php - 7 个带有内连接的查询到 1 个查询，mysql

Question

我有一个问题，如何将总共 7 个查询变成 1 个查询？减少数据库查询？我有 1 个表包含所有内容，但 suggest 是一个“分隔符”，每个建议我必须加载 33 行并按日期线排序，我考虑过使用任何内部联接...等，但我认为这不是一个办法吗？如我错了请纠正我。感谢您的帮助!谢谢!!

这是mysql查询1

$query = DB::query("SELECT t.*, d.did AS dingid, d.id AS dingpid, f.id AS bookmark, f.uid AS buid 
        FROM ".DB::table('comeing_tao')." AS t 
        LEFT JOIN ".DB::table('comeing_tao_ding')." AS d ON t.id = d.id AND d.uid = ".$_G['uid']." 
        LEFT JOIN ".DB::table('comeing_tao_fans')." AS f ON t.id = f.id AND f.uid = ".$_G['uid']." 
        WHERE t.suggest = 0 AND t.state = 1 ORDER BY dateline DESC LIMIT 33");

这是mysql查询2

$query = DB::query("SELECT t.*, d.did AS dingid, d.id AS dingpid, f.id AS bookmark, f.uid AS buid 
            FROM ".DB::table('comeing_tao')." AS t 
            LEFT JOIN ".DB::table('comeing_tao_ding')." AS d ON t.id = d.id AND d.uid = ".$_G['uid']." 
            LEFT JOIN ".DB::table('comeing_tao_fans')." AS f ON t.id = f.id AND f.uid = ".$_G['uid']." 
            WHERE t.suggest = 1 AND t.state = 1 ORDER BY dateline DESC LIMIT 33");

这是mysql查询3

$query = DB::query("SELECT t.*, d.did AS dingid, d.id AS dingpid, f.id AS bookmark, f.uid AS buid 
        FROM ".DB::table('comeing_tao')." AS t 
        LEFT JOIN ".DB::table('comeing_tao_ding')." AS d ON t.id = d.id AND d.uid = ".$_G['uid']." 
        LEFT JOIN ".DB::table('comeing_tao_fans')." AS f ON t.id = f.id AND f.uid = ".$_G['uid']." 
        WHERE t.suggest = 2 AND t.state = 1 ORDER BY dateline DESC LIMIT 33");

这是mysql查询3

$query = DB::query("SELECT t.*, d.did AS dingid, d.id AS dingpid, f.id AS bookmark, f.uid AS buid 
        FROM ".DB::table('comeing_tao')." AS t 
        LEFT JOIN ".DB::table('comeing_tao_ding')." AS d ON t.id = d.id AND d.uid = ".$_G['uid']." 
        LEFT JOIN ".DB::table('comeing_tao_fans')." AS f ON t.id = f.id AND f.uid = ".$_G['uid']." 
        WHERE t.suggest = 3 AND t.state = 1 ORDER BY dateline DESC LIMIT 33");

这是mysql查询4

$query = DB::query("SELECT t.*, d.did AS dingid, d.id AS dingpid, f.id AS bookmark, f.uid AS buid 
        FROM ".DB::table('comeing_tao')." AS t 
        LEFT JOIN ".DB::table('comeing_tao_ding')." AS d ON t.id = d.id AND d.uid = ".$_G['uid']." 
        LEFT JOIN ".DB::table('comeing_tao_fans')." AS f ON t.id = f.id AND f.uid = ".$_G['uid']." 
        WHERE t.suggest = 4 AND t.state = 1 ORDER BY dateline DESC LIMIT 33");

依此类推...我跳过以下3个查询，因为除了t.suggest = 5、t.suggest = 6、t.suggest = 7之外，其他都是完全相同的

目标是将所有查询集中在一起，然后使用数组。

Answer 1

您可以使用OR

(t.suggest=3 OR t.suggest=4 OR ...) AND ...

如果您希望每个建议最多包含 33 条记录，请尝试 UNION

(SELECT ...) UNION (SELECT ...) UNION (SELECT ...) ...