Question

如何在选择查询中添加值？在我的代码下面我有5金，2银和1铜。 1金等于3点，1银是2点。我的问题是如何在选择查询中为此计算？有可能吗？

这是我的php脚本

$sql = "SELECT team.shortcut,
            SUM(IF(rank = 1, 1, 0)) AS gold, 
            SUM(IF(rank = 2, 1, 0)) AS silver,
            SUM(IF(rank = 3, 1, 0)) AS bronze
        FROM team 
            INNER JOIN academicorg ON team.shortcut = academicorg.shortcut 
        GROUP BY shortcut 
        ORDER BY gold DESC, silver DESC, bronze DESC";                         

$con = mysqli_connect($server_name,$mysql_user,$mysql_pass,$db_name); 

$result = mysqli_query($con,$sql);
$response = array();
while($row=mysqli_fetch_array($result))
{
    array_push($response, array("shortcut"=>$row[0], "gold"=>$row[1], "silver"=>$row[2], "bronze"=>$row[3], "total"=>$row[4]));
}

echo json_encode (array("news_response"=>$response));
mysqli_close($con);
?>

这是我得到的

    {"news_response":[{"shortcut":"aaa","gold":"6","silver":"3","bronze":"5","total":null},

{"shortcut":"bbb","gold":"6","silver":"3","bronze":"4","total":null}

这就是我想要的。我得到的总数是29，因为金等于3分，银是2，青铜是1。

 {"news_response":[{"shortcut":"aaa","gold":"6","silver":"3","bronze":"5","total"29},

{"shortcut":"bbb","gold":"6","silver":"3","bronze":"4","total":28}

Answer 1

我很想尝试这个。它未经测试我没有符合要求的数据库

SELECT team.shortcut,
    SUM(IF(rank = 1, 1, 0)) AS gold, 
    SUM(IF(rank = 2, 1, 0)) AS silver,
    SUM(IF(rank = 3, 1, 0)) AS bronze,
    SUM(IF(rank = 1, 3, IF(rank = 2, 2, IF(rank = 3, 1, 0)))) as total
FROM team 
    INNER JOIN academicorg ON team.shortcut = academicorg.shortcut 
GROUP BY shortcut 
ORDER BY gold DESC, silver DESC, bronze DESC

在php中的select查询中添加值

1 个答案: