计算并在PHP的不同列中显示多个mysql行值

Question

我已经使用以下mysql语句来计数和检索mysql数据库中的多个数据库行：

SELECT COUNT(*) FROM leavesrecords WHERE leavetype IN ('Casual','Annual','Medical') and empno = '202'  GROUP BY leavetype

但是此查询的输出仅包含一列，列名称为COUNT(*) 。 该查询返回三个计数值，我希望这些值分配PHP变量。 以下我用来获取值的PHP代码：

include 'database.php';
$query = "SELECT COUNT(*) FROM leavesrecords WHERE leavetype IN ('Casual','Annual','Medical') and empno = '202'  GROUP BY leavetype; ";
$stmt = $con->prepare($query);
$stmt->execute();

$num = $stmt->rowCount();
if($num>0){
    while ($row = $stmt->fetch(PDO::FETCH_ASSOC)){
        extract($row);
        echo $row["leavesrecords"];
    } 
}
else{ echo "";}
?>

但是这段代码在没有显示单个值的情况下抛出了错误。 我想将这三个mysql计数分配给三个mysql变量。 我怎样才能做到这一点？

Answer 1

您应该为thr列结果使用别名

 SELECT COUNT(*) as my_count 
 FROM leavesrecords 
 WHERE leavetype IN ('Casual','Annual','Medical') and empno = '202'  GROUP BY leavetype

并使用别名的密钥引用该列

if($num>0){
    while ($row = $stmt->fetch(PDO::FETCH_ASSOC)){
        // extract($row);
        echo $row["my_count"];
    } 
}

您不应使用表名而是使用正确的列名来引用行内容

Answer 2

考虑在查询中使用IF或CASE语句返回单行：

 SELECT sum(if(leavetype='Casual',1,0)) AS `CasualCount`,
    sum(if(leavetype='Annual',1,0)) AS `AnnualCount`,
    sum(if(leavetype='Medical',1,0)) as `MedicalCount`
FROM leaverecords
WHERE leavetype IN ('Casual','Annual','Medical') and empno = '202';

现在，您将有1行包含3个字段。