正如标题所说...当用户单击 Java Swing 应用程序中的按钮时,我尝试使用以下代码执行 PHP 脚本:
URL url = new URL( "http://www.mywebsite.com/my_script.php" );
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
conn.connect();
但什么也没发生...有什么不对 ?
我认为您错过了下一步,如下所示:
InputStream is = conn.getInputStream();
HttpURLConnection
基本上只打开套接字connect
为了做某事,你需要做一些事情,比如打电话getInputStream()
或者更好getResponseCode()
URL url = new URL( "http://google.com/" );
HttpURLConnection conn = (HttpURLConnection) url.openConnection();
if( conn.getResponseCode() == HttpURLConnection.HTTP_OK ){
InputStream is = conn.getInputStream();
// do something with the data here
}else{
InputStream err = conn.getErrorStream();
// err may have useful information.. but could be null see javadocs for more information
}
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