为什么我的查询未定义此值？

Question

我有一个数据库查询，显示一个国家鸟的列表。 我选择了一个脚本，该脚本添加了很酷的细节：可以为代表多个国家的鸟类添加星号，星号等。

但是，我必须在我的代码或数据库表中进行了一些更改以使其破坏。 现在，我收到错误消息未定义索引：星标 。 谁能发现问题？

我将查询粘贴到phpMyAdmin> SQL中，它似乎可以正常工作，显示代表国家鸟的国家的行...

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，它仅列出了各种鸟类的拉丁文或学名。 看来上面的查询没有与下面的循环中的脚本“连接”。 有谁预感到问题是什么？

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

我认为您的左T.Stars使Stars字段为null尝试将查询字段中的T.Stars替换为

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

如果字段为空，这将设置默认值0

Answer 2

您忘记将T.Latin放在查询的选定部分，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