输出报告为rtf格式 [英] Output report to rtf format

问题描述

嗨！


此行不会在Windows 2000中执行，即使它在Windows XP中运行良好。


DoCmd.OutputTo acReport，strReportName，acFormatRTF，strRTFName


调试所有参数时似乎没问题。 strRTFName具有存在的目录的

位置，strReportName是当前数据库中确实存在的

对象的名称。是否有任何

的原因，为什么这在Win2000中不起作用？


我将不胜感激任何帮助


Vegard

解决方案

Vegard Villmones写道：

嗨！

此行不会在Windows 2000中执行，即使它在Windows XP中运行良好。

DoCmd.OutputTo acReport，strReportName，acFormatRTF，strRTFName
对象的名称。有什么原因导致它在Win2000中不起作用吗？

我会感激任何帮助

Vegard

它怎么不工作？你收到任何错误信息吗？

没有。当我运行调试器时，它跳转到它应该的行（在

错误转到K）并打印我的错误消息。现在你可能会认为我忘了

一个退出子语句，但我没有。

有没有办法获得更多关于错误发生原因的信息？猜猜我不是

a pro :)

***通过Developersdex发送 http://www.developersdex.com ***

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否。当我运行调试器时，它跳转到它应该的行（在

错误转到K）并打印我的错误消息。现在你可能会认为我忘了

一个退出子语句，但我没有。

有没有办法获得更多关于错误发生原因的信息？猜猜我不是

a pro :)

***通过Developersdex发送 http://www.developersdex.com ***

不要只是参加USENET ......获得奖励！

Hi!

This line wont execute in Windows 2000, even though it worked fine in
Windows XP.

DoCmd.OutputTo acReport, strReportName, acFormatRTF, strRTFName

When debugging all the arguments seem fine. strRTFName has the
location of a directory that exists, strReportName is the name of an
object which indeed exists in the current database. Is there any
reason why this shouldn''t work in Win2000?

I''d appreciate any help

Vegard

解决方案

Vegard Villmones wrote:

Hi!

This line wont execute in Windows 2000, even though it worked fine in
Windows XP.

DoCmd.OutputTo acReport, strReportName, acFormatRTF, strRTFName

When debugging all the arguments seem fine. strRTFName has the
location of a directory that exists, strReportName is the name of an
object which indeed exists in the current database. Is there any
reason why this shouldn''t work in Win2000?

I''d appreciate any help

Vegard

How is it not working? are you getting any error messages?

No. When I run the debugger it jumps to the line it''s supposed to (on
error goto K) and prints my error message. Now you might think I forgot
an exit sub-statement, but I haven''t.
Is there any way to get more info on why the error occurs? Guess I''m not
a pro :)
*** Sent via Developersdex http://www.developersdex.com ***
Don''t just participate in USENET...get rewarded for it!

No. When I run the debugger it jumps to the line it''s supposed to (on
error goto K) and prints my error message. Now you might think I forgot
an exit sub-statement, but I haven''t.
Is there any way to get more info on why the error occurs? Guess I''m not
a pro :)
*** Sent via Developersdex http://www.developersdex.com ***
Don''t just participate in USENET...get rewarded for it!

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