如何解决如何根据另一个php下拉列表中的选择自动填充drodown?
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">×</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);
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