如何根据所选教师列出车辆

Question

我有一个列表框，当用户选择教师时显示车辆列表。我只想根据用户选择的教师列出车辆。

<td>No</td>
<td>Registration No</td>

<?php
     $count = 0;
     $db = mysql_connect('localhost','root') 
     or die ("unable to connect");

 mysql_select_db('fyp',$db) or die ("able to select");
 $sql="SELECT * FROM vehicle_record";
  $result = mysql_query($sql) or die ("Query failed!");
 while($row = mysql_fetch_array($result)){
       $count = $count + 1;
?>

<select name="faculty" >
               <option value="" selected>-- Please Select --</option>
               <option>City Campus</option>
               <option>MFI</option>
               <option>BMI</option>
               <option>MSI</option>
               <option>MIAT</option>
               <option>MICET</option>
               <option>MIMET</option>
               <option>RCMP</option>
             </select> 

Answer 1

我无法完全理解您的问题，但我假设您希望在提交表单后选择车辆设施，您将 javascript 用于此目的：

          <select name="faculty" id="faculty">
           <option value="" selected>-- Please Select --</option>
           <option>City Campus</option>
           <option>MFI</option>
           <option>BMI</option>
           <option>MSI</option>
           <option>MIAT</option>
           <option>MICET</option>
           <option>MIMET</option>
           <option>RCMP</option>
         </select>

         <script type="text/javascripot">
          document.getElementById("faculty").value = "<?php echo $_POST['faculty']?>";
         </script>

但是，如果您的意思是如何显示来自数据库的设施，那么您可以像这样:

<select name="faculty" id="faculty">
<?php
  while($row = mysql_fetch_array($result)){
   echo '<option value="'.$row['field_name'].'">'.$row['field_name'].'</option>';
  }
?>
</select>

Answer 2

我想你有 2 个选择，一个用于教师，一个用于车辆

<select name="faculty" id="faculty" onchange="getVehicles()">
....
</select>


<select name="vehicle" id="vehicle">
</select>

Javascript 代码

function getVechiles()
{                                       
   $.getJSON("getVechiles.php",{faculty: $("select#faculty").val()}, function(j){
   if(j.length>0)
   {

        var options = '';
        for (var i = 0; i < j.length; i++) {
                options += '<option class="' + j[i].langClass + '" value="' + j[i].value + '">' + j[i].text + '</option>';
        }
        $("#vechile").html(options);

    }

    });
}

和 getVechiles.php

    SELECT * FROM TABLE
    while($row = mysql_fetch_array($result)){
      $js="{value:".row['id'].",text:\"".$row["field"]."\"}";
      $rjson[] =  $js;

    }

   if(is_array($rjson))
   {
     echo '[' . implode(',', $rjson) . ']';
   }    

Answer 3

<?php
$count = 0;
$i_faculty = $_POST['ifaculty'];
$db = mysql_connect('localhost','root') or die ("unable to connect");

mysql_select_db('fyp',$db) or die ("able to select");
$sql="SELECT * FROM vehicle_record WHERE faculty ='".$i_faculty."' ";
$result = mysql_query($sql) or die ("Query failed!");
while($row = mysql_fetch_array($result))
{
    $count = $count + 1;
    ?>
    <select name="ifaculty">
       <option value="" selected="selected">-- Select Faculty --</option>
       <option>City Campus</option>
       <option>MFI</option>
       <option>BMI</option>
       <option>MSI</option>
       <option>MIAT</option>
       <option>MICET</option>
       <option>MIMET</option>
       <option>RCMP</option>
    </select>