如何根据另一个php下拉列表中的选择自动填充drodown？

问题描述

我已经阅读了一些类似的问题，但是没有找到解决方案。我正在尝试根据另一个下拉菜单的选择获取一个下拉菜单。第一个下拉列表是学校名称，选择后应该会获取该特定学校下的用户。

数据库中有两个表。第一个具有名为“学校名称”的“学校名称”列，另一个名为“人”的表具有一个名为“学校”的列，该列是外键，也具有名字和姓氏，应在选择时获取。

我正在参考本教程（https://makitweb.com/how-to-autopopulate-dropdown-with-ajax-pdo-and-php/）

我尝试了以下代码：

queries.PHP

class Queries {
public static function getSchool() {   
    $dbUser = "xxx";
    $dbPass = "xxxx";
    $dbConn = "(DESCRIPTION = (ADDRESS = (PROTOCOL=TCP)(HOST=xxxx)(PORT=1521))(CONNECT_DATA=(SID=xxxx)))";
    $conn = oci_connect($dbUser,$dbPass,$dbConn);
    $sql = oci_parse($conn,"SELECT  a.School,a.SchoolName FROM SchoolName a WHERE a.Formdisplay = 'Y' ORDER BY a.SchoolName");
        return $sql;   
}  
}

form.PHP

<?PHP
require_once './functions/queries.PHP';
$getschool = new Queries();
?>

<div class="container">
<div class="panel panel-default">
    <div class="panel-body">
        <div id="addroles" class="hide" role="alert">
            <button type="button" class="close" data-dismiss="alert" aria-label="Close"><span aria-hidden="true">&times;</span></button>
            <div id="resultRoleContent"></div>
        </div>  
             <form class="cmxform" action ='functions/processform.PHP' id="Form1" method="post">
           <legend> Form</legend>
            <label for="addname">Please Select School</label>
           <select class="form-control" name="school" id="school">
                <?PHP
                $nameslist = $getschool->getSchool();
                oci_execute($nameslist,OCI_DEFAULT);
                while ($row = oci_fetch_array($nameslist,OCI_ASSOC+OCI_RETURN_NULLS)) {
                    echo '<option value="' . $row['SCHOOLNAME'] . '">' . $row['SCHOOLNAME']. '</option>';
                }
                ?>
           </select>
            <label for="addname">Please Select Name</label>
           <select class="form-control" name="names" id="names">
                <?PHP
                
              
                ?>
           </select>
        </form>

    </div>
   </div>       
</div>

<script>
$(document).ready(function(){
 $('#school').change(function(){
 var schoolname = $(this).val();
 $('#names').find('option').not(':first').remove();

 // AJAX request
 $.ajax({
   url: 'getUsers.PHP',type: 'post',data: {request: 1,primaryschool: schoolpropername},dataType: 'json',success: function(response){
     var len = response.length;
     for( var i = 0; i<len; i++){
       var id = response[i]['primaryschool'];
       var firstname = response[i]['firstName'];
       var lastname = response[i]['lastName'];

       $("#names").append("<option value='"+id+"'>"+firstname+"</option>");

     }
   }
 });

  });
  });
</script>

getUsers.PHP

<?PHP
$dbUser = "xxxx";
$dbPass = "xxxx";
$dbConn = "(DESCRIPTION = (ADDRESS = (PROTOCOL=TCP)(HOST=xxxx)(PORT=1521))(CONNECT_DATA=(SID=xxxx)))";
$conn = oci_connect($dbUser,$dbConn);
$request = 0;
if(isset($_POST['request'])){
   $request = $_POST['request'];
}
if($request == 1){
   $schoolname = $_POST['school'];
   $sql =oci_parse($conn,"SELECT * FROM person ");
   oci_bind_by_name($sql,':primaryschool',$schoolname);
   $result = oci_execute($sql);
   $response = array();
   foreach($result as $row){
       $response[] = array(
    "firstname" => $row['firstname']
      );
   }
   echo json_encode($response);
    exit;
 }

在上面的代码中，我收到以下三个错误：

PHP注意：未定义索引：getusers.PHP中的学校 2.PHP警告：oci_bind_by_name（）：ORA-01036：getusers.PHP中的非法变量名称/编号 3.PHP警告：oci_bind_by_name（）：ORA-01036：getusers.PHP中的非法变量名称/编号

解决方法

发生未定义错误，因为学校价值没有进入$_POST数组。打印$ _post数组，并检查要获取哪个school值的索引。
您没有在查询的WHERE子句中提及列名来获取特定学校的详细信息。因此，您将收到该警告。您的查询应如下所示。

假设表中的列名称为schoolname。

$sql =oci_parse($conn,"SELECT * FROM person WHERE  schoolname = :primaryschool");
oci_bind_by_name($sql,':primaryschool',$schoolname);