;
//echo '<pre>'.print_r($_POST,true).'</pre>';
die;
}
}
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8">
<script type="text/javascript" src="http://ajax.aspnetcdn.com/ajax/jQuery/jquery-1.7.2.min.js" charset="utf-8"></script>
<script type="text/javascript">
function update(){
$.post('./<?php echo basename(__FILE__)?>',
{
M1: $("#M1").is(':checked'),
M2: $("#M2").is(':checked')
},
function(data) {
$('#result').replaceWith('<div id="result"><h1>The Result:</h1>'+ data +'</div>');
});
}
</script>
</head>
<body>
<form method="POST" action="<?php echo basename(__FILE__)?>">
Permissions:
<input type="checkbox" id="M1" name="M1" value="M1" onChange="update()"/>M1
<input type="checkbox" id="M2" name="M2" value="M2" onChange="update()"/>M2
<input type="submit" value="Szukaj" />
</form>
<p id='result'><?php echo isset($result)?$result:null;?></p>
</body>
</html>
您应该使用PHP MySQL函数从数据库中检索所需的数据，然后通过PHP而不是javascript显示这些数据。
特别要看一看：mysql_fetch_assoc-这是一个完整的工作示例。