2
こんにちは私のデータテーブルで助けが必要です。私の問題は、ここで見ているように、選択したドロップダウンでschoolyear_idの値を持つ卒業生だけに基づいてデータを表示するようにデータテーブルに設定することです。Codeigniter：Select ID値に基づいてデータテーブルを変更する
今のところ、これはドロップダウンで私の変数です。
<select class="form-control" style="width:115px" name="schoolyearData" id="schoolyearData">
<?php foreach ($schoolyearData as $schoolyear)
{
?>
<option value="<?php echo $schoolyear['schoolyear_id']; ?>"> <?php echo $schoolyear['schoolyear_date']; ?> </option>
<?php
}
?>
</select>
さて、これはDataTableのとAjaxのための私のカスタムJSです。
var manageRegistrationTable;
$(document).ready(function() {
manageRegistrationTable = $("#manageRegistrationTable").DataTable({
'ajax' : 'registration/fetchRegistrationData',
'order' : []
});
});
これは、Ajaxによるアクセスになります私のコントローラ機能です。今モデルの
public function fetchRegistrationData($registrationId = null)
{
if($registrationId) {
$registrationData = $this->model_registration->fetchRegistrationData($registrationId);
echo json_encode($registrationData); // USED FOR EDITING
}
else {
$registrationData = $this->model_registration->fetchRegistrationData();
$result = array('data' => array());
$x = 1;
foreach ($registrationData as $key => $value) {
$button = '<!-- Single button -->
<div class="btn-group">
<button type="button" class="btn btn-default dropdown-toggle" data-toggle="dropdown" aria-haspopup="true" aria-expanded="false">
Action <span class="caret"></span>
</button>
<ul class="dropdown-menu">
<li><a type="button" class="btn btn-default" style="text-align:left;" data-toggle="modal" data-target="#viewRegistraionModal" onclick="viewRegistration('.$value['registration_id'].')"> <i class="glyphicon glyphicon-user"></i> View</a></li>
<li><a type="button" class="btn btn-default" style="text-align:left;" data-toggle="modal" data-target="#updateRegistrationModal" onclick="editRegistration('.$value['registration_id'].')"> <i class="glyphicon glyphicon-edit"></i> Edit</a></li>
<li><a type="button" class="btn btn-default" style="text-align:left;" data-toggle="modal" data-target="#removeRegistrationModal" onclick="removeRegistration('.$value['registration_id'].')"> <i class="glyphicon glyphicon-trash"></i> Remove</a></li>
</ul>
</div>';
$photo = ' <img src="../'.$value['student_image'].'" alt="Photo" class="img-circle candidate-photo" width="95" height="95"/>';
$result['data'][$key] = array(
$x,
$photo,
$value['student_lastname'],
$value['student_firstname'],
$value['student_middlename'],
$value['schoolyear_date'],
$value['gradelevel_name'],
$value['section_name'],
$value['registration_dateadded'],
$button
);
$x++;
} // /froeach
echo json_encode($result);
} // /else
}
。
public function fetchRegistrationData($registrationId = null)
{
if($registrationId) {
$sql = "SELECT * FROM tbl_registration
INNER JOIN tbl_student ON tbl_student.student_id = tbl_registration.student_id
INNER JOIN tbl_schoolyear ON tbl_schoolyear.schoolyear_id = tbl_registration.schoolyear_id
INNER JOIN tbl_gradelevel ON tbl_gradelevel.gradelevel_id = tbl_registration.gradelevel_id
INNER JOIN tbl_section ON tbl_section.section_id = tbl_registration.section_id WHERE registration_id = ?";
$query = $this->db->query($sql, array($registrationId));
return $query->row_array();
// USED FOR EDIT
}
else {
$sql = "SELECT * FROM tbl_registration
INNER JOIN tbl_student ON tbl_student.student_id = tbl_registration.student_id
INNER JOIN tbl_schoolyear ON tbl_schoolyear.schoolyear_id = tbl_registration.schoolyear_id
INNER JOIN tbl_gradelevel ON tbl_gradelevel.gradelevel_id = tbl_registration.gradelevel_id
INNER JOIN tbl_section ON tbl_section.section_id = tbl_registration.section_id ORDER BY schoolyear_date DESC";
$query = $this->db->query($sql);
return $query->result_array();
}
}
私は、これはのonchange関数または何を必要としていることを考えていますか？ そして、どのように私はschoolyear_idの値をデータ型ajaxに渡しますか。