In one of the website, on footer there is a contact info with phon number , after clicking on that phone number 1 popup comes which is not aler (poup- <Make a call from - skype>) as below
enter image description here
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This is my issue: I have a text field item in a normal page (P7_PRODUCT), in this page there is a button that opens a modal page and, in the modal page there is another text field item (P7_PRODUCT). I would like that when I write a product in P7_PRODUCT, the same product appears in P8_PRODUCT. I tried with several dynamic actions, but none of them worked.
Here's one option.
on page 7, there are: P7_PRODUCT text item and P7_GOTO_8 button
let button Submit the page
create branch
name it e.g. "Goto 8"
it should redirect to page 8
in "Link Builder", set P8_PRODUCT item to &P7_PRODUCT. (note leading ampersand and trailing dot)
run page 7
enter something into the P7_PRODUCT item
press the button
modal dialog page opens and its P8_PRODUCT contains value you entered on page 7
A good blog to read is this one, by John Snyders from the apex development team. As he explains, it's not advised to branch to a modal dialog - instead you should link to it.
Here is an example based on techniques described in [this] (https://jeffkemponoracle.com/2022/02/reusable-region-as-a-modal-page/) blog.
I have 2 pages: Page 108 (normal page) and page 109 (modal page). Page 108 has a region (Region 1) with page item P108_NEW and a button called TO_MODAL. Page 109 has page item P109_NEW. When a user enters a value in P108_NEW and clicks the button, page 109 should open with the value from P108_NEW shown in P109_NEW.
The technique generates the url of the modal page and then redirects to it.
Give the region on page 108 a static id: "region1". This is used if you want to create a dynamic action on close of the dialog.
Create a new hidden item on page 108 (P108_URL_TO_MODAL1) to hold the value of the modal page url. Source type: Expression, source:
apex_page.get_url(
p_page => 109,
p_triggering_element => '$(''#region1'')'
)
Make sure to set "Used" to "Always, replacing any existing value in session state".
Create a dynamic action on click of button on page 108.
Add an action to the dynamic action of type pl/sql to set the value on page 109.
Action 1: Execute Server-side Code, Pl/SQL Code:
:P109_NEW := :P108_NEW;
Values to submit: P108_NEW
Action 2: Execute Javascript code:
apex.navigation.redirect("&P108_URL_TO_MODAL1.");
This should be it.
Is there a way to let a help text region (APEX 5.0) show the help text of another page?
I created a help page 0 (as modal dialog) in APEX 5.0, which is reachable by clicking a Navigation Bar List Entry.
The page itself has only one Region, which shall display the Help Text of the Page it was called from. The ID of the calling Page is transmitted into a hidden item via URL.
It is possible. You can get the Help Text of a page anywhere in your application. You can use this query as a source of your item that will display the help text:
select HELP_TEXT
from APEX_APPLICATION_PAGES
where APPLICATION_ID = yourappid and
PAGE_ID = pagenumber_of_the_helptext_you_want_toget
I'm doing some UI automation testing at work and I'm not quite sure how to do this
I have a form that asks for a user's name and date of birth before they can click a button to the next page. If they enter anything with invalid characters (a name of '1111' or a DOB in the future) the input boxes are highlighted in red and an error message appears. How can I test that the 'continue' button is clickable only once valid input is entered?
Actually, in my project, there are many many "type-of-button".
It can be an input-tage , span-tag or even image!
So in my opinion, you can't verify that button can be clickable.
You should verify what system does after you click on this button for example: page is loaded, image is uploaded...
Hope it helpful :)
First scenario :
Username: Valid
Password: Valid
button is displayed
use the driver.FindElement(By.Id("id/xpath of the button ")).Displayed returns true (C# code )
for java there is similar display
else
you enter the invalid scenario
button is not displayed
the above code would give false
Hope this clears your doubt
so I'm running a Runescape Private Server and my java code for voting in it is if the link = www.clearpkz.webs.com/votevote?Username= + username entered is active, then it allows users to do ::claim to get a reward. But how can I make it so the Submit button to enter their username only displays if they click on every image first? Or can I only make it display once every 24 hours even if the page is refreshed?
Here is my current code:
http://pastebin.com/rfQeVkfT
Here is my voting website: www.clearpkz.webs.com/vote
You must disable with jQuery a submit function and check if user clicked on image then you can enable a button for submit.
Also:
1. If you show page> $("input[type=submit]").attr("disabled", "disabled");
2. You need one variable for click test. if user clicked on image COUNTER++
3. If COUNTER > 0 then $("input[type=submit]").removeAttr("disabled");
this is supposed to be simple
i have two pages A and B what i want to do is add a button (a hyperlink or image will do) in page A that when clicked go to page B
how can i do that ?
thanks
If you don't want a button, simply use a html href to f?p=&APP_ID.:<your page number to link to>:&SESSION.
Go to page 2! takes you to your page number 2...
There are many ways. Perhaps the simplest is to create an Apex button using the Create Button wizard. Follow the instructions and when you get to the "Action When Clicked" page, choose "Redirect to Page in this Application" as the Action and enter the Page number you want to go to when the button is clicked. You can also enter other details such as Request, Clear Cache, Set These Items.