1、今天用json_encode()编辑数据，但是发现怎么都无法将数据格式成json，后来才发现json_encode编辑的内容必须是utf8的。。。

2、数据库联合查询select t.*,p.`wyear` as year2,p.price as price2 from  tbl_w   as   t LEFT JOIN tbl_wineprice pon t.wi_id = p.wiid where t.wine_id='$wineid' and p.`wineyear`=$year

这句话含义为  从t表和p表联合查询，t表为主，查询 t表的全部字段，和p表的wyear，price字段，联合查询的条件是t.wi_id = p.wiid ，并且t.wine_id='$winid' and p.`wiyear`=$year

1.left join

sql语句如下:

select * from A

left join B

on A.aID = B.bID

结果如下:

aID               aNum                          bID                  bName

1                   a20050111                1                      2006032401

2                   a20050112                2                     2006032402

3                   a20050113                3                     2006032403

4                   a20050114                4                     2006032404

5                   a20050115                NULL              NULL

(所影响的行数为 5 行)

结果说明:

left join是以A表的记录为基础的,A可以看成左表,B可以看成右表,left join是以左表为准的.

换句话说,左表(A)的记录将会全部表示出来,而右表(B)只会显示符合搜索条件的记录(例子中为: A.aID = B.bID).

B表记录不足的地方均为NULL.

2.right joinsql语句如下:

select * from A

right join B

on A.aID = B.bID

结果如下:

aID               aNum                          bID                  bName

1                   a20050111                1                      2006032401

2                   a20050112                2                     2006032402

3                   a20050113                3                     2006032403

4                   a20050114                4                     2006032404

NULL           NULL                          8                     2006032408

(所影响的行数为 5 行)

结果说明:

仔细观察一下,就会发现,和left join的结果刚好相反,这次是以右表(B)为基础的,A表不足的地方用NULL填充.

3.inner joinsql语句如下:

select * from A

innerjoin B

on A.aID = B.bID

结果如下:

aID               aNum                          bID                  bName

1                   a20050111                1                      2006032401

2                   a20050112                2                     2006032402

3                   a20050113                3                     2006032403

4                   a20050114                4                     2006032404

结果说明:         很明显,这里只显示出了 A.aID = B.bID的记录.这说明inner join并不以谁为基础,它只显示符合条件的记录.  还有就是inner join 可以结合where语句来使用 如：   select * from A innerjoin B on A.aID = B.bID where b.bname='2006032401' 这样的话 就只会放回一条数据了