嗨，我正在尝试选择发布的最后五个主题。我本以为这个问题会奏效，但如果有人能帮我的话，我真的被这个问题卡住了。
$query = yasDB_select("SELECT * FROM forumposts INNER JOIN forumtopics ON forumposts.id=forumtopics.id ORDER by forumposts.date DESC limit 5");
我发现您的查询存在一些问题。
合适的解决方案：
SELECT DISTINCT(forumtopics.name)
FROM forumtopics LEFT JOIN forumposts ON forumposts.topicid = forumtopics.id
ORDER BY forumposts.id DESC LIMIT 5;
如果你对有关主题的更复杂的数据感兴趣，你可以做以下事情：
SELECT * FROM (SELECT DISTINCT(forumtopics.id) FROM forumtopics LEFT
JOIN forumposts ON forumposts.topicid = forumtopics.id ) topics LEFT
JOIN forumtopics ON topics.id = forumtopics.id ORDER BY topics.id DESC
LIMIT 5 ;
我使用了以下测试模式：
CREATE TABLE forumposts ( id int, topicid int, content text );
CREATE TABLE forumtopics ( id int, name varchar(200) );
INSERT INTO forumtopics SET id = 1, name = "1st topic"; INSERT INTO
forumtopics SET id = 2, name = "2nd topic"; INSERT INTO forumtopics
SET id = 3, name = "3rd topic"; INSERT INTO forumtopics SET id = 4,
name = "4th topic"; INSERT INTO forumtopics SET id = 5, name = "5th
topic"; INSERT INTO forumtopics SET id = 6, name = "6th topic";
INSERT INTO forumposts SET id = 1, topicid=1, content = "1st post";
INSERT INTO forumposts SET id = 2, topicid=3, content = "2nd post";
INSERT INTO forumposts SET id = 3, topicid=3, content = "3rd post";
INSERT INTO forumposts SET id = 4, topicid=4, content = "4th post";
INSERT INTO forumposts SET id = 5, topicid=5, content = "5th post";
INSERT INTO forumposts SET id = 6, topicid=6, content = "6th post";
INSERT INTO forumposts SET id = 7, topicid=6, content = "7th post";
您可以在上测试查询http://sqlfiddle.com.