Question

伙计们我试图在这里实际创建像facebook这样的Feed，但是我得到了这个错误，我无法弄清楚我哪里出错了。这是代码：

$sql3="select u.update_id, u.update_body,u.account_name,u.os_id,u.author,u.time,u.title,"
            . "c.comment_body, c.os_id,c.author,c.time"
            . "from updates as u, comment_update as c "
            . "where c.os_id=u.update_id and u.account_name = ':session' and u.type in ('a','c') and u.account_name=':friend' and u.type =('a'|'c') order by u.time asc,c.time desc";
     $stmth=$conn->prepare($sql3);
   $stmth->execute(array(":session"=>$_SESSION['uname'],":friend"=>$friend));
    $status_reply= $stmth->fetchAll(PDO::FETCH_ASSOC);

这是错误代码： -

致命错误：带有消息'SQLSTATE [42000]的未捕获异常'PDOException'：语法错误或访问冲突：1064 SQL语法中有错误;查看与您的MySQL服务器版本对应的手册，以便在'/ u，comment_update附近使用正确的语法，其中c.os_id = u.update_id和u.account_name ='：sessi'在/ opt / lampp的第1行' /htdocs/project-chg/status&comments.php:28堆栈跟踪：＃0 /opt/lampp/htdocs/project-chg/status&comments.php(28）：PDOStatement-＆gt; execute（Array）＃1 / opt / lampp / htdocs / project-chg / example1.php（30）：include（'/ opt / lampp / htdo ...'）＃opt {lamp}抛出/ opt / lampp / htdocs / project-chg / status＆amp; ;评论.php在第28行

任何帮助将不胜感激。

Answer 1

您的代码：

$sql3="select u.update_id, u.update_body,u.account_name,u.os_id,u.author,u.time,u.title,"
            . "c.comment_body, c.os_id,c.author,c.time"
            . "from updates as u, comment_update as c "
            . "where c.os_id=u.update_id and u.account_name = ':session' and u.type in ('a','c') and u.account_name=':friend' and u.type =('a'|'c') order by u.time asc,c.time desc";

在
将('a'|'c')更改为('a','c')
考虑使用逗号语法的JOIN语法instad
您可能需要移除'周围的:friend和:session
整个WHERE clause conditions对我来说很奇怪u.account_name = :session and u.account_name=:friend同一列??? +加倍u.type条件

结果：

$sql3="select u.update_id, u.update_body,u.account_name,u.os_id,u.author,u.time,u.title,"
            . "c.comment_body, c.os_id,c.author,c.time"
            . " from updates as u JOIN comment_update as c ON c.os_id=u.update_id "
            . "where u.account_name = :session and u.type in ('a','c') and u.account_name=:friend and u.type =('a','c') order by u.time asc,c.time desc";

如何纠正“SQLSTATE [42000]：语法错误或访问冲突：1064”错误

1 个答案: