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我在这里问了一个问题，非常感谢Gordon Linoff对其 SQL 部分的完整回答。

问题

我有一个 sql 表（使用 PRIMARY KEY = [ ip & id ] ），如下所示：

ip | id | visits

我想从该表中生成一个 php 数组，如下所示：

product_rating_for_ip = array(
  ip=>array(
    id=>rating, id=>rating, ...
  ),
  ip=>array(
    id=>rating, id=>rating, ...
  ),
  .
  .
  .
);

例如：

product_rating_for_ip = array(
  "78.125.15.92"=>array(
    1=>0.8,2=>0.2,3=>1.0
  ),
  "163.56.75.112"=>array(
    1=>0.1,2=>0.3,3=>0.5,8=>0.1,12=>1.0
  ),
  .
  .
  .
);

Gordon Linoff 建议了这个运行良好的 SQL 查询：

SQL 查询

select v.ip, group_concat(concat(v.id, ':', v.visits / iv.maxvisits)) as ratings
from visit v join
  (SELECT ip, id, visits, max(visits) as maxvisits
  FROM visit
  GROUP BY ip
  ) iv
 on v.ip = iv.ip
group by v.ip;

产生这个： 

+-------------+--------------------------+
|ip           |ratings                   |
+-------------+--------------------------+
|154.18.72.15 |2:0.1667,1:1.0000,3:0.1667|  
|162.16.18.22 |1:1.0000,3:0.3750,2:0.1250|  
|142.128.1.14 |2:1.0000,1:0.2500,3:0.5000|  
|78.15.92.131 |1:0.1429,3:1.0000,2:0.2857|  
+-------------+--------------------------+

问题总结

如何将此结果集放入此 PHP 数组中？

    product_rating_for_ip = array(
      ip=>array(
        id=>rating, id=>rating, ...
      ),
      ip=>array(
        id=>rating, id=>rating, ...
      ),
      .
      .
      .
    );

我有问题的答案

我认为我应该这样做（可能不正确或效率低下）：

$stmt = mysqli_prepare($con, "select ...");
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt,$ip,$ratings);
while (mysqli_stmt_fetch($stmt)){
  $value_array = explode(",",$ratings);
  product_rating_for_ip = array_fill_keys($ip, $value_array);
}
4

1 回答 1

2 
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt,$ip,$ratings);
$product_rating_for_ip = array();
while (mysqli_stmt_fetch($stmt)){
    $value_array = explode(",",$ratings);
    foreach ($value_array as $key => $value){
        $parts = explode(':', $value);
        $product_rating_for_ip[$ip][$parts[0]] = $parts[1];
    }
}
于 2013-07-28T15:27:43.933 回答