Синтаксическая ошибка (отсутствует оператор) в выражении запроса - MS-Access

Question

Это не приводит к синтаксическим ошибкам, но дает ошибку неправильных аргументов:

SELECT left(
  [aname],
 IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,
   Len(FinalForgotten.aname),
  IIF(instr([aname], ",") = 0,
    InStr(1,[aname]," ")-1,
    InStr(1,[aname],",")-1)))
&  ","  &

right(
   [aname],
 IIF(instr([aname], " " = 0 AND instr([aname], ",") = 0,
   Len(FinalForgotten.aname),
    IIF(instr([aname], ",") = 0,
      Len(aname)-InStr(1,[aname]," "),
      Len(aname)-InStr(1,[aname],",")))))
 &  " " &

defense_final.middle_initial AS fullname INTO FinalForgottenWithMiddle
FROM FinalForgotten INNER JOIN defense_final ON (right(FinalForgotten.aname,
 IIF(instr([aname], ",") = 0,
 Len(FinalForgotten.aname)-InStr(1,FinalForgotten.[aname]," "),
 Len(FinalForgotten.aname)-InStr(1,FinalForgotten.[aname],",")
  )
 )=defense_final.first_name) AND (left(FinalForgotten.aname,
IIF(instr([aname], ",") = 0,
  InStr(1,FinalForgotten.[aname]," ")-1,
  InStr(1,FinalForgotten.[aname],",")-1
    )
)=defense_final.last_name);

Итак, добавив дополнительные аргументы, я теперь получаю синтаксическую ошибку «Отсутствует»,] или «Элемент в выражении запроса»:

   SELECT left(
  [aname],
 IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,
   Len(FinalForgotten.aname),
  IIF(instr([aname], ",") = 0,
    InStr(1,[aname]," ")-1,
    InStr(1,[aname],",")-1)))
&  ","  &

right(
   [aname],
 IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,
   Len(FinalForgotten.aname),
    IIF(instr([aname], ",") = 0,
      Len(aname)-InStr(1,[aname]," "),
      Len(aname)-InStr(1,[aname],","))))
 &  " " &

defense_final.middle_initial AS fullname INTO FinalForgottenWithMiddle
FROM FinalForgotten INNER JOIN defense_final ON 

left(FinalForgotten.aname,
   IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,
      Len(FinalForgotten.aname),
      IIF(instr([aname], ",") = 0,
        InStr(1,FinalForgotten.[aname]," ")-1,
        InStr(1,FinalForgotten.[aname],",")-1)))

 =defense_final.last_name AND 

 right(FinalForgotten.aname,
      IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,
      Len(FinalForgotten.aname),
       IIF(instr([aname], ",") = 0,
        Len(FinalForgotten.aname)-InStr(1,FinalForgotten.[aname]," "),
        Len(FinalForgotten.aname)-InStr(1,FinalForgotten.[aname],","))))

=defense_final.first_name;

Есть обходной путь?

Спасибо за ответ.

Answer 1

Было бы проще поместить эту логику в функции VBA, чтобы ваш запрос выглядел следующим образом

SELECT GetFullNameWithMiddle([aname]) AS fullname
INTO FinalForgottenWithMiddle
FROM FinalForgotten INNER JOIN defense_final
    ON GetLastName([aname])=defense_final.last_name
    AND GetFirstName([aname])=defense_final.first_name

Поскольку вы говорите, что не хотите использовать VBA, попробуйте это. Я вложил каждый

InStr(x, y) = 0

в скобках, поэтому

IIf(InStr(x, y) = 0 AND InStr(p, q) = 0, g, h)

становится

IIf((InStr(x, y) = 0) AND (InStr(p, q) = 0), g, h)

Внесение этого изменения позволило мне переключиться на сетку проектирования запросов, что означает, что анализатору SQL это понравилось:

SELECT left(
    [aname],
    IIF((instr([aname], " ") = 0) AND (instr([aname], ",") = 0),
    Len(FinalForgotten.aname),
    IIF((instr([aname], ",") = 0),
    InStr(1,[aname]," ")-1,
    InStr(1,[aname],",")-1)))

    &  ","  &

right(
    [aname],
    IIF((instr([aname], " ") = 0) AND (instr([aname], ",") = 0),
    Len(FinalForgotten.aname),
    IIF((instr([aname], ",") = 0),
      Len(aname)-InStr(1,[aname]," "),
      Len(aname)-InStr(1,[aname],","))))

    &  " " &

defense_final.middle_initial AS fullname INTO FinalForgottenWithMiddle
FROM FinalForgotten INNER JOIN defense_final ON 

left(FinalForgotten.aname,
    IIF((instr([aname], " ") = 0) AND (instr([aname], ",") = 0),
    Len(FinalForgotten.aname),
    IIF((instr([aname], ",") = 0),
    InStr(1,FinalForgotten.[aname]," ")-1,
    InStr(1,FinalForgotten.[aname],",")-1)))

=defense_final.last_name AND 

right(FinalForgotten.aname,
    IIF((instr([aname], " ") = 0) AND (instr([aname], ",") = 0),
    Len(FinalForgotten.aname),
    IIF((instr([aname], ",") = 0),
    Len(FinalForgotten.aname)-InStr(1,FinalForgotten.[aname]," "),
    Len(FinalForgotten.aname)-InStr(1,FinalForgotten.[aname],","))))

=defense_final.first_name;

Answer 2

SELECT left( [aname], Таким образом [aname] является текстовым полем.

 IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,

Если это условие истинно, вы хотите, чтобы длина левой строки была:

 FinalForgotten.aname,

Какиене является целым числомВы хотите:

SELECT left(
  [aname],
 IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,
   Len(FinalForgotten.aname),

Answer 3

 IIF(instr([aname], " " = 0 AND instr([aname], ",") = 0,

в строке 12 должно быть

 IIF(instr([aname], " ") = 0 AND instr([aname], ",") = 0,

Answer 4

строка 10

right(

не имеет соответствующей правой круглой скобки