试试这个：

SELECT 
    distinct user_id
FROM 
    table
WHERE user_id not in (SELECT DISTINCT user_id FROM table WHERE is_published = 1)

你可以用 not exists :

SELECT t.* 
FROM t 
WHERE NOT EXISTS (SELECT 1 
                  FROM t t2 
                  WHERE t2.user_id = t.user_id AND
                        t2.is_published = 1
                 );

例如，所有用户至少有两个条目。此查询将返回具有一个未发布条目的用户。目前尚不清楚这是否可取。否则，修改查询就相当简单了。

例如。：
注意：不要在表/列标识符中使用空格。我已相应地修改了列名。

SELECT x.* 
  FROM table1 x 
  LEFT
  JOIN table1 y
    ON y.user_id = x.user_id
   AND y.is_published = 1
 WHERE x.is_published = 0
   AND y.id IS NULL;

使用嵌套查询：

SELECT DISTINCT `id`, 
                `user_id`, 
                `post_title`, 
                `is published`
FROM table1
WHERE user_id NOT IN (SELECT DISTINCT user_id FROM table1 WHERE `is published` = 1)
AND user_id  NOT IN (SELECT DISTINCT T1.user_id
                     FROM Table1 T1
                     INNER JOIN Table2 T2
                     ON T1.id=T2.table1_id)

或使用联接：

SELECT T1.*
FROM
Table1 T1
INNER JOIN
(
SELECT user_id,COUNT(*) AS CNT
FROM TABLE1 
GROUP BY user_id
) T2
ON T1.user_id=T2.user_id
INNER JOIN
(
SELECT user_id,COUNT(*) AS CNT FROM Table1
WHERE `is published` = 0
GROUP BY user_id
) T3
ON T2.user_id=T3.user_id
AND T2.CNT=T3.CNT
WHERE T1.user_id
NOT IN (
        SELECT T1.user_id
        FROM Table1 T1
        INNER JOIN Table2 T2
        ON T1.id=T2.table1_id
       )

输出

id  user_id post_title  is published
1   9332    some title  0
3   9332    some title  0

演示
http://sqlfiddle.com/#!2012年9月25日

简单的方法是使用 SUBQUERY 与 GROUP BY 以及 HAVING 条款。我们可以返回重复的行，但是没有 is_published = 1 然后与外部查询连接，返回匹配id的记录

SELECT t.id,
    t.user_id,
    t.post_title,
    t.is_published 
FROM temp t
INNER JOIN (SELECT user_id
            FROM temp
            GROUP BY user_id HAVING SUM(is_published) = 0) u ON u.user_id = t.user_id
ORDER BY t.id

输出：

id  user_id post_title  is_published
1   9332    some title  0
3   9332    some title  0
7   3221    some title  0
8   3221    some title  0