Question

我为php / mysql / jquery三个链式选择框找到this插件。这适用于这种方法：选择第一个框是可见的！选择任何option onload selectbox 2之后，选择任何option onload last selectbox（3）之后。现在我的问题是：选择方框2＆amp;从prevuise选择框中选择option后，隐藏3并显示（可见）。我需要显示（显示）所有三个选择框，然后选择option将结果显示到选择框中。

PHP功能：

<?php
//**************************************
//     Page load dropdown results     //
//**************************************
function getTierOne()
{
    $result = mysql_query("SELECT DISTINCT tier_one FROM three_drops") 
    or die(mysql_error());

      while($tier = mysql_fetch_array( $result )) 

        {
           echo '<option value="'.$tier['tier_one'].'">'.$tier['tier_one'].'</option>';
        }

}

//**************************************
//     First selection results     //
//**************************************
if($_GET['func'] == "drop_1" && isset($_GET['func'])) { 
   drop_1($_GET['drop_var']); 
}

function drop_1($drop_var)
{  
    include_once('db.php');
    $result = mysql_query("SELECT DISTINCT tier_two FROM three_drops WHERE tier_one='$drop_var'") 
    or die(mysql_error());

    echo '<select name="drop_2" id="drop_2">
          <option value=" " disabled="disabled" selected="selected">Choose one</option>';

           while($drop_2 = mysql_fetch_array( $result )) 
            {
              echo '<option value="'.$drop_2['tier_two'].'">'.$drop_2['tier_two'].'</option>';
            }

    echo '</select>';
    echo "<script type=\"text/javascript\">
$('#wait_2').hide();
    $('#drop_2').change(function(){
      $('#wait_2').show();
      $('#result_2').hide();
      $.get(\"func.php\", {
        func: \"drop_2\",
        drop_var: $('#drop_2').val()
      }, function(response){
        $('#result_2').fadeOut();
        setTimeout(\"finishAjax_tier_three('result_2', '\"+escape(response)+\"')\", 400);
      });
        return false;
    });
</script>";
}


//**************************************
//     Second selection results     //
//**************************************
if($_GET['func'] == "drop_2" && isset($_GET['func'])) { 
   drop_2($_GET['drop_var']); 
}

function drop_2($drop_var)
{  
    include_once('db.php');
    $result = mysql_query("SELECT * FROM three_drops WHERE tier_two='$drop_var'") 
    or die(mysql_error());

    echo '<select name="drop_3" id="drop_3">
          <option value=" " disabled="disabled" selected="selected">Choose one</option>';

           while($drop_3 = mysql_fetch_array( $result )) 
            {
              echo '<option value="'.$drop_3['tier_three'].'">'.$drop_3['tier_three'].'</option>';
            }

    echo '</select> ';
    echo '<input type="submit" name="submit" value="Submit" />';
}
?>

JS＆amp;索引PHP：

include('db.php');
 include('func.php');

<script type="text/javascript">
$(document).ready(function() {
    $('#wait_1').hide();
    $('#drop_1').change(function(){
      $('#wait_1').show();
      $('#result_1').hide();
      $.get("func.php", {
        func: "drop_1",
        drop_var: $('#drop_1').val()
      }, function(response){
        $('#result_1').fadeOut();
        setTimeout("finishAjax('result_1', '"+escape(response)+"')", 400);
      });
        return false;
    });
});

function finishAjax(id, response) {
  $('#wait_1').hide();
  $('#'+id).html(unescape(response));
  $('#'+id).fadeIn();
}
function finishAjax_tier_three(id, response) {
  $('#wait_2').hide();
  $('#'+id).html(unescape(response));
  $('#'+id).fadeIn();
}
</script>

<body>
<p>
<form action="" method="post">

    <select name="drop_1" id="drop_1">

      <option value="" selected="selected" disabled="disabled">Select a Category</option>

      <?php getTierOne(); ?>

    </select> 

    <span id="wait_1" style="display: none;">
    <img alt="Please Wait" src="ajax-loader.gif"/>
    </span>
    <span id="result_1" style="display: none;"></span>
    <span id="wait_2" style="display: none;">
    <img alt="Please Wait" src="ajax-loader.gif"/>
    </span>
    <span id="result_2" style="display: none;"></span> 

</form>
</p>
<p>
<?php if(isset($_POST['submit'])){
    $drop = $_POST['drop_1'];
    $drop_2 = $_POST['drop_2'];
    $drop_3 = $_POST['drop_3'];
    echo "You selected a ";
    echo $drop_3." ".$drop." ".$drop_2;
}
?>

感谢任何帮助.u可以通过我的意思告诉我任何e.x和演示。

链式选择框jquery php

0 个答案: