將sql變量從INNER JOIN SQL查詢傳遞給PHP腳本

Question

我有以下INNER JOIN查詢：

SELECT  b.*, c.date2
FROM    (
            SELECT a.work, a.amount, 
                   COUNT(*) totalCount, 
                   SUM(Amount) totalAmount
            FROM work_times a WHERE Organisation=?
            GROUP BY a.work, a.amount
        ) b
        INNER JOIN
        (
            SELECT a.work, a.amount, DATE_FORMAT(Date,'%D %M %Y') date2,
                    date
            FROM work_times a
        ) c ON b.work = c.work and b.amount=c.amount
ORDER BY b.work, b.totalCount, c.date

你可以看到它在行動上SQL搗鼓一個示例表在這里 。

我的目標是返回以下內容：

5 consultancy sessions @ £50 each: £250

1st February 2013
8th February 2013
15th February 2013
22nd February 2013
1st March 2013

3 therapy sessions @ £40 each: £120

2nd February 2013
9th February 2013
16th February 2013

2 therapy sessions @ £20 each: £40

3rd February 2013
10th February 2013

但是使用以下PHP：

$stmt->bind_param("s", $name1);
$stmt->execute();
$stmt->store_result();  
$stmt->bind_result($work,$amount,$count,$total_group,$date);

while ($stmt->fetch()) {

        if ($count>1) {
           echo $count." ".$work."s @ &pound;".$amount." each<br><br>";
           echo date("jS F Y",strtotime($date))."<br><br>";
           $total_work=$total_work+$total_group;
        }
        else {
           echo $count." ".$work." @ &pound;".$amount."<br><br>";
           echo date("jS F Y",strtotime($date))."<br><br>";
           $total_work=$total_work+$total_group;
        }

        }

我為每一行獲得一行，而不是分組，即：

5 Consultancy Sessions @ £50.00

1st February 2013

5 Consultancy Sessions @ £50.00

8th February 2013

5 Consultancy Sessions @ £50.00

15th February 2013

...etc

我不知道如何修改我的PHP以獲得所需的輸出。

當前輸出

5 Consultancy Sessions @ £50.00

1st February 2013

8th February 2013

15th February 2013

22nd February 2013

1st March 2013

2nd February 2013

9th February 2013

16th February 2013

3rd February 2013

10th February 2013

Answer 1

問題似乎在於你要為每一行調用“頭”。 因此，您應首先檢查它是否已被調用。 我希望以下內容可以幫助您：

$stmt->bind_param("s", $name1);
$stmt->execute();
$stmt->store_result();  
$stmt->bind_result($work,$amount,$count,$total_group,$date);

$last_work = "";
while ($stmt->fetch()) {
    if($work != $last_work || $amount != $last_amount){
        if ($count>1) {
           echo "<br>".$count." ".$work."s @ &pound;".$amount." each<br><br>";

        }
        else {
           echo "<br>".$count." ".$work." @ &pound;".$amount."<br><br>";
        }
        $last_work = $work;
        $last_amount = $amount;
    }
    echo date("jS F Y",strtotime($date))."<br>";
    $total_work=$total_work+$total_group;
}

我將echo date和$total_work外面，因為它們在兩種情況下均被調用（ $count >1和else ）