我想從我的Java程序中調用一些PHP腳本,這樣我就可以查詢/修改我的數據庫。我對數據庫完全陌生,並不知道發生了什麼。從Java調用PHP
我想從java內部調用下面的腳本。
腳本
<?php
include('test.php');
define("HOST", "localhost");
define("USERNAME", "xxxx");
define("PASSWORD", "xxxx");
define("DATABASE", "project_database");
mysql_connect(HOST, USERNAME, PASSWORD);
mysql_select_db(DATABASE);
?>
Java代碼
public static String excutePost(String targetURL, String urlParameters)
{
URL url;
HttpURLConnection connection = null;
try {
//Create connection
url = new URL(targetURL);
connection = (HttpURLConnection)url.openConnection();
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type",
"application/x-www-form-urlencoded");
connection.setRequestProperty("Content-Length", "" +
Integer.toString(urlParameters.getBytes().length));
connection.setRequestProperty("Content-Language", "en-US");
connection.setUseCaches (false);
connection.setDoInput(true);
connection.setDoOutput(true);
//Send request
DataOutputStream wr = new DataOutputStream (
connection.getOutputStream());
wr.writeBytes (urlParameters);
wr.flush();
wr.close();
//Get Response
InputStream is = connection.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(is));
String line;
StringBuffer response = new StringBuffer();
while((line = rd.readLine()) != null) {
response.append(line);
response.append('\r');
}
rd.close();
return response.toString();
} catch (Exception e) {
e.printStackTrace();
return null;
} finally {
if(connection != null) {
connection.disconnect();
}
}
}
應該採取什麼目標網址是?
我提醒你「詢問」按鈕是用來提問的。你想知道什麼?此外,除非您有不切實際的期望,否則您的[tag:php],[tag:sql]和[tag:database]標記完全不相關... –
謝謝您提醒我! – TomSelleck