Question

这是我收到的错误。

错误详情：

SQL查询的解析失败。请检查SQL语法。一些技巧：在第58行第1列解析错误。遇到：ALL（这个“全部”是一个紧接着闭括号。）

这是我的查询。

SELECT  ALL "TR-ID" .   "TR-ID" ,   "Site List Data"    .   "TR-ID",
                                "Issue number from Issue Log",
                        "Project Risks" .   
                                "Suvey Docs/Photos Uploaded to WS Repo",
                        "IPACK" .   
                                "Submit for MIROR code",
                        "Forecasts" .   
                                 "Equipment Ordered Planned",
                                 "Calix Blocker"
FROM                                
(SELECT         "TR-ID"     FROM        "Site List Data"        UNION
SELECT      "TR-ID"     FROM        "Project Risks"     UNION
SELECT      "TR-ID"     FROM        "IPACK"     UNION
SELECT      "TR-ID"     FROM        "Forecasts")        
ALL     "TR-ID"                     LEFT OUTER JOIN
"Site List Data"    ON ALL IDS. "TR-ID" =   "Site List Data"    .   "TR-ID"     LEFT OUTER JOIN
"Project Risks" ON ALL IDS. "TR-ID" =   "Project Risks" .   "TR-ID"     
LEFT OUTER JOIN
"IPACK" ON ALL IDS. "TR-ID" =   "IPACK" .   "TR-ID"     
LEFT OUTER JOIN
"Forecasts" ON ALL IDS. "TR-ID" =   "Forecasts" .   "TR-ID"

Answer 1

您只能在all之后立即使用select关键字，不得在每个联接表后重复此关键字。但是你根本不需要all关键字。

Answer 2

All是mysql中的关键字。

我认为你只是想对你的子查询进行别名，因此空间就是问题所在。我建议用方括号中的空格包装名称（并删除引号），例如[所有TR-IDS]。

您还使用了两个不同的名称：ALL TR-ID和ALL IDS。试试这个：

SELECT [ALL TR-IDS].[TR-ID], [Site List Data].[TR-ID], [Issue number from Issue Log], [Project Risks].[Suvey Docs/Photos Uploaded to WS Repo], IPACK.[Submit for MIROR code], Forecasts.[Equipment Ordered Planned], [Calix Blocker] FROM
(SELECT [TR-ID] FROM [Site List Data] UNION SELECT [TR-ID] FROM [Project Risks] UNION SELECT [TR-ID] FROM IPACK UNION SELECT [TR-ID] FROM Forecasts) [ALL TR-IDS] 
LEFT OUTER JOIN [Site List Data] ON [ALL TR-IDS].[TR-ID] = [Site List Data].[TR-ID] 
LEFT OUTER JOIN [Project Risks] ON [ALL TR-IDS].[TR-ID] = [Project Risks].[TR-ID] 
LEFT OUTER JOIN IPACK ON [ALL TR-IDS].[TR-ID] = IPACK.[TR-ID] 
LEFT OUTER JOIN Forecasts ON [ALL TR-IDS].[TR-ID] = Forecasts.[TR-ID]

Answer 3

＆＃34;全选＆＃34;是一个有效的命令，但＆＃34;所有＆＃34;是默认值并假定如果没有给出（即＆＃34;选择所有＆＃34;或＆＃34;选择不同＆＃34;都有效，使用＆＃34;选择＆＃34;本身与＆＃相同34;选择所有＆＃34;）

因为＆＃34;所有＆＃34;在SQL命令中使用，你应该避免它的名称。同样在MySQL中使用反引号而不是像这样的双引号：

SELECT
      `ALL_TR_IDS`.`TR-ID`
    , `Site List Data`.`TR-ID`
    , `Issue number from Issue Log`
    , `Project Risks`.`Suvey Docs/Photos Uploaded to WS Repo`
    , IPACK.`Submit for MIROR code`
    , Forecasts.`Equipment Ordered Planned`
    , `Calix Blocker`
FROM (
      SELECT
            `TR-ID`
      FROM `Site List Data`
      UNION
            SELECT
                  `TR-ID`
            FROM `Project Risks`
      UNION
            SELECT
                  `TR-ID`
            FROM IPACK
      UNION
            SELECT
                  `TR-ID`
            FROM Forecasts
) `ALL_TR_IDS`
      LEFT OUTER JOIN `Site List Data` ON `ALL_TR_IDS`.`TR-ID` = `Site List Data`.`TR-ID`
      LEFT OUTER JOIN `Project Risks` ON `ALL_TR_IDS`.`TR-ID` = `Project Risks`.`TR-ID`
      LEFT OUTER JOIN IPACK ON `ALL_TR_IDS`.`TR-ID` = IPACK.`TR-ID`
      LEFT OUTER JOIN Forecasts ON `ALL_TR_IDS`.`TR-ID` = Forecasts.`TR-ID`

ps：如果您停止在名称中使用空格，它将使您的SQL生活更轻松。

为什么“全部”会破坏我的查询？

3 个答案: