计算并在PHP的不同列中显示多个mysql行值

Question

I have using following mysql statement to count and retrieve multiple database rows in mysql database:

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

But the output of this query only contain one column and the column name is COUNT(*) . This query returns three counted values and I want those value to assign PHP variables. following PHP code I used to get values:

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 "";}
?>

But this code throwing me an error without showing single value. I want to assign those three mysql counts to three mysql variables. how can I do that?

Answer 1

You should use an alias for thr column result

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

and refer to the column using the alias name for key

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

You should not refer to the rows contens using the tablename but a proper column name

Answer 2

Consider using an IF or CASE statement in your query to return a single row:

 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';

Now you will have 1 row with three fields.