[英]Open android application by click on website button?
I have android application which contain html webpage with one single button. 我有一个包含一个按钮的html网页的android应用程序。 The html file is placed into the asset folder in android. html文件被放置在android中的asset文件夹中。 I open html file into the webview. 我将html文件打开到webview中。 Now, i what want to open the another application which is install in android device by click on the html page button. 现在,我想通过单击html页面按钮打开另一个安装在android设备中的应用程序。
Can you please anyone help me how i will do that and also give me the some reference from where i will get some help? 能否请任何人帮助我我将如何做,并从我能获得帮助的地方给我一些参考?
I think @Sk Maniruddin is asking how to open another app from a webpage. 我认为@Sk Maniruddin正在询问如何从网页打开另一个应用程序。 And the solution to that is using an intent, as the link. 解决此问题的方法是使用意图作为链接。
<a href="intent://scan/#Intent;scheme=zxing;package=com.google.zxing.client.android;end"> Take a QR code </a>
for more information you can look at the doc: https://developer.chrome.com/multidevice/android/intents#example 有关更多信息,请查看文档: https : //developer.chrome.com/multidevice/android/intents#example
Firstly set your onclik method of button in html like this. 首先像这样在html中设置按钮的onclik方法。
<input type="submit" name="submit" id="submit_id" onclick="click.performClick();" />
enable javascript your webview 启用javascript您的webview
webView.getSettings().setJavaScriptEnabled(true);
after that you can detect this click by this code: 之后,您可以通过以下代码检测到此点击:
webView.addJavascriptInterface(new Object() {
@JavascriptInterface
public void performClick() {// detected click
Intent i = new Intent(FromActivity.This, DestinationActivity.class);
startActivity(i);
}
}, "click");
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