為什么我的查詢未定義此值？

Question

我有一個數據庫查詢，顯示一個國家鳥的列表。 我選擇了一個腳本，該腳本添加了很酷的細節：可以為代表多個國家的鳥類添加星號，星號等。

但是，我必須在我的代碼或數據庫表中進行了一些更改以使其破壞。 現在，我收到錯誤消息未定義索引：星標 。 誰能發現問題？

我將查詢粘貼到phpMyAdmin> SQL中，它似乎可以正常工作，顯示代表國家鳥的國家的行...

SELECT GS.N, GS.IDArea SymArea, GS.IDSymbol, GS.URL, GS.Title,
GS.PageKind3, GS.Symbol, GS.Latin, GS.Desig, GS.DesigGen, GS.DesigGroup, GS.Date,
GG.N, GG.IDArea, GG.Name, GG.Type, GG.IDParent, GG.IDParent2, GG.Parent, GG.Parent2,    GG.IDReg, 
T.Stars 
FROM gs AS GS 
LEFT JOIN gw_geog AS GG ON GG.IDArea = GS.IDArea 

LEFT JOIN ( 
 SELECT Latin, COUNT(Latin) as Stars 
 FROM gs 
 GROUP BY Latin
) as T ON GS.Latin = T.Latin 

WHERE GS.DesigGen = 'bird' AND GG.Type = 'nat' AND GS.Symbol != '' 
OR GS.DesigGen = 'bird' AND GG.Type = 'dep' AND GS.Symbol != '' 
GROUP BY GS.IDArea ORDER BY GS.N

最重要的值可能是GS.Latin，它僅列出了各種鳥類的拉丁文或學名。 看來上面的查詢沒有與下面的循環中的腳本“連接”。 有誰預感到問題是什么？

while ($row = $stm->fetch())
{

  switch (TRUE)
 { 
  case ($row['Stars'] == 2): 
  $star_rating = ' (2)'; 
  break; 
  case ($row['Stars'] == 3): 
  $star_rating = ' (3)'; 
  break; 
  case ($row['Stars'] == 4): 
  $star_rating = ' (4)'; 
  break; 
  case ($row['Stars'] > 5): 
  $star_rating = ' (5 or more)'; 
  break; 
  default: 
  $star_rating = ''; 
  break; 
 }

}

Answer 1

我認為您的左T.Stars使Stars字段為null嘗試將查詢字段中的T.Stars替換為

COALESCE(T.Stars,0) AS `Stars`

如果字段為空，這將設置默認值0

Answer 2

您忘記將T.Latin放在查詢的選定部分，例如

SELECT GS.N, GS.IDArea SymArea, GS.IDSymbol, GS.URL, GS.Title,
GS.PageKind3, GS.Symbol, GS.Latin, GS.Desig, GS.DesigGen, GS.DesigGroup, GS.Date,
GG.N, GG.IDArea, GG.Name, GG.Type, GG.IDParent, GG.IDParent2, GG.Parent, GG.Parent2,    GG.IDReg, 
T.Stars, T.Latin
FROM gs AS GS 
LEFT JOIN gw_geog AS GG ON GG.IDArea = GS.IDArea 

LEFT JOIN ( 
 SELECT Latin, COUNT(Latin) as Stars 
 FROM gs 
 GROUP BY Latin
) as T ON GS.Latin = T.Latin 

WHERE GS.DesigGen = 'bird' AND GG.Type = 'nat' AND GS.Symbol != '' 
OR GS.DesigGen = 'bird' AND GG.Type = 'dep' AND GS.Symbol != '' 
GROUP BY GS.IDArea ORDER BY GS.N