SELECT m.id,count(*) as qq FROM xmx_users m LEFT JOIN xmx_users n on m.id = n.recommend_id GROUP BY id 这条sql ,查询出来的结果即使没有符合条件的也显示数量为1。表:CREATE TABLE IF NOT EXISTS `xmx_users ` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT, `recommend_id` int(10) unsigned NOT NULL DEFAULT '0', PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;测试数据:INSERT INTO `xmx_users ` (`id`,`recommend_id`) VALUES(2, 0),(3, 2),(5, 2),(6, 0),(7, 5),(8, 0),(9, 0),(10, 0),(11,5),(13, 7);
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