Question

好的，所以我现在已经在这几个小时了，不管我做什么，它都行不通。我已经阅读了文档页面，甚至通过示例逐字尝试，但仍然不起作用：

$db = mysqli_connect($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，我根本不知道如何修复它：

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，给出了什么？

var_dump($return)返回NULL值

我该如何解决这个问题？我只想在$return变量中返回1行，以便我可以用它做一些事，arggg！

Answer 1

您正在选择6列但您只在$return

中绑定mysqli_stmt_bind_result($stmt, $return);

尝试

mysqli_stmt_bind_result($stmt, $artid, $title, $body, $tagid, $vis, $tagtitle );

如何使用mysqli绑定此1变量？

1 个答案: