结果排序数

Question

这里是我的代码

$getlast = mysql_query("SELECT * FROM `offersdone` group by uid"); 
$content .= "<table><tr><th>Username</th><th>Total Offers</th></tr>"; 
while($last = mysql_fetch_array($getlast)){ 
    $getuser = mysql_query("SELECT * FROM `users` WHERE ID='".$last['uid']."'"); 
    if (($user = mysql_fetch_array($getuser)){ 
     $getlast1 = mysql_query("SELECT * FROM `offersdone` WHERE uid='".$user['ID']."' "); 
     $num1 = mysql_num_rows($getlast1); 
     $content .= "<tr><td>" . $user['username'] . " </td><td> " . $num1 . "</td></tr>"; 
    } 
} 

结果就这样产生了：

Username Total Offers 
martin 15 
dragan 1 
mile1986 1 
naramix 7 
rrrob 55 
sha09hh 17 
shakest 1 

Answer 1

$sql = "SELECT 
      u.username, 
      count(*) as count 
     FROM 
      offersdone o 
     LEFT JOIN 
      users u 
      ON u.ID = o.uid 
     GROUP BY 
      o.uid 
     ORDER BY 
      count(*) DESC"; 
$getlast = mysql_query($sql); 
$content .= "<table><tr><th>Username</th><th>Total Offers</th></tr>"; 
while($last = mysql_fetch_array($getlast)){ 
     $content .= "<tr><td>" . $last['username'] . " </td><td> " . $last['count'] . "</td></tr>"; 
} 

这应该选择这两个用户名，并提供了一个查询和已排序数量按优惠的数量计算。

演示：

mysql> SELECT * FROM offersdone; 
+-----+ 
| uid | 
+-----+ 
| 1 | 
| 1 | 
| 2 | 
| 2 | 
| 2 | 
| 2 | 
+-----+ 
6 rows in set (0.00 sec) 

mysql> SELECT * FROM users; 
+----+----------+ 
| ID | username | 
+----+----------+ 
| 1 | A  | 
| 2 | B  | 
+----+----------+ 
2 rows in set (0.00 sec) 

mysql> SELECT u.username, COUNT(*) as count 
    -> FROM offersdone o 
    -> LEFT JOIN users u ON u.ID = o.uid 
    -> GROUP BY o.uid 
    -> ORDER BY COUNT(*) DESC; 
+----------+-------+ 
| username | count | 
+----------+-------+ 
| B  |  4 | 
| A  |  2 | 
+----------+-------+ 
2 rows in set (0.00 sec) 

Answer 2

添加到您的查询 “ORDER BY总供应ASC”;