下拉列表中未显示的建筑物和房间列表

时间：2012-01-27 15:51:17

标签： php

以下是数据库中的“房间”表格：

   Room          Building            Capacity
    CW5/10        Canalside West        50
    CW4/09        Canalside West        40
    CW2/08        Canalside West        40
    CW4/10        Canalside West        25
    CE1/03        Canalside East        40

以下是完整代码：

           <?php

foreach (array('courseid','building') as $varname) {
    $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : '';
  }

if (isset($_POST['submit'])) {
    $query = "
                 SELECT cm.CourseId, cm.ModuleId, 
                 c.CourseName,
                 m.ModuleName
                 FROM Course c
                 INNER JOIN Course_Module cm ON c.CourseId = cm.CourseId
                 JOIN Module m ON cm.ModuleId = m.ModuleId
                 WHERE
                 (c.CourseId = '".mysql_real_escape_string($courseid)."')
                 ORDER BY c.CourseName, m.ModuleId
                 ";

    $num = mysql_num_rows($result = mysql_query($query));

    if($num ==0){
        echo "<p>Sorry, No Course was found with this Course ID '$courseid'</p>";
    } else { 

        $dataArray = array();

        session_start();

        while ($row = mysql_fetch_array($result)) { 
            $dataArray[$row['CourseId']]['CourseName'] = $row['CourseName']; 
            $dataArray[$row['CourseId']]['Modules'][$row['ModuleId']]['ModuleName'] = $row['ModuleName']; 

$_SESSION['idcourse'] = $row['CourseId'];
$_SESSION['namecourse'] = $row['CourseName'];

    }

     foreach ($dataArray as $courseId => $courseData) {

          $output = ""; 

          $output .= "<p><strong>Course:</strong> " . $courseId .  " - "  . $courseData['CourseName'] . "</p>";

       $moduleHTML = ""; 
       $moduleHTML .= '<select name="module" id="modulesDrop">'.PHP_EOL;
       $moduleHTML .= '<option value="">Please Select</option>'.PHP_EOL;      
            foreach ($courseData['Modules'] as $moduleId => $moduleData) {        

            $moduleHTML .= "<option value='".$moduleId.' '.$moduleData['ModuleName']."'>" . $moduleId . " - " . $moduleData['ModuleName'] ."</option>".PHP_EOL;        
  } 
            }
            $moduleHTML .= '</select>';

      echo $output;

//上面是课程和模块，下面是建筑物和房间（所有代码都在一个<?php?>标签中。

 $sql="SELECT Building, Room FROM Room WHERE Building = '".$building."'";

    $sqlresult = mysql_query($sql);

    $buildings = array(); // easier if you don't use generic names for data

    while($sqlrow = mysql_fetch_array($sqlresult))
    {
        // you need to initialise your building array cells
        if (!isset($buildings[$sqlrow['Building']])) {
            $buildings[$sqlrow['Building']] = array('Rooms' => array());
        }

        // you can add the room to the building 'Rooms' array
        $buildings[$sqlrow['Building']]['Rooms'][] = $sqlrow['Room'];
    }


    $buildingHTML = ""; 
    $buildingHTML = "<form action=\"\" method=\"post\">";
    $buildingHTML .= '<select name="buildings" id="buildingssDrop" onchange="document.getElementById(\'dropDownForm\').submit()">'.PHP_EOL;
    $buildingHTML .= '<option value="">Please Select</option>'.PHP_EOL; 

    foreach ($buildings as $building => $buildingData) {      
        $buildingHTML .= "<option value='".$building."'>" . $building . "</option>".PHP_EOL;        
    }
    $buildingHTML .= '</select>';
    $buildingHTML .= '</form>';

    $roomHTML = ""; 
    $roomHTML .= '<select name="rooms" id="roomsDrop">'.PHP_EOL;
    $roomHTML .= '<option value="">Please Select</option>'.PHP_EOL; 

    if(isset($_POST['buildings'])){
        $buildingname = $_POST['buildings'];
    foreach ($buildings[$buildingname]['Rooms'] as $roomId => $roomData) {        
        $roomHTML .= "<option value='".$roomId."'>" . $roomId . "</option>".PHP_EOL;        
    }
}
    $roomHTML .= '</select>';


        ?>

现在在脚本的上半部分，它控制课程和模块。用户输入一个CourseId，如果错误，它会显示一条消息，说明找不到课程，如果正确则会显示课程名称，然后会显示一个下拉菜单，其中包含属于的模块列表那个课程。这很好。

代码的下半部分是问题所在。假设显示2个下拉菜单，一个下拉列表将显示建筑物列表，在第二个下拉列表中，它将显示从第一个下拉菜单中属于所选建筑物的房间列表。此代码显示2个下拉菜单，两个菜单都只显示“请选择”选项，而不显示任何其他选项。

