Mysqli排名语法

Question

我有一个mysqli的问题。当我尝试在php内使用它时，我给了我一个错误，但是当我在某些database application中执行查询时，它完美地工作。这让我对

SET @rank=0; SELECT (@rank := @rank+1) 

错误在那里我可以执行它在mysql yog正确。有没有其他的选择，我怎么才能使它工作？

下面的代码：

function get_rank($branch,$cat){ 
    global $connection; 
    $result = array(); 
    $rank = 0; 
    if ($statement = $connection->prepare("SET @rank=0; SELECT (@rank := @rank+1) AS rank, a.branch_code_id, SUM(b.amount), c.category FROM sales_add_h AS a INNER JOIN sales_add_i AS b ON a.id = b.sales_h_id INNER JOIN control_panel_item_create AS c ON b.item_code_id = c.id WHERE a.branch_code_id = ? AND c.category = ? GROUP BY c.category, a.branch_code_id, b.amount ORDER BY SUM(b.amount) DESC")) {   
     $statement->bind_param("is",$branch,$cat); 
     $statement->execute(); 
     $statement->bind_result($a,$b,$c,$d); 
     while ($row = $statement->fetch()) { 
      array_push($result,array($a,$b,$c,$d)); 
     }  
     $statement->close(); 
    } else { 
     printf("Errormessage: %s\n", $connection->error); 
     echo " error in SQL Statement."; 
    } 
    return $result; 
} 

Answer 1

您正在执行2查询您的代码，并且不与PHP函数工作。但是你可以整合你set在select

SELECT (@rank := @rank+1) AS rank, a.branch_code_id, SUM(b.amount), c.category 
FROM sales_add_h AS a, 
    (select @rank := 0) r 
INNER JOIN sales_add_i AS b ON a.id = b.sales_h_id 
INNER JOIN control_panel_item_create AS c ON b.item_code_id = c.id 
WHERE a.branch_code_id = ? 
AND c.category = ? 
GROUP BY c.category, a.branch_code_id, b.amount 
ORDER BY SUM(b.amount) DESC 

Answer 2

而且我的评论，尝试一下本作的SQL

SELECT (@rank := @rank+1) AS rank, a.branch_code_id, SUM(b.amount), c.category 
FROM sales_add_h AS a 
INNER JOIN sales_add_i AS b ON a.id = b.sales_h_id 
INNER JOIN control_panel_item_create AS c ON b.item_code_id = c.id 
CROSS JOIN (SELECT @rank=0) Sub1 
WHERE a.branch_code_id = ? 
AND c.category = ? 
GROUP BY c.category, a.branch_code_id, b.amount 
ORDER BY SUM(b.amount) DESC