好的,所以我在这里检查了网和其他问题,我很难过。我从这里发布的问题尝试了一个javascript解决方案,但我认为它不喜欢填充<option>
的MySQL。我会复制我所拥有的所有代码,包括我的javascript。
SCRIPT:
<script>
$(function() {
$('#groups').on('change', function() {
var val = $(this).val();
var sub = $('#sub_groups');
$('option', sub).filter(function() {
if (
$(this).attr('data-group') === val || $(this).attr('data-group') === 'SHOW'
) {
$(this).show();
} else {
$(this).hide();
}
});
});
$('#groups').trigger('change');
});
</script>
PHP第一次下拉:
<select class="form-control" id="groups">
<?php
$sql = "SELECT BoilerBrand FROM boilerbrands";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['ID']."'>".$row['BoilerBrand']."</option>";
}
?>
</select>
PHP第二次下拉
<select class="form-control" id="sub_groups">
<option data-group='SHOW' value="0">Model</option>
<?php
$sql = "SELECT * FROM boilermodels";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
echo "<option data-group='".$row['BoilerBrand']."' value='".$row['BoilerGC']."'>".$row['BoilerModel']."</option>";
}
?>
</select>
非常感谢任何帮助!
谢谢:)
答案 0 :(得分:1)
我通常这样做的方式是隐藏/显示我删除/添加它们的选项。我相信如果你隐藏option
s,那么select
输入仍然可以具有该值。
SCRIPT:
<script>
$(function(){
<?php
$sql = "SELECT * FROM boilermodels";
$result = mysql_query($sql);
$models = array();
while ($row = mysql_fetch_array($result)) {
$models[$row['BoilerBrand']][] = $row;
}
/* should look like
$models = [];
$models[1][] = ['ModelID'=>'1','BoilerBrand'=>'1','BoilerModel'=>'240E','BoilerGC'=>'47-777-77','BoilerImage'=>'47-777-77.jpg' ];
$models[1][] = ['ModelID'=>'3','BoilerBrand'=>'1','BoilerModel'=>'290D','BoilerGC'=>'11-111-11','BoilerImage'=>'11-111-11.jpg' ];
$models[2][]= ['ModelID'=>'2','BoilerBrand'=>'2','BoilerModel'=>'250E','BoilerGC'=>'47-777-77','BoilerImage'=>'47-777-77.jpg' ];
*/
?>
var _boilermodels = '<?php echo json_encode($models); ?>';
var jsonBoilerModels = JSON.parse(_boilermodels);
console.log(jsonBoilerModels);
$('#groups').on('change', function(){
var $this = $(this);
var val = $this.val();
var sub = $('#sub_groups');
sub.find('option').remove();
var appendList = [];
$.each(jsonBoilerModels[val],function(key,value){
appendList.push('<option value="'.concat(value['BoilerGC'], '">', value['BoilerModel'], '</option>'));
});
sub.append(appendList);
});
$('#groups').trigger('change');
});
</script>
第一次下拉:
<select class="form-control" id="groups">
<?php
$sql = "SELECT ID ,BoilerBrand FROM boilerbrands";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result)) {
echo "<option value='".$row['ID']."'>".$row['BoilerBrand']."</option>";
}
?>
</select>
第二次下拉:
<select class="form-control" id="sub_groups">
<option value="">Select A Model</option>
</select>