3

我有以下 INNER JOIN 查询：

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

您可以在此处的 SQL fiddle 上的示例表上看到它的运行情况。

我的目标是返回以下内容：

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;
        }

        }

我为每一行得到一行，而不是分组，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

而且我不确定如何修改我的 PHP 以获得所需的输出。

电流输出

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
4

1 回答 1

1

问题似乎在于您正在为每一行调用“头部”。因此，您应该首先检查它是否已被调用。我希望以下内容可以帮助您：

$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;
}

我将 theecho date和移到了$total_work外面，因为它们在两种情况下都被平等地调用了 ($count >1和else)

于 2013-02-22T13:01:23.717 回答