我在数据库中使用了这个sql语句，它使用count和union all

SELECT tem.book_id, COUNT( * )
    FROM(SELECT book_id FROM borrowdetails
    WHERE borrowdetails.borrow_status = 'returned'
    UNION ALL
    SELECT book_id from reserve) as tem
    GROUP BY book_id
    ORDER BY book_id DESC

然后我用这段代码命名php表中的id

while($row=mysql_fetch_array($user_query)){
    $id=$row['book_id'];
    $book_query = mysql_query("select * from book where book_id = '$id'")or die(mysql_error());
}

它工作得很好，但是它没有用相同的名称统一数据，因为它有不同的id，就像这样

+------+------+-------+
|  id  | name | count |
+------+------+-------+
|1     |A     |   3   | the id here is not shown at the table in PHP just   
+------+------+-------+ inside the database but the A is duplicate.
|2     |A     |   1   |
+------+------+-------+
|3     |B     |   2   |
+------+------+-------+

我想要的输出应该是这样的，而表中没有显示id。

+------+-------+
| name | count |
+------+-------+
|A     |   4   |
+------+-------+ this is the table that should be shown in PHP
|B     |   2   |
+------+-------+

正如您所看到的，名称a的count变为4，因为我还要加上两个a的count（*）。
为了达到预期的结果，我该怎么做？
顺便说一句，这是使用的数据库表。

+------+------+     +---------+---------+       +-------+-------+   
|  book table |     |borrowdetails table|       | reserve table |   
+------+------+     +---------+---------+       +-------+-------+   
|  id  | name |     |brw_dt_id| book_id |       |res_id |book_id|
+------+------+     +---------+---------+       +-------+-------+   
|1     |A     |     |    1    |    2    |       |   1   |   1   |
+------+------+     +---------+---------+       +-------+-------+
|2     |A     |     |    2    |    3    |       |   2   |   1   |
+------+------+     +---------+---------+       +-------+-------+
|3     |B     |     |    3    |    3    |       |   3   |   1   |
+------+------+     +---------+---------+       +-------+-------+