OpenForm WhereCondition不起作用 [英] OpenForm WhereCondition not working

问题描述

我现在已经坚持了一段时间了。

我现在有这个代码

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解决方案

嗯，很困惑为什么你有两个字段绑定到同一个字段。

通常在这种情况下，一个字段将被命名为Apps。另一个例如"文本1"因为Access不允许表单上的两个字段具有相同的名称。


弹出窗口要求应用程序值表示字段Apps。找不到，因此您需要检查formcontrols的名称（请参阅属性窗口的其他选项卡下的Name属性。）


Nic ; o）

对不起我必须解释得很糟糕。


按钮位于名为List1的表单上，并有一个名为App_Name的字段。因此，当按下按钮时，我希望它打开一个名为Form1的表单，其中字段Apps应显示与App_Name中相同的值。它们都绑定到表中的相同字段，但它们的格式不同。

我使用的名称是正确的，我已经仔细检查了所有这些...

对不起我必须解释得很糟糕。


按钮位于名为List1的表单上，并且有一个名为APP_NAME。因此，当按下按钮时，我希望它打开一个名为Form1的表单，其中字段Apps应显示与App_Name中相同的值。它们都绑定到表中的相同字段，但它们的格式不同。

我使用的名称是正确的，我已经仔细检查了所有这些...

展开 | 选择 | Wrap | 行号

I have been stuck for some time on this now.
I have this code at the moment

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解决方案

Hmm, puzzled why you have two fields bound to the same field.
Normally when this is the case, one field will be named "Apps" and the other e.g. "Text1" as Access won''t allow two fields on a form to have the same name.

The popup asking for the "Apps" value is an indication that the field "Apps" isn''t found, thus you''ll need to check the names of the formcontrols (See the Name property under the "Other" tab of the property window.)

Nic;o)

Sorry I must have explained it badly.

The button is locates on the form called List1 and has a field called App_Name. So when the button is pressed I want it to open a form called Form1 in which field Apps should display the same value as was in App_Name. They both are bound to the same field in table but they are on different forms.
The names I am using are correct, I have double checked all of them...

Sorry I must have explained it badly.

The button is locates on the form called List1 and has a field called App_Name. So when the button is pressed I want it to open a form called Form1 in which field Apps should display the same value as was in App_Name. They both are bound to the same field in table but they are on different forms.
The names I am using are correct, I have double checked all of them...

Expand|Select|Wrap|Line Numbers

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