Kev*_*n S 3 java http httpurlconnection
我想打开一个 URL 并向其提交以下参数,但似乎只有在我的代码中添加 BufferedReader 时它才有效。这是为什么?
Send.php 是一个脚本,它将向我的数据库添加用户名和时间。
以下代码不起作用(它不会向我的数据库提交任何数据):
final String base = "http://awebsite.com//send.php?";
final String params = String.format("username=%s&time=%s", username, time);
final URL url = new URL(base + params);
final HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestProperty("User-Agent", "Agent");
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.connect();
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但这段代码确实有效:
final String base = "http://awebsite.com//send.php?";
final String params = String.format("username=%s&time=%s", username, time);
final URL url = new URL(base + params);
final HttpURLConnection connection = (HttpURLConnection) url.openConnection();
connection.setRequestProperty("User-Agent", "Agent");
connection.setRequestMethod("POST");
connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
connection.connect();
final BufferedReader in = new BufferedReader(new InputStreamReader(connection.getInputStream()));
String line;
while ((line = in.readLine()) != null) {
System.out.println(line);
}
connection.disconnect();
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据我所知。当您调用该connect()函数时,它只会创建连接。
您至少需要调用getInputStream()或getResponseCode()来提交连接,以便 url 指向的服务器能够处理该请求。
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