2

好的，所以我已经在这几个小时了，无论我做什么，它都不起作用。我已经阅读了几页文档，甚至逐字逐句地尝试了示例，但仍然无法正常工作：

$db = mysqli_connect($db_server, $db_user, $db_passwd, $db_name);

if (mysqli_connect_errno())
{
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$db_query = "SELECT ca.id_article, ca.title, ca.body, ca.id_tag, ca.visible, ct.title AS tag_title
            FROM cdfi_articles AS ca
            INNER JOIN cdfi_tags AS ct ON(ct.id_tag = ca.id_tag)
            WHERE ca.id_article = ?
            LIMIT 1";
;
if ($stmt = mysqli_prepare($db, $db_query)) {
    mysqli_stmt_bind_param($stmt, "i", $_REQUEST['article']);

    mysqli_stmt_execute($stmt);
    mysqli_stmt_bind_result($stmt, $return);
    mysqli_stmt_fetch($stmt);

    var_dump($return);

}

mysqli_close($db);

发生以下错误，我完全不知道如何解决：

Warning: mysqli_stmt_bind_result() [function.mysqli-stmt-bind-result]: Number of bind variables doesn't match number of fields in prepared statement in ... filepath on line {line number}

bind variables不匹配？我只使用 1 个变量，查询中只有 1 个问号，什么给出？

var_dump($return)返回NULL值

我怎样才能解决这个问题？我只想在$return变量中返回那 1 行，以便我可以用它做点什么，arggg！

4

1 回答 1

3

$return您正在选择 6 列，但您只绑定mysqli_stmt_bind_result($stmt, $return);

尝试

mysqli_stmt_bind_result($stmt, $artid, $title, $body, $tagid, $vis, $tagtitle );
于 2013-09-16T12:22:52.467 回答