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如何从URL Scheme iPhone再次打开应用程序

[英]How to open an application again from URL Scheme iPhone

I am trying to implement URL Schemes in my iPhone. 我正在尝试在我的iPhone中实现URL方案。

Actually i Want to open a transaction page in my iPhone application. 实际上,我想在iPhone应用程序中打开交易页面。 It opens in some browser lets say i opened http:www.xxxx.com. 它在某些浏览器中打开,可以说我打开了http://www.xxxx.com。 Now it has some php code written at backend. 现在,它在后端编写了一些PHP代码。 There is a submit button. 有一个提交按钮。 I submit , it process the php code and do obtain some response. 我提交,它处理php代码并获得一些响应。

Now I want to open the application again and get that response. 现在,我想再次打开应用程序并获得该响应。 How can I call my application from my php code of that webpage I called and pass that response? 我如何从该网页的php代码中调用我的应用程序并传递该响应?

you must add url scheme in your application. 您必须在应用程序中添加url方案。 Go in Xcode to your target -> info->URL Types and create an url scheme like "myapp". 在Xcode中进入目标-> info-> URL Types,然后创建一个类似于“ myapp”的URL方案。 In your app delegate implement the - canOpenUrl... method and return yes when you want your app to launch. 在您的应用程序委托中,实现canOpenUrl...方法,并在您希望启动应用程序时返回yes。

On your php side, you will need to tell the browser to open the address myapp://dataTopasstoYourApplicationThatYouCanParseInTheCanOpenURLMethodInYourAppDelegate 在PHP方面,您将需要告诉浏览器打开地址myapp:// dataTopasstoYourApplicationThatYouCanParseInTheCanOpenURLMethodInYourAppDelegate

basically you do this with some javascript like location.somethingIDontRemember... = myapp://... or in PHP with header location 基本上,您可以使用一些JavaScript来完成此操作,例如location.somethingIDontRemember... = myapp://...或在具有标头位置的PHP中

<?php
header('Location: myapp://whatYouWant');
exit;
?>

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