小编pro*_*_24的帖子

我有两个组合框,当另一个组合框数据被更改时,我需要填充相同的信息.

但是,当我使用.append()方法时,很明显它只执行最新的追加调用.这是代码:

    $('.sc-days').change(function () {
        var scedo = $('.sc-edo');
        var sclpo = $('.sc-lpo');
        var selected = $(".sc-days option:selected");
        scedo.append(selected);
        sclpo.append(selected);
    });

运行此选项时,只会填充"sclpo".你可以在这里看到结果.http://jsfiddle.net/DF42s/

我试图从我的mysqli结果中构建一个json对象.我该怎么做呢 目前它不会创建一个json看对象.

这是我的代码:

$result = $dataConnection->prepare("SELECT id, artist, COUNT(artist) AS cnt FROM {$databasePrefix}users GROUP BY artist ORDER BY cnt DESC LIMIT 0 , 30");
$result->execute();
if($result->error)
{
die("That didn't work. I get this: " . $result->error);
}
$result->bind_result($id, $artist, $count);
$data = array();
while($result->fetch()){
$data[] = '{ id :'.$id.', artist :'.$artist.', count :'.$count.'}';
}
echo json_encode($data);
$dataConnection->close();

我想要一个数据对象,如:

{"id":"27","artist":"myArtist","count":"29"},....