doujiao2443
doujiao2443
2015-03-10 11:12

jQuery / Colorbox - 在弹出窗口中将一个java变量传递给PHP变量

Will you please please tell me how to get java variable "data_id" value in PHP Variable "$gal_id" i m not able pass the value of java variable into PHP Variable so please also let me know if you find the solution

JS IN Head

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script src="js/jquery.colorbox.js"></script>
<script type="text/javascript">
$(document).ready(function(){
//Examples of how to assign the Colorbox event to elements

$(".inline").colorbox({inline:true, width:"70%", height:"500px;"});
});
</script>

HTML

<li><a class='inline' href="#inline_content" data-id="1">ABC</a></li>
<li><a class='inline' href="#inline_content" data-id="2">DEF</a></li>
<li><a class='inline' href="#inline_content" data-id="3">XYZ</a></li>

POPUP Box

<script type="text/javascript">
$('.inline').click(function(){
   var data_id = $(this).data('id');
   $('#idvalue').html(""+data_id);
});
</script>

<div style='display:none'>
    <div id='inline_content'>
    <span id="idvalue"></span><!--we'll get here data_id value of clicked list in output -->
    <?php
    $gal_id = ""; // Will you please please tell me how to get java variable "data_id" value here from above js
    $check = mysql_query("select * from img where id = '$gal_id'");
    while ($run = mysql_fetch_array($check)){

    .
    .
    .
    }?>

</div>
</div>
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2条回答

  • douye2020 douye2020 6年前

    try this edit... I have integrated ajax with your code

    <script type="text/javascript">
    $('.inline').click(function(){
       var data_id = $(this).data('id');
       $('#idvalue').html(""+data_id);
     $.ajax({
             type: "GET",
             url: "ajax.php?gal_id="+data_id,
             success: function(data) {
                 $('#images').html(data);
             }
         }); 
    });
    </script>
    
    <div style='display:none'>
        <div id='inline_content'>
        <span id="idvalue"></span>
    <span id="images"></span>
    
    
    </div>
    </div>
    

    create a new file ajax.php

    and put this code there...

    <?php
    //whatever content you will echo here will be send to the main page and put to #images span tag...
    //put your connection string here.. to connect the database
            $gal_id = $_GET['gal_id']; 
            $check = mysql_query("select * from img where id = '$gal_id'");
            while ($run = mysql_fetch_array($check)){
    
            .
            .
            .
            }?>
    
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  • dongshixingga7900 dongshixingga7900 6年前

    If I understood your question, you can't do it this way. You can check this other question that can surely help you: How to pass JavaScript variables to PHP?

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