user2990084 user2990084 - 2 months ago 36
Android Question

Open url in webview - phonegap

I would like to know how can I open an url in the app context of embed webview. Currently this demo will open a new tab in external browser, so, not what I am expected. I am using google.com just for testing.

Summary, I am looking for a functional demo.

<?xml version="1.0" encoding="UTF-8"?>

<!-- config.xml reference: https://build.phonegap.com/docs/config-xml -->
<widget xmlns = "http://www.w3.org/ns/widgets"
xmlns:gap = "http://phonegap.com/ns/1.0"
xmlns:android = "http://schemas.android.com/apk/res/android"
id = "com.xxx.xxxxx"
version = "1.0.0">

<preference name="stay-in-webview" value="true" />

<access origin="*" browserOnly="true" subdomains="true" />

<content src="index.html" />

<allow-navigation href="https://google.com/*" />

<gap:plugin name="cordova-plugin-whitelist" source="npm" version="~1" />
<gap:plugin name="org.apache.cordova.inappbrowser" />
<gap:plugin name="org.apache.cordova.splashscreen" />

<preference name="phonegap-version" value="cli-5.4.1" />
<preference name="permissions" value="none"/>
<preference name="target-device" value="universal"/>
<preference name="fullscreen" value="true"/>

</widget>





<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/index.css" />
</head>
<body>
<div>
<script type="text/javascript" charset="utf-8">
document.addEventListener("deviceready", onDeviceReady, false);

function onDeviceReady() {
window.location.href = 'https://google.com';
}
</script>
</div>
<script type="text/javascript" src="cordova.js"></script>
</body>
</html>


Update:
Complete xml file:
https://codeshare.io/Vw3Fl

Answer

try :

window.open('https://google.com', '_self ', 'location=yes');

instead of :

window.location.href = 'https://google.com';

This will use the InAppBrowser, and use _self as target.

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