How do add a button to go back to another page in oracle apex interactive grid

问题

I want to add a customized button in oracle apex interactive grid to allow users to go back to the previous page. Can someone explain how I can do this in javascript? Thanks

回答1:

You can use history.back() or if you want to go multiple steps back history.go(-2)

回答2:

Without javascript (recommended):

Create your button, let's say BTN_PREVIOUS, and a hidden page item, P1_PREVIOUS_PAGE
Create a new branch:

Point: After Porcessing

Behavior Type: Page Identified by Item (Show only)

Item: Select your hidden item created on step 1 (P1_PREVIOUS_PAGE)

Server-side Condition: When Button Pressed, BTN_PREVIOUS

Now for every page that calls this one, you have to set the item P1_PREVIOUS_PAGE with the number of the calling page. There's innumerable ways to to this (through declarative list entries, URL, javascript), I recommend you to read https://docs.oracle.com/cd/E59726_01/doc.50/e39147/concept_url.htm

Obs: if you do the steps above, make sure that every other process, computation and branches have a server side condition, because this method relies on submitting the page and you don't want your "Previous" button triggering any other logic that you have already created.

With javascript:

Create your button, let's say BTN_PREVIOUS;
Create a dynamic action

When: Event: Click

Selection Type: Button

Button: BTN_PREVIOUS

Inside the dynamic action create a true action:

Action: Execute JavaScript code

Code: history.back()

Although the javascript way is simpler, it's always preferable to use what the tool (APEX) has to offer.

来源：https://stackoverflow.com/questions/57103876/how-do-add-a-button-to-go-back-to-another-page-in-oracle-apex-interactive-grid

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javascript

oracle-apex