Question

我正在尝试在我带有复选框的表单的同一页面上显示来自mysql的数据。问题是如何编写将显示它的js脚本。代码是：

<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>

Answer 1

您可以使用jquery表单插件解决您的问题，它可以帮助您提交表单，而无需重新加载页面并在同一页面显示目标页面的返回。只需按照说明操作：

首先下载此jquery表单plugin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:(我从你的另一篇帖子here获得了以下代码，虽然没有检查过）

 <?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，您可以学习jquery表单插件的网站。

Answer 2

下面是一个伪示例，显示如何使用jQuery执行此操作，这也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个工作，所以我不会包含它。只需复制并粘贴。

<?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>

Answer 3

您应该使用PHP MySQL functions从数据库中检索所需的数据，然后通过PHP显示它们，而不是javascript。

特别要看一下：mysql_fetch_assoc - 有一个完整的例子。

使用jquery显示mysql查询结果

3 个答案: