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我尝试使用Ajax/PHP为我的页面加载动态加载的选择菜单。但jQuery UI插件可以防止加载动态加载的数据。所以当我改变第一个选择菜单时我什么都看不到。jquery UI.js可以阻止加载动态加载的选择菜单
我的代码是这样的。
<script>
$(document).ready(function($) {
var list_target_id = 'list-target'; //first select list ID
var list_select_id = 'list-select'; //second select list ID
var initial_target_html = '<option value="">-Select-</option>'; //Initial prompt for target select
$('#'+list_target_id).html(initial_target_html); //Give the target select the prompt option
$('#'+list_select_id).change(function(e) {
//Grab the chosen value on first select list change
var selectvalue = $(this).val();
//Display 'loading' status in the target select list
$('#'+list_target_id).html('<option value="">Loading...</option>');
if (selectvalue == "") {
//Display initial prompt in target select if blank value selected
$('#'+list_target_id).html(initial_target_html);
} else {
//Make AJAX request, using the selected value as the GET
$.ajax({url: 'loadcity.php?svalue='+selectvalue,
success: function(output) {
//alert(output);
$('#'+list_target_id).html(output);
},
error: function (xhr, ajaxOptions, thrownError) {
alert(xhr.status + " "+ thrownError);
}});
}
});
});
</script>
<form method="post">
<div class="select-country">
<label>District</label>
<select name="list-select" id="list-select">
<option value="">Please select..</option>
<?php
$sel_dis2 = mysql_query("SELECT * FROM district", $connection);
confirm_query($sel_dis2);
while($dis2 = mysql_fetch_assoc($sel_dis2)){
?>
<option value="<?php echo $dis2["id_district"]; ?>"><?php echo $dis2["district"]; ?></option>
<?php
}
?>
</select>
</div>
<div class="select-state">
<label>City</label>
<select name="list-target" id="list-target"></select>
</div>
</form>
我测试了没有包含jQuery UI的代码,它工作正常。但是我想为这个页面添加jQuery UI。这是php文件
<?php
$connection = mysqli_connect("localhost", "user", "password", "database");
$selectvalue = mysqli_real_escape_string($connection, $_GET['svalue']);
mysqli_select_db($connection, "database");
$result = mysqli_query($connection, "SELECT city.id_city, city.city FROM city WHERE city.district_id = '$selectvalue'");
echo '<option value="">-Select-</option>';
while($row = mysqli_fetch_array($result))
{
echo '<option value="'.$row['id_city'].'">' . $row['city'] . "</option>";
//echo $row['drink_type'] ."<br/>";
}
mysqli_free_result($result);
mysqli_close($connection);
?>
我没加loadcity。 PHP文件在这里。它适用于php和mysql数据库。但是当我包含jQuery UI时,它不起作用。 –
这些示例同时使用jQuery和jQuery UI。他们是否应该做更多的事情? – junkfoodjunkie
是的,我想使用两者。他们需要使用该网站的用户界面。 –