傳遞SQL變量連接SQL查詢到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一個示例表上的SQL小提琴here。

我的目標是返回如下：

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，使用下面的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; 
     } 

     } 

我收到一條線的每一行，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 

而我不知道如何修改我的PHP來獲得所需的輸出。

電流輸出

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

這個問題似乎是，我們在調用「頭」每一行的事實。因此，您應該先檢查它是否已被調用。我希望下面的可以幫助你：

$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; 
} 

我感動的echo date和$total_work外面他們被在兩種情況下（$count >1和else）