Mysql與CASE選擇不檢索其他列值？

作爲我從我的問題得到的答案，它促使我堅持應用我的代碼出於某種原因，爲什麼它沒有奏效。Mysql與CASE選擇不檢索其他列值？

這是我在桌上的一部分數據。

enter image description here

原始代碼：

SELECT AGE_RANGE, COUNT(*) FROM (
    SELECT CASE 
     WHEN YearsOld BETWEEN 0 AND 5 THEN '0-5' 
     WHEN YearsOld BETWEEN 6 AND 10 THEN '6-10' 
     WHEN YearsOld BETWEEN 11 AND 15 THEN '11-15' 
     WHEN YearsOld BETWEEN 16 AND 20 THEN '16-20' 
     WHEN YearsOld BETWEEN 21 AND 30 THEN '21-30' 
     WHEN YearsOld BETWEEN 31 AND 40 THEN '31-40' 
     WHEN YearsOld > 40 THEN '40+' 
     END AS 'AGE_RANGE' 
     FROM (
      SELECT YEAR(CURDATE())-YEAR(DATE(birthdate)) 'YearsOld' 
      FROM MyTable 
     ) B 
) A 
GROUP BY AGE_RANGE

，這裏是結果。

enter image description here

我想要做的是，我想補充另一列這將有多少人指望誰是在這方面這將是位置，你會看到在其中包括珀斯，西澳大利亞州，森伯裏，維多利亞等。

首先試圖解決我的問題

正如你會看到下面，我已經添加了位置，和COUNT（位置）上文獲得位置的名稱和多少的位置，是重複計數在桌子裏。

SELECT AGE_RANGE, COUNT(*), location, COUNT(location) loc FROM (
    SELECT CASE 
     WHEN YearsOld BETWEEN 0 AND 5 THEN '0-5' 
     WHEN YearsOld BETWEEN 6 AND 10 THEN '6-10' 
     WHEN YearsOld BETWEEN 11 AND 15 THEN '11-15' 
     WHEN YearsOld BETWEEN 16 AND 20 THEN '16-20' 
     WHEN YearsOld BETWEEN 21 AND 30 THEN '21-30' 
     WHEN YearsOld BETWEEN 31 AND 40 THEN '31-40' 
     WHEN YearsOld > 40 THEN '40+' 
     END AS 'AGE_RANGE', 'location' 
     FROM (
      SELECT YEAR(CURDATE())-YEAR(DATE(birthday)) 'YearsOld' 
      FROM event_participants 
     ) B 
) A 
FROM event_participants WHERE location <> '' GROUP BY location HAVING loc >= 1 ORDER BY loc DESC LIMIT 5

哪些結果..

enter image description here

顯然，從神的幫助聽力不工作了，以好..

我的第二次嘗試來解決這個問題

SELECT AGE_RANGE, COUNT(*), location FROM (
    SELECT CASE 
     WHEN YearsOld BETWEEN 0 AND 5 THEN '0-5' 
     WHEN YearsOld BETWEEN 6 AND 10 THEN '6-10' 
     WHEN YearsOld BETWEEN 11 AND 15 THEN '11-15' 
     WHEN YearsOld BETWEEN 16 AND 20 THEN '16-20' 
     WHEN YearsOld BETWEEN 21 AND 30 THEN '21-30' 
     WHEN YearsOld BETWEEN 31 AND 40 THEN '31-40' 
     WHEN YearsOld > 40 THEN '40+' 
     END AS 'AGE_RANGE', 'location' 
     FROM (
      SELECT YEAR(CURDATE())-YEAR(DATE(birthday)) 'YearsOld' 
      FROM event_participants 
     ) B 
) A 
GROUP BY AGE_RANGE

和結果是...

enter image description here

而不是顯示國家的名字，只顯示位置。

好吧..

