我试图根据其他下拉列表的值来填充“Reigons”下拉列表,这是使用php和javascript的“城市”,代码确实工作时将附加区域名称和ID添加到select标签中它返回 尽管它们存在于数据库中! 注意:SQL查询没有问题,因为当我在wampserver
上执行它时,它可以正常工作这是我的DB(HWP):
CREATE TABLE `cities` (
`city_id` int(5) NOT NULL auto_increment,
`city_name` varchar(50) NOT NULL,
PRIMARY KEY (`city_id`),
UNIQUE KEY `city_name` (`city_name`));
CREATE TABLE `reigons` (
`city_id` int(5) NOT NULL,
`reigon_id` int(5) NOT NULL auto_increment,
`reigon_name` varchar(50) NOT NULL,
PRIMARY KEY (`reigon_id`),
UNIQUE KEY `reigons_name` (`reigon_name`));
INSERT INTO `cities` (`city_id`, `city_name`) VALUES
(3, 'ALJAZERA'),
(1, 'Khartoum'),
(2, 'RED SEA');
INSERT INTO `reigons` (`city_id`, `reigon_id`, `reigon_name`) VALUES
(1, 1, 'Khartoum'),(1, 2, 'Bahri'),(1, 3, 'Um dorman'),
(2, 4, 'Read sea North'),(2, 5, 'Read sea South'),(3, 6, 'Baant'),
(3, 7, 'Soriba');
Home.php:
<?php
$con=mysql_connect("localhost", "root", "");
$db=mysql_select_db("hwp",$con);
$query2 = mysql_query("select * from cities;") or die (mysql_error());?>
<select name="city" id="city" onchange="get_reg(this.value)">
<option value="defult"> -Select City- </option>
<?php
while ($f2=mysql_fetch_array($query2)) {
echo ('<option value="'.$f2['city_id'].'">'. $f2['city_name'].'</option>');}
?>
</select>
<script type="text/javascript">
function get_reg(city)
{
$.ajax({
type: "POST",
url: 'getreg.php',
data: city,
success: function(html) {
$("#reigon").html(html);
}});
}
</script>
<select name="reigon" id="reigon">
<option value="defult"> -Select Reig- </option>
</select>
getreg.php:
<?php
mysql_connect("localhost","root","");
mysql_select_db("hwp");
$city= mysql_real_escape_string($_POST["city"]);
$query1 = mysql_query("select * from reigons where city_id= '$city'") or die(mysql_error());
while ($f1 = mysql_fetch_array($query1))
{
echo '<option value="', $f1['reigon_id'] ,'">', $f1['reigon_name'],'</option>';
}
?>
请帮助。