PHP＆MySQL - 你的SQL語法問題有錯誤

Question

我在下面得到這個錯誤，想知道它是什麼？我能做些什麼來解決它？

 
You have an error in your SQL syntax; check the manual that corresponds 
to your MySQL server version for the right syntax to use near 
    ') AND articles_comments.user_id IN()' 
at line 6 

這是我的MySQL代碼。

$u_id2 = array(); 
$dbc = mysqli_query($mysqli,"SELECT users.user_id FROM users WHERE users.active IS NOT NULL OR users.deletion = 1"); 
if(!$dbc){ 
    print mysqli_error($mysqli); 
} else { 
    while($row = mysqli_fetch_assoc($dbc)) { 
     $u_id2[] = $row['user_id']; 
    } 
} 

$bad_com_id2 = array(); 
$dbc = mysqli_query($mysqli,"SELECT articles_comments.comment_id 
          FROM articles_comments 
          WHERE articles_comments.user_id IN(" . implode(",", $u_id2) . ") 
          AND articles_comments.parent_comment_id = 0"); 
if(!$dbc){ 
    print mysqli_error($mysqli); 
} else { 
    while($row = mysqli_fetch_assoc($dbc)) { 
     $bad_com_id2[] = $row['comment_id' ]; 
    } 
} 

$com_id2 = array(); 
$ucom_id2 = array(); 
$dbc = mysqli_query($mysqli,"SELECT articles_comments.comment_id, articles_comments.user_id FROM articles_comments WHERE articles_comments.article_id = '" . $article_id . "'"); 
if(!$dbc){ 
    print mysqli_error($mysqli); 
} else { 
    while($row = mysqli_fetch_assoc($dbc)) { 
     $com_id2[] = $row['comment_id' ]; 
     $ucom_id2[] = $row['user_id']; 
    } 
} 

$comments_to_be_displayed2 = array_diff($ucom_id2, $u_id2); 

$new_com_id2 = array(); 
$dbc = mysqli_query($mysqli,"SELECT articles_comments.comment_id 
          FROM articles_comments 
          INNER JOIN users ON articles_comments.user_id = users.user_id 
          INNER JOIN users_articles ON articles_comments.article_id = users_articles.id 
          WHERE articles_comments.article_id = '" . $article_id . "' 
          AND articles_comments.comment_id IN(" . implode(",", $com_id2) . ") 
          AND articles_comments.user_id IN(" . implode(",", $comments_to_be_displayed2) . ")"); 
if(!$dbc){ 
    print mysqli_error($mysqli); 
} else { 
    while($row = mysqli_fetch_assoc($dbc)) { 
     $new_com_id2[] = $row['comment_id' ]; 
    } 
} 

Answer 1

IN()是你的問題。你需要有價值觀。

你既可以測試是否有價值觀和排除PHP的情況，當你生成的查詢字符串或使用一招：

IN(XYZ, " . implode(",", $u_id2) . ")

，其中XYZ是，你一定會永遠的值用於user_id（如果該列是標準自動增量主鍵，則可能爲-1）。

看到這裏供參考：http://dev.mysql.com/doc/refman/5.0/en/comparison-operators.html#function_in