Hi, I am building an Android app. I have a Webview component in which I can successfully load my site and follow the links. I need to do the following and need your help -
I reach a specific URL say http://abc.com/story/. This webpage opens in the webview and it has some links in it. I need to open the Links of that page to open in the Phone’s Browser with some notification like “The link will open out of the App. Do you want to open the link in Chrome etc.?”- followed by two buttons- “Yes” and “Cancel”. If the user taps “Yes” the link opens in external Browser. Tapping “Cancel” puts the user in the webview and does not open the link.
Please help.
on yes , use this block it helps you,
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A little more samiu8336. Please guide to make the Notification and Buttons.
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there 2 methods for that, first is simple with notifier, like this, on yes do your action
second is make a dailouge in designer custom, and with notifier load in initialize like this, on action show, and set 2 button of your choice in that and on that button click do your work,
try these
where are you using this if else?
use like this, and after yes when user click on page then it goes for external browser, not directly,
And no Message or Buttons showing.
your link is not valid, try to add a label on screen and set label text to webview1 current url, and check what link is showing that on that page, i think that must contain, www in page, check with label
Using www is not mandatory.
Any progress on this front? In another app I have used the same format of URL https://amarclass.in and the App was working find without the www part.
FANTASTIC!!! GREAT. Please tell me where to get the Label block?
Label is not important i just used label to check web link, not important for you
Imran can you please share the aio with me? Don’t know why but I can’t produce the results.
ok wait, let me create that
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you work is on screen 2, check there, on first button is your link, try that
testing_web.aia (28.0 KB)
Thank You Imran. Thank You very much