我正在寻找一个mysql查询(使用php)一次一个地查看每个表,并且只显示与t_lastmod表中的键日期不匹配的所有结果。我不知道怎么说ON tablename = field value。
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
t_lastmod
table_name | lastmod
t_one | 2012-01-01 01:00:00
t_two | 2012-01-02 01:00:00
t_three | 2012-01-01 02:00:00
查询结果为:
t_one => 2 | Tom | 2012-01-02 01:00:00
t_one => 3 | Sue | 2012-03-01 02:00:00
t_two => 4 | cat | 2012-01-01 01:00:00
t_two => 6 | fish | 2012-03-01 02:00:00
t_three => 8 | pear | 2012-01-02 01:00:00
t_three => 9 | grape | 2012-03-01 02:00:00
到目前为止我的代码(在JOIN t_lastmod ON上需要帮助......)
$tables = array('t_one', 't_two', 't_three');
foreach ($tables AS $table) {
$query = " select $table.* from $table JOIN t_lastmod ON $table = t_lastmod.TABLE_NAME WHERE $table.lastmod != t_lastmod.lastmod ";
}
答案 0 :(得分:1)
select $table.*
from $table
JOIN t_lastmod ON '$table' = t_lastmod.TABLE_NAME
WHERE $table.lastmod != t_lastmod.lastmod "