dongsou0083 2017-05-18 09:45
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在Android中调用网址。 无需等待任何回复

In my project I need to call a URL .

This urls, looks like: http://myip:80/increaseCounter.php

I got a Xampp (MAC) with mysql. The functionallity of increaseCounter.php, is retrive the value from the table, increase it in 1, and update it. The url I tested in localhost, pointing from another device via Chrome, and works perfect.

But know I'm trying to do just a call to this link in Android, I don't need to recover any data or similar, just call it.

I tried to do:

private void trackUrl(String href) {
    HttpURLConnection con = null;
    try {
        URL url = new URL(href);
        con = (HttpURLConnection) url.openConnection();
        con.connect();
    } catch (Exception e) {
        Log.d("APP", e.getMessage());
    } finally {
        if (con != null) {
            con.disconnect();
        }
    }
}

But is no working.No connection log in xampp, and the data in mysql ,is not updated.

=======

<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools"
    package="my.aswsome.app"
    android:versionCode="15">

    <uses-permission android:name="android.permission.CAMERA" />
    <uses-permission android:name="android.permission.FLASHLIGHT" />

    <uses-feature
        android:name="android.hardware.camera"
        android:required="false" />
    <uses-feature
        android:name="android.hardware.camera.autofocus"
        android:required="false" />

    <uses-permission android:name="android.permission.INTERNET" />
.....
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2条回答 默认 最新

  • dongyou9373 2017-05-18 10:30
    关注

    Try

    int code = urlConnection.getResponseCode();
    if(code==200){ 
        Log.i("TAG","success");
    } 
    

    instead of con.connect();

    https://developer.android.com/reference/java/net/URLConnection.html#connect()
    .connect(); only opens the connect but does nothing after. So by doing .getResponseCode() you actually do a request.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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