我正在寻找一个mysql查询（使用php），一次遍历每个表，只显示与cross_check.lastmod 中不同的结果
t_one
guid | name | lastmod
1 | Joe | 2012-01-01 01:00:00
2 | Tom | 2012-01-02 01:00:00
3 | Sue | 2012-03-01 02:00:00
t_two
guid | pet | lastmod
4 | cat | 2012-01-01 01:00:00
5 | dog | 2012-01-02 01:00:00
6 | fish | 2012-03-01 02:00:00
t_three
guid | fruit | lastmod
7 | orange | 2012-01-01 01:00:00
8 | pear | 2012-01-02 01:00:00
9 | grape | 2012-03-01 02:00:00
交叉检查
guid | lastmod
1 | 2012-01-01 01:00:00
2 | 2012-01-02 01:00:00
3 | 2012-01-01 02:00:00
4 | 2012-01-01 01:00:00
5 | 2012-01-02 01:00:00
6 | 2012-01-01 02:00:00
7 | 2012-01-01 01:00:00
8 | 2012-01-02 01:00:00
9 | 2012-01-01 02:00:00
查询结果为：
t_one => 3 | Sue | 2012-03-01 02:00:00
t_two => 6 | fish | 2012-03-01 02:00:00
t_three => 9 | grape | 2012-03-01 02:00:00
到目前为止我的代码（需要表格交叉引用）
$tables = array('t_one', 't_two', 't_three');
foreach ($tables AS $table) {
$query = sprintf("SELECT * FROM '%s'",
mysql_real_escape_string($table));
$result = mysql_query($query);
}
仅cross_check表中的JOIN。
SELECT t_one.*
FROM t_one
JOIN cross_check ON t_one.guid=cross_check.guid
WHERE t_one.lastmod != cross_check.lastmod
不过，每个表仍然有一个查询（您可以随意使用联合，但这不会减少查询的核心数量）。
如果你想把它们组合成一个，你可以使用UNIONs:
[query for t_one]
UNION
[query for t_two]
UNION
[query for t_three]
我觉得这并不比做三个不同的查询更有效率（尽管这是一种直觉，但未经证实的说法）。
你也可以这样做（尽管这只在t_*和guid之间没有交叉的情况下有效；如果有t_one会更受欢迎，那么t_two）：
SELECT COALESCE(t_one.guid,t_two.guid,t_three.guid) AS guid
FROM cross_check
LEFT JOIN t_one ON t_one.guid=cross_check.guid
LEFT JOIN t_two ON t_two.guid=cross_check.guid
LEFT JOIN t_three ON t_three.guid=cross_check.guid
WHERE (t_one.lastmod IS NOT NULL AND t_one.lastmod != cross_check.lastmod)
OR (t_two.lastmod IS NOT NULL AND t_two.lastmod != cross_check.lastmod)
OR (t_three.lastmod IS NOT NULL AND t_three.lastmod != cross_check.lastmod)
如果您想获取与id相反的标签，也可以放置COALESCE(t_one.name, t_two.pet, t_three.fruit) AS label。
尝试使用此查询，当然要替换foreach循环中的表名。
SELECT origin.guid as guid FROM t_one as origin, cross_check as cross WHERE cross.lastmod != origin.lastmod;