Как добавить третий снимок этого кода. Я не смог понять это. Благодаря!!!
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html> <head> <script type="text/javascript" src="jqueryui/js/jquery-ui-1.8.15.custom.min.js"></script> <script type="text/javascript" src="jqueryui/js/jquery-1.6.2.min.js"></script> <script type="text/javascript"> $(document).ready(function(){ $("select#category").change(function(){ var id = $("select#category option:selected").attr('value'); $.post("select_type.php", {id:id}, function(data){ $("select#type").html(data); }); }); }); </script> </head> <body> <?php include("select.class.php"); ?> <form id="select_form"> Choose a category:<br /> <select id="category"> <?php echo $opt->ShowCategory(); ?> </select> <br /><br /> choose a type:<br /> <select id="type"> <option value="0">choose...</option> </select> <br /><br /> choose third drop down: <br /> <select id="third_table"> <option value="0">choose...</option> </select> <br /><br /> <input type="submit" value="confirm" /> </form> <div id="result"></div> </body> </html> ************************************************************************************************* Это класс, я уже добавил код, который запрашивает третью таблицу базы данных
<?php class SelectList { protected $conn; public function __construct() { $this->DbConnect(); } protected function DbConnect() { include "db_config.php"; $this->conn = mysql_connect($host,$user,$password) OR die("Unable to connect to the database"); mysql_select_db($db,$this->conn) OR die("can not select the database $db"); return TRUE; } public function ShowCategory() { $sql = "SELECT * FROM category"; $res = mysql_query($sql,$this->conn); $category = '<option value="0">choose...</option>'; while($row = mysql_fetch_array($res)) { $category .= '<option value="' . $row['id_cat'] . '">' . $row['name'] . '</option>'; } return $category; } public function ShowType() { $sql = "SELECT * FROM type WHERE id_cat=$_POST[id]"; $res = mysql_query($sql,$this->conn); $type = '<option value="0">choose...</option>'; while($row = mysql_fetch_array($res)) { $type .= '<option value="' . $row['id_type'] . '">' . $row['name'] . '</option>'; } return $type; } public function ShowThird() { $sql = "SELECT * FROM thirdtable WHERE id_type=$_POST[id2]"; $res = mysql_query($sql,$this->conn); $third = '<option value="0">choose...</option>'; while($row = mysql_fetch_array($res)) { $third .= '<option value="' . $row['id_third'] . '">' . $row['name'] . '</option>'; } } } $opt = new SelectList(); ?> ************************************************************************************************* select_type.php <?php include "select.class.php"; echo $opt->ShowType(); ?>
Ваш jquery изменился на:
<script type="text/javascript"> $(document).ready(function(){ $("select#category").change(function(){ var id = $("select#category option:selected").attr('value'); $.post("select_type.php?event=type", {id:id}, function(data){ $("select#type").html(data); }); }); $("select#type").change(function(){ var id = $("select#type option:selected").attr('value'); $.post("select_type.php?event=third_table", {id:id}, function(data){ $("select#third_table").html(data); }); }); }); </script>
И конец вашего файла select_type.php изменился на:
...... <?php include "select.class.php"; switch ($_GET['event']) { case 'third_table': echo $opt->ShowThird(); break; default: echo $opt->ShowType(); break; } ?>
Надеюсь, это поможет вам
HTML и PHP
<h4>Search By Regions</h4> <select id="country" name="country"> <option value="">--</option> <? $result1 = mysql_query("SELECT * FROM country ORDER BY name"); while($row1 = mysql_fetch_array($result1)) { echo '<option value="'.$row1['name'].'">'.$row1['name'].'</option>'; } ?> </select> <select id="oblast" name="oblast"> <option value="">--</option> <? $result2 = mysql_query("SELECT oblast.name, country.name AS myCountry FROM oblast,country WHERE oblast.country_id = country.id ORDER BY oblast.name"); while($row2 = mysql_fetch_array($result2)) { echo '<option value="'.$row2[0].'" class="'.$row2['myCountry'].'">'.$row2[0].'</option>'; } ?> </select> <select id="gorod" name="gorod"> <option value="">--</option> <? $result3 = mysql_query("SELECT gorod.name, oblast.name AS myOblast FROM gorod,oblast WHERE gorod.region_id = oblast.id ORDER BY gorod.name"); while($row3 = mysql_fetch_array($result3)) { echo '<option value="'.$row3[0].'" class="'.$row3['myOblast'].'">'.$row3[0].'</option>'; } ?> </select>
И не забудьте включить это
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> <script src="/template/js/jquery.chained.min.js"></script> <script> $(document).ready(function() { $("#oblast").chained("#country"); $("#gorod").chained("#oblast"); }); </script>