長話短說，代碼有什麼問題？任何解決這個東西？

這就像，我需要的結果是，

AGE_RANGE COUNT(*) location   total number of same location 

16-20  4  Name of location     # 

21-30  45  Name of location     # 

31-40  79  Name of location     # 

40+  102  Name of location     #

來源

2013-06-27 Wesley Brian Lachenal

請使用SQL小提琴:) –

如果'location'是一個列，請不要用單引號，'AS'AGE_RANGE'，位置'來包裝它。<<刪除它並查看它是否有效。 –

這應該工作：想想每個子查詢的，因爲它是自己的表，所以你只需要確保選擇您需要的所有列。（CASE WHEN THEN END AS）實際上只是一個列值。您可能還想在地點上進行分組，以便在不同地點獲得相同的年齡段。（例。11-15在加拿大和美國11-15）

SELECT AGE_RANGE, COUNT(*), A.location FROM (
     SELECT CASE 
      WHEN YearsOld BETWEEN 0 AND 5 THEN '0-5' 
      WHEN YearsOld BETWEEN 6 AND 10 THEN '6-10' 
      WHEN YearsOld BETWEEN 11 AND 15 THEN '11-15' 
      WHEN YearsOld BETWEEN 16 AND 20 THEN '16-20' 
      WHEN YearsOld BETWEEN 21 AND 30 THEN '21-30' 
      WHEN YearsOld BETWEEN 31 AND 40 THEN '31-40' 
      WHEN YearsOld > 40 THEN '40+' 
      END AS 'AGE_RANGE', B.location 
      FROM (
       SELECT YEAR(CURDATE())-YEAR(DATE(birthday)) 'YearsOld', 
       location /* << just missing this select */ 
       FROM event_participants 
      ) B 
    ) A 
    GROUP BY A.location, AGE_RANGE

來源

2013-06-27 05:43:41

嗯，我不知道怎麼說這個，但是，我在發佈這個問題之前就已經對我過去的嘗試做了什麼，在這裏調試，直到我瘋了足以再次發佈。那麼，出於某種原因，位置COLUMN的價值是「位置」而不是國家的名稱。最上面的圖片上的前三列名稱是「性別」「生日」「位置」它與代碼上的位置列匹配，但它只顯示結果'位置'。 –

稍微修改了代碼，您可能需要在A子查詢中使用B.location而不是'location'。另外請確保您將位置列添加到B子查詢的選擇中。 –

哦！有效！儘管如此，我期望的結果卻是不同的。儘管如此，我得到了我需要的所有要求，能夠顯示具有相應年齡的所有值，現在我可以自己管理這些值。再次感謝您再次幫助我！ :) –

這是不正確的： -

SELECT AGE_RANGE, COUNT(*), location FROM

當您使用GROUP BY在選擇屬性列表必須是適用於聚合函數必須成爲GROUP BY條款的一部分。

在下面你的例子不能顯示位置，因爲人在age-group的名稱可能屬於不同的位置： -

AGE_RANGE COUNT(*) location   total number of same location 

16-20  4  Name of location     # 

21-30  45  Name of location     #

來源

2013-06-27 05:45:59

試試這個

SELECT AGE_RANGE, COUNT(*), location FROM (
     SELECT CASE 
      WHEN YearsOld BETWEEN 0 AND 5 THEN '0-5' 
      WHEN YearsOld BETWEEN 6 AND 10 THEN '6-10' 
      WHEN YearsOld BETWEEN 11 AND 15 THEN '11-15' 
      WHEN YearsOld BETWEEN 16 AND 20 THEN '16-20' 
      WHEN YearsOld BETWEEN 21 AND 30 THEN '21-30' 
      WHEN YearsOld BETWEEN 31 AND 40 THEN '31-40' 
      WHEN YearsOld > 40 THEN '40+' 
      END AS 'AGE_RANGE', 'location' 
      FROM (
       SELECT YEAR(CURDATE())-YEAR(DATE(birthday)) 'YearsOld' 
       FROM event_participants 
      ) B 
    ) A 
    GROUP BY location, AGE_RANGE

來源

2013-06-27 05:54:54 Napster

Mysql與CASE選擇不檢索其他列值？

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