正如一些专家指出的那样，他们认为问题是$building变量包含null。它是如何工作的，以便$ building变量能够从查询中检索建筑物列表？

您可以查看应用程序here输入'info101'作为courseId并提交，您将看到所有功能都出现。转到模块下拉列表，您将看到与课程相关的模块。如果您查看建筑物和房间下拉菜单，它们都只显示“请选择”。

2 个答案:

答案 0 :(得分：2)

您需要更改PHP代码才能实现此方案。如果您希望在更改建筑物名称时动态填充房间，您可以使用jquery。在脚本中包含jQUery库。您需要执行ajax调用以动态填充列表。

我只从建筑部分添加。以上休息是一样的。在foreach循环中从数组中删除构建。将当前的PHP代码更改为此。

PHP代码：

 $sql = "SELECT DISTINCT Building FROM Room";

 $sqlresult = mysql_query($sql);

 $buildings = array(); // easier if you don't use generic names for data

 $buildingHTML = ""; 
 $buildingHTML = "<form action=\"\" method=\"post\">";
 $buildingHTML .= '<select name="buildings" id="buildingssDrop" onchange="getRooms();">'.PHP_EOL;
 $buildingHTML .= '<option value="">Please Select</option>'.PHP_EOL; 

 while($sqlrow = mysql_fetch_array($sqlresult))
 {
     $building = $sqlrow['Building'];
     $buildingHTML .= "<option value='".$building."'>" . $building . "</option>".PHP_EOL; 
  }

  $buildingHTML .= '</select>';
  $buildingHTML .= '</form>';

  $roomHTML = ""; 
  $roomHTML .= '<select name="rooms" id="roomsDrop">'.PHP_EOL;
  $roomHTML .= '<option value="">Please Select</option>'.PHP_EOL; 
  $roomHTML .= '</select>';

  echo $buildingHTML; 
  echo $roomHTML;

您需要为动态人口添加脚本。

JS CODE：

function getRooms() {
    var building = jQuery("#buildingssDrop").val();
    jQuery('#roomsDrop').empty();
    jQuery('#roomsDrop').html('<option value="">Please Select</option>');
    jQuery.ajax({
          type: "post",
          url:  "rooms.php",
          data: "building="+building,
          success: function(response){
              jQuery('#roomsDrop').append(response);
          }
        });
 }

这里的rooms.php是您需要为动态数据填充调用的ajax文件：

rooms.php 

$building = isset($_POST['building']) ? $_POST['building'] : '';
$sql = "SELECT Room FROM Room WHERE Building ='".$building."'";

$sqlresult = mysql_query($sql);


$roomHTML  = ""; 

while($sqlrow = mysql_fetch_array($sqlresult))
{
     $room = $sqlrow['Room'];
     $roomHTML .= "<option value='".$room."'>" . $room . "</option>".PHP_EOL; 
}


echo $roomHTML;

希望这会对你有所帮助

答案 1 :(得分：0)

您确定已设置建筑物的后期价值。

因为你告诉sql输出没有显示构建参数的值，所以肯定$ building将为null。您需要将post值传递给$ building

您需要像这样更改代码：

if(isset($_POST['buildings'])){
    $building = $_POST['buildings'];
} else {
   $building = '';
}

$sql="SELECT Building, Room FROM Room WHERE Building = '".$building."'";

$sqlresult = mysql_query($sql);

$buildings = array(); // easier if you don't use generic names for data

while($sqlrow = mysql_fetch_array($sqlresult))
{
    // you need to initialise your building array cells
    if (!isset($buildings[$sqlrow['Building']])) {
        $buildings[$sqlrow['Building']] = array('Rooms' => array());
    }

    // you can add the room to the building 'Rooms' array
    $buildings[$sqlrow['Building']]['Rooms'][] = $sqlrow['Room'];
}


$buildingHTML = ""; 
$buildingHTML = "<form action=\"\" method=\"post\">";
$buildingHTML .= '<select name="buildings" id="buildingssDrop" onchange="document.getElementById(\'dropDownForm\').submit()">'.PHP_EOL;
$buildingHTML .= '<option value="">Please Select</option>'.PHP_EOL; 

foreach ($buildings as $building => $buildingData) {      
    $buildingHTML .= "<option value='".$building."'>" . $building . "</option>".PHP_EOL;        
}
$buildingHTML .= '</select>';
$buildingHTML .= '</form>';

$roomHTML = ""; 
$roomHTML .= '<select name="rooms" id="roomsDrop">'.PHP_EOL;
$roomHTML .= '<option value="">Please Select</option>'.PHP_EOL; 

if(isset($_POST['buildings'])){
    $buildingname = $_POST['buildings'];
     foreach ($buildings[$buildingname]['Rooms'] as $roomId => $roomData) {        
           $roomHTML .= "<option value='".$roomId."'>" . $roomId . "</option>".PHP_EOL;        
     }
}
$roomHTML .= '</select>';


echo $buildingHTML;

echo $roomHTML;
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