Question

以上是选择以上之后

1.Html文件

这里我有3个组合框。第二个组合框取决于第一个组合框，第三个组合框应取决于第二个组合框。

我试图显示第二个查询（我的意思是单独使用php），它完全正常工作。

简单的值不是从第二个组合框传递到php文件。（数据：＆＃39; sub =＆＃39; + SUB，//值不通过）。 $ rowCount = mysql_num_rows（$ query）; // $ rowcount在第二个组合框中始终为NULL。

    <html><!DOCTYPE html>
    <html xmlns="http://www.w3.org/1999/xhtml">
    <head>
    <style type="text/css">
    .select-boxes{width: 280px;text-align: center;}
    select {
        background-color: #F5F5F5;
        border: 1px double #FB4314;
        color: #55BB91;
        font-family: Georgia;
        font-weight: bold;
        font-size: 14px;
        height: 39px;
        padding: 7px 8px;
        width: 250px;
        outline: none;
        margin: 10px 0 10px 0;
    }
    select option{
        font-family: Georgia;
        font-size: 14px;
    }
    </style>
    <script src="jquery.min.js"></script>
    <script type="text/javascript">

    $(document).ready(function(){
        $('#sem').on('change',function(){
            var SEM = $(this).val();
            if(SEM){
                $.ajax({
                    type:'POST',
                    url:'sta.php',
                    data:'sem='+SEM,
                    success:function(html){
                        $('#sub').html(html);
                        $('#staff').html('<option value="">Select sub first</option>'); 
                    }
                }); 
            }else{
                $('#sem').html('<option value="">Select batch first</option>');
                $('#sub').html('<option value="">Select sem first</option>'); 
            }
        });
        $('#sub').on('change',function(){
            var SUB = $(this).val();
            if(1){
                $.ajax({
                    type:'POST',
                    url:'sta.php',
                    data:'sub='+SUB,  //VALUE IS NOT PASSING
                    success:function(html){
                        $('#staff').html(html);
                    }
                }); 
            }else{
                $('#staff').html('<option value="">Select sub first</option>'); 
            }
        });
    }); 
    </script>
    </head>
    <body>
        <div class="select-boxes">
       <form name="signup" method="POST" action="feedform.php">
    <br><br><br><br>
   <select id="sem" name="sem">
    <option value="" selected>Select current semester</option>
    <option value="1">1</option>
    <option value="2">2</option>
    <option value="3">3</option>
    <option value="4">4</option>
    </select><br><br><br>
        <select name="sub" id="sub">
            <option value="">Select semester first</option>
        </select>

        <select name="staff" id="staff">
            <option value="">Select sub first</option>
        </select>
        </form>
        </div>
    </body>
    </html>

2.sta.php

<?php
include("../Mine/connection.php");
if(isset($_POST["sem"]) && !empty($_POST["sem"]))
{
    $query = mysql_query("SELECT course_name FROM subject WHERE semester =".$_POST['sem']);

    //Count total number of rows
    $rowCount = mysql_num_rows($query);

    if($rowCount > 0){
        echo '<option value="">Select course</option>';
        while($row = mysql_fetch_assoc($query)){ 
            echo '<option value="'.$row['course_name'].'">'.$row['course_name'].'</option>';
        }
    }else{
        echo '<option value="">course not available</option>';
    }
}

if(isset($_POST["sub"]) && !empty($_POST["sub"]))
{
    $query = mysql_query("SELECT staff_name FROM subject WHERE course_name = ".$_POST['sub']);

    //Count total number of rows
    $rowCount = mysql_num_rows($query);

    if($rowCount > 0){
        echo '<option value="">Select Staff</option>';
        while($row = mysql_fetch_assoc($query)){ 
            echo '<option value="'.$row['staff_name'].'">'.$row['staff_name'].'</option>';
        }
    }else{
        echo '<option value="">Staff not available</option>';
    }
}

?>

值没有从一个动态下拉框传递到php中的另一个动态下拉框

0 个答案: