我有一张桌子：
| s_id | s_date | s_movie | s_hour | s_price |
| 1 | 2015-11-11 | The lord of the rings | 11:00 | 5 |
| 2 | 2015-11-11 | The lord of the rings | 11:00 | 4 |
| 3 | 2015-11-11 | The lord of the rings | 11:00 | 4 |
| 4 | 2015-11-11 | Harry Potter | 12:00 | 5 |
| 5 | 2015-11-11 | Harry Potter | 12:00 | 13 |
| 6 | 2015-11-11 | Harry Potter | 12:00 | 13 |
| 7 | 2015-11-11 | Harry Potter | 12:00 | 5 |
结果：
$result = mysql_query("
SELECT *
, count(s_price) as uniq_p
, SUM(s_price) as uniq_all
from table
group
by s_movie
, s_price
order
by s_movie desc
");
HTML表格显示：
<b>Date Movie Price Sum</b>
2015-11-11 The lord.. 2 x 4 8
2015-11-11 The lord.. 1 x 5 5
2015-11-11 Harry.. 2 x 5 10
2015-11-11 Harry.. 2 x 13 26
如何加入这样的结果？
<b>Date Movie Price_1 Price_2 Sum</b>
2015-11-11 The lord.. 2 x 4 1 x 5 13
2015-11-11 The Harry..2 x 5 2 x 13 36
这是一个msql示例： http://sqlfiddle.com/#!2/2c0f4/2
和PHP代码：
<table border="1" style="width:50%">
<thead>
<tr>
<td>Date</td>
<td>Movie</td>
<td>Price</td>
<td>Sum</td>
</tr>
</thead>
<? while ($row = mysql_fetch_array($result)):?>
<tr>
<td><?= $row["s_date"]; ?></td>
<td><?= $row["s_movie"]; ?></td>
<td><?= $row["uniq_p"]; ?> x <?= $row["s_price"]; ?></td>
<td><?= $row["uniq_all"]; ?></td>
</tr>
<? endwhile ?>
</table>