Question

尝试使用Undertow。这里有一个简单的例子：

public class MyExample {

private static int SIMPLE_HANDLER_CALL = 0;
private static int LIBRE_OFFICE_CALL = 0;

public static void main(String[] args) {

    Undertow server = Undertow.builder()
            .addHttpListener(3333, "localhost")
            .setHandler(new SimpleHttpHandler())
            .build();

    server.start();
}

private static class SimpleHttpHandler implements HttpHandler{
    @Override
    public void handleRequest(HttpServerExchange exchange) throws Exception {
        System.out.println("---------------------------------------------");
        System.out.println("start handleRequest()");
        exchange.getResponseHeaders().put(Headers.CONTENT_TYPE, "text/plain");
        connectToPg(exchange, "msgPrev = " + Integer.toString(SIMPLE_HANDLER_CALL) + "; msgCur = " + Integer.toString(++SIMPLE_HANDLER_CALL));
        System.out.println("end handleRequest()");
    }
}

private static void connectToPg(HttpServerExchange exchange, String msg){
    try(
            Connection connection = DriverManager.getConnection("jdbc:postgresql://10.10.2.158:5432/myDb", "myUser", "myPass");
            Statement st = connection.createStatement();
    )
    {
        ResultSet rs = st.executeQuery("select count(*) as CNT from event.event");

        java.util.Date now = new java.util.Date();
        while(rs.next()){
            int cnt = rs.getInt("CNT");
            System.out.print("cnt = " + cnt);
            exchange.getResponseSender().send("Date = " + now + "; cnt = " + cnt);
        }

        System.out.println("rs = " + rs.toString());
        System.out.println("msg = " + msg);
    }
    catch (Exception ex){
        System.out.println(ex.getMessage());
    }
    finally {
        System.out.println("end connectToPg()");
    }
}
}

有效。我输入

http://localhost:3333/

在浏览器和处理程序中创建查询到Postgres并接收答案。但是，处理程序重复2次！我在控制台中得到了这样的输出：

---------------------------------------------
start handleRequest()
cnt = 12rs = org.postgresql.jdbc4.Jdbc4ResultSet@12eeffa4
msg = msgPrev = 0; msgCur = 1
end connectToPg()
end handleRequest()
---------------------------------------------
start handleRequest()
cnt = 12rs = org.postgresql.jdbc4.Jdbc4ResultSet@17b98b20
msg = msgPrev = 1; msgCur = 2
end connectToPg()
end handleRequest()

为什么叫它2次？

Answer 1

我建议打印（或注销）交换参数以查看正在处理的请求类型。

System.out.println(exchange.getRequestURL());

一个选项是第二个请求是针对/favicon.ico的。您还可以尝试以更加可控的方式提出请求 - 即。使用其中一个REST客户端甚至卷曲。

取消多个处理程序调用

1 个答案: