拆分和比较记录时 MS-Access 中的联接操作中的语法错误

Question

此语句出现上述错误消息：

 SELECT f.fullname INTO SummaryJudgment_FinalForgottenWithMiddle
FROM (
(SELECT Left([aname],InStr(1,[aname],",")-1)) As lastname FROM
SummaryJudgment_FinalForgotten) & " " & (SELECT
RIGHT([aname],InStr(1,[aname],",")+1)) As firstname FROM
SummaryJudgment_FinalForgotten) & " " & (SELECT
summary_judgment.middle_initial AS middlename FROM summary_judgment)
) AS fullname
FROM SummaryJudgment_FinalForgotten AS f INNER JOIN summary_judgment
AS s ON f.lastname = s.last_name && f.firstname = s.first_name;

基本上这就是两张表的样子（请注意，它们的字段多于 1 个，其中不同字段的姓氏或名字可能相似）：

SummaryJudgment_FinalForgotten (table)
aname (field)
Leventhal,Raymond (data)


summary_judgment (table)
first_name(field)
Raymond (data)
last_name (field)
Leventhal (data)
middle_initial (field)
P (data)

最终，我正在尝试创建一个类似于 SummaryJudgment_FinalForgotten 但中间首字母来自 附加的summary_judgment： Leventhal，雷蒙德 P

Answer 1

您不需要编写 3 个选择语句来将值连接到一个字段中。

select left(...) & right(...) & initial AS fullname INTO SummaryJudgment_FinalForgottenWithMiddle
 FROM SummaryJudgment_FinalForgotten

Answer 2

您是否尝试使用两个 & 符号来表示逻辑与？

FROM SummaryJudgment_FinalForgotten AS f INNER JOIN summary_judgment
AS s ON f.lastname = s.last_name && f.firstname = s.first_name;

我认为这对于 Access 的 Jet/ACE 数据库引擎是不合法的。尝试使用 AND 关键字代替 &&。

OTOH，我想知道你是否可以做一些更简单的事情。

SELECT last_name & "," & first_name & " " & middle_initial AS fullname
INTO SummaryJudgment_FinalForgottenWithMiddle
FROM summary_judgment;

Answer 3

这行得通：

SELECT left([aname],InStr(1,[aname],",")-1) & " " 
     & right([aname],Len(aname)-InStr(1,[aname],",")) & " "
     & summary_judgment.middle_initial AS fullname 
INTO SummaryJudgment_FinalForgottenWithMiddle
FROM SummaryJudgment_FinalForgotten, summary_judgment;

虽然你可能想要这个：

SELECT left([aname],InStr(1,[aname],",")-1) & ", " 
     & right([aname],Len(aname)-InStr(1,[aname],",")) & " "
     & summary_judgment.middle_initial AS fullname 
INTO SummaryJudgment_FinalForgottenWithMiddle
FROM SummaryJudgment_FinalForgotten, summary_judgment;

第二个版本在你的姓氏后面加上逗号。请注意，Right 从右侧计算 ，这就是为什么您必须从长度中减去 InStr 值。

编辑：

我上面给出的代码适用于您的示例数据——每个表中的一行。有了更多的行，它给出了 (LastName, FirstName) x MiddleInitial 的叉积。我突然想到可能是这种情况，所以我回到我的测试并添加了第二行——它是正确的。于是我试着写了连接表达式......

访问不喜欢这样：

... ON left([aname],InStr(1,[aname],",")-1) = last_name ...

它会抛出错误“不支持连接表达式”。改成这样：

... ON (trim((left(SummaryJudgment_FinalForgotten.aname,InStr(1,[aname],",")-1))=trim(summary_judgment.last_name))) ...

导致运行并创建表的查询，但不创建任何行（在我添加“trim”调用以尝试修复它之前也是如此）。

所以我尝试为所有出现的名称指定表。不高兴——直到我意识到我在进行错误的比较（从最后到最后派生和从先到后派生——哎呀）。

在上面的 SELECT ... INTO 中使用以下 FROM 子句 是否 可以正常工作：

FROM 
  SummaryJudgment_FinalForgotten INNER JOIN 
  summary_judgment ON 
    ((left(SummaryJudgment_FinalForgotten.aname,InStr(1,SummaryJudgment_FinalForgotten.[aname],",")-1))=summary_judgment.last_name) AND 
    ((right(SummaryJudgment_FinalForgotten.aname,Len(SummaryJudgment_FinalForgotten.aname)-InStr(1,SummaryJudgment_FinalForgotten.[aname],","))=summary_judgment.first_name));

既然我从第一个到最后一个加入（因为表格之间没有重复），它甚至可能在没有每个字段的完整限定的情况下正常工作，但证明它确实有效，我完成。