我在wordpress工作。您将看到两个查询，一个是select查询，另一个是insert查询。我希望在运行插入查询后，在页面上获得更新的结果。

选择查询有一个带有提交按钮的表单标记。。。提交时，它将拾取id或当前获取的行，并在另一个表中以相同的id执行插入。但在此之后，单击“提交我的页面”不会显示数据库中的更新值或插入值。我希望你们能在这方面提出建议。。。谢谢

wordpress页面模板中的PHP代码

        $sql = "SELECT 1user.uid, 1user.username, 1user.competition, 1user.path, Sum(votes.votes) AS votessum FROM 1user LEFT JOIN votes on 1user.uid=votes.uid GROUP BY 1user.username, 1user.competition";

$results = $wpdb->get_results($sql) or die(mysql_error());

foreach( $results as $result ) {
echo \'<form action="" method="post">\';
echo "<img src=\'$result->path\' width=\'150\' height=\'150\' >" . \'<br><br>\';
echo "<input name=\'id\' type=\'hidden\' value=\'$result->uid\'>";
echo "<input name=\'comp\' type=\'hidden\' value=\'$result->competition\'>";
echo $result->username.\'<br>\';

echo $result->votessum.\'<br>\';
echo "<input style=\'margin-bottom:30px;\' value=\'vote\' name=\'submit\' type=\'submit\'/></form>";    

}

if(isset($_POST[\'submit\'])){
global $wpdb;
$votes = 1;
$competition = $_POST[\'comp\'];
$uid = $_POST[\'id\'];
//$uid = get_current_user_id(); 

echo \'id of image = \'.$_POST[\'id\'];
echo \'<br>\'.\'competition is\'.$_POST[\'comp\'];
if($wpdb->insert(
        \'votes\',
        array(
                \'votes\' => $votes,
                \'competition\' => $competition,
                \'uid\' => $uid


            )
) == false) wp_die(\'Database Insertion failed\'); else echo \'Database insertion successful<p />\';


}