Oracle APEX:从经典报告上的链接列调用内联弹出窗口 [英] Oracle APEX: calling an inline popup from a link column on a classic report

问题描述

我试图从链接列调用 javascript，这将通过将该列的目标 URL 设置为:

I am attempting to call javascript from a link column, that will open an inline popup by setting target URL for that column to:

javascript:$s("P3_ITEM","#COLUMN1#");openModal("MY_INLINE_POPUP");

并且我收到一个 javascript 错误:未捕获的错误:无法在初始化之前调用对话框上的方法；试图调用方法 'open'.我做错了什么?

and I am getting a javascript error: Uncaught Error: cannot call methods on dialog prior to initialization; attempted to call method 'open'. What am I doing wrong?

推荐答案

当您未将尝试调用的区域的模板设置为模态/内联对话框时，会出现此错误..

This error occurs when you do not set the template of the region which you are trying to call as modal/inline dialog..

您需要将区域MY_INLINE_POPUP"的模板更改为内联对话框，您可以通过转到区域设置 -> 外观 -> 模板，然后选择内联对话框...

You need to change the template of your region "MY_INLINE_POPUP" to Inline Dialog and you can do this by going to Region Settings -> Appearance -> Template and then select Inline Dialog...

另外一个提示:在创建内联对话框时，您还需要注意另外一个区域设置，即区域布局.在Region Layout里面，位置必须是Inline Dialogs..

One additional tip: while creating an inline dialog, you also need to take care of one more Region setting and i.e. Region Layout. Inside Region Layout, the position must be Inline Dialogs..

这篇关于Oracle APEX:从经典报告上的链接列调用内联弹出窗口的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！