PHP/MySql報告使用distinct和group？

Question

我試圖寫一個MySQL表一個PHP報告不過我有點卡住我應該尋找到正確的命令......

表結構如下：

| Int | employeeID | lessonID | date  

| 1 | 15110  | 50  | 2010-10-18 
| 2 | 15110  | 51  | 2010-10-18 
| 3 | 15110  | 52  | 2010-10-18 
| 4 | 15110  | 50  | 2010-10-18 
| 5 | 13211  | 51  | 2010-10-18 
| 6 | 13211  | 52  | 2010-10-18 
| 7 | 11542  | 60  | 2010-10-18 

因此，每次打開課程時，都會添加一個新行，其中包含employeeID和lessonID。

我想創建會顯示訪問過誰lessonID 50，51和52

例如，報告將看起來像人的報告：

| employeeID | lessonID 50 | lessonID 51 | lessonID 52 | 

    15110   Y     Y    Y 
    13211   N     Y    Y 

我有什麼至今我的SQL查詢：

SELECT DISTINCT employeeID, lessonID 

FROM elearning.el_lessonmetrics 

WHERE lessonID = 50 OR lessonID = 51 OR lessonID = 52; 

其中給出：

| employeeID | lessonID | 
    15110   50 
    15110  51 
    15110  52 
    13211  50 
    13211  51 

任何人都可以幫助我如何將這些結果形成上述報告結構？

感謝

Answer 1

你要尋找的是被稱爲pivot table。

有幾個可用的教程在互聯網上爲MySQL：

• WikiBooks
• Mysql Tech-Resources
• Well House
• EHow

現在，只要您的具體問題，它取決於在ha之前是否知道課程的數量ND。如果是這樣，你可以只參加在自己每課：

SELECT a.employeeID, 
    CASE WHEN j1.lessonID IS NULL THEN 'N' ELSE 'Y' END AS `lessionID50`, 
    CASE WHEN j2.lessonID IS NULL THEN 'N' ELSE 'Y' END AS `lessionID51`, 
    CASE WHEN j3.lessonID IS NULL THEN 'N' ELSE 'Y' END AS `lessionID52` 
FROM elearning.el_lessonmetrics AS a 
    LEFT JOIN elearning.el_lessonmetrics AS j1 ON a.employeeID = j1.employeeID 
    LEFT JOIN elearning.el_lessonmetrics AS j2 ON a.employeeID = j2.employeeID 
    LEFT JOIN elearning.el_lessonmetrics AS j3 ON a.employeeID = j3.employeeID 
WHERE (j1.lessonID = 50 OR j1.lessonID IS NULL) 
    AND (j2.lessonID = 51 OR j2.lessonID IS NULL) 
    AND (j3.lessonID = 52 OR j3.lessonID IS NULL) 
GROUP BY a.employeeID 

如果你不知道的教訓數量，還是有大量的人則此方法將無法正常使用（和你很可能會關閉後處理結果的PHP構建你的表）好......

例如：

SELECT DISTINCT employeeID, lessonID 
FROM elearning.el_lessonmetrics 
WHERE lessonID IN (50, 51, 52); 

然後，在PHP

$employees = array(); 
$lessons = array(); 
while ($row = getRowFromDatabase()) { 
    $lessons[$row['lessonID']] = true; 
    $employees[$row['employeeID']][$row['lessonID']] = true; 
} 
foreach ($employees as &$employee) { 
    foreach ($lessons as $lessonId => $status) { 
     if (!isset($employee[$lessonId])) { 
      $employee[$lessonId] = false; 
     } 
    } 
} 

這會導致類似的數組：

array (
    15110 => array(
     50 => true, 
     51 => true, 
     52 => true, 
    ), 
    13211 => array(
     50 => true, 
     51 => true, 
     52 => false, 
    ), 
)