Question

这是我的问题：

SELECT qa.id,
       qa.subject,
       qa.category cat,
       qa.keywords tags,
       qa.body_html,
       qa.amount,
       qa.author_id author,
       qa.visibility,
       qa.date_time,
       COALESCE(u.reputation, 'N') reputation,
       COALESCE(CONCAT(u.user_fname, ' ', u.user_lname), 'ناشناس') name,
       COALESCE(u.avatar, 'anonymous.png') avatar,

  (SELECT COALESCE(sum(vv.value),0)
   FROM votes vv
   WHERE qa.id = vv.post_id
     AND 15 = vv.table_code) AS total_votes,

  (SELECT COALESCE(sum(vt.total_viewed),0)
   FROM viewed_total vt
   WHERE qa.id = vt.post_id
     AND 15 = vt.table_code
   LIMIT 1) AS total_viewed
FROM qanda qa
INNER JOIN qanda_tags qt ON qt.qanda_id = qa.id
INNER JOIN tags ON tags.id = qt.tag_id
LEFT JOIN users u ON qa.author_id = u.id
AND qa.visibility = 1
WHERE qa.type = 0 tags.name = :t
ORDER BY qa.date_time DESC
LIMIT :j,
      11;

它抛出了这个错误：

致命错误：未捕获PDOException：SQLSTATE [42000]：语法错误或访问冲突：1064您的SQL语法有错误;检查与您的MariaDB服务器版本对应的手册在＆＃39; tags.name =？附近使用的语法？订购qa.date_time desc LIMIT ？，11＆＃39;在C：\ xampp \ htdocs \ myweb \ others \ questions.php：第18行堆栈跟踪：＃0 C：\ xampp \ htdocs \ myweb \ others \ questions.php（136）： PDO->准备（＆＃39; SELECT qa.id，q ...＆＃39;）＃1 C：\ XAMPP \ htdocs中\ MyWeb即可\人\ questions.php（293）：问题＆gt;索引（＆＃39; tags.name =：t1＆＃39;，＆＃39; \ tINNER加入qan ......＆＃39;）＃2 C：\ xampp \ htdocs \ myweb \ application \ other.php（21）：questions-＆gt; tagged（）＃3 C：\ xampp \ htdocs \ myweb \ index.php（149）：require_once（＆＃39; C：\ xampp \ htdocs ...＆＃39;）＃4 {main}引入第136行的C：\ xampp \ htdocs \ myweb \ others \ questions.php

我真的不知道问题是什么。错误消息显示其附近tags.name = ?，但它对我来说完全正确。

有人问题是什么？

Answer 1

替换

WHERE qa.type = 0 tags.name = :t

与

WHERE qa.type = 0 AND tags.name = :t

语法错误或访问冲突

1 个答案: