所以按照这个问题(how to plug a TCP-IP client server in a spring MVC application), 我成功地将网关连接到我的 Spring REST Controller 中。但是,我对下一步该去哪里感到困惑。这是我要实现的目标:
1) 当某个路由遇到 POST 请求时,打开从 POST 传递的到某个 IP 的连接(或使用已经使用此 IP 打开的连接)并发送消息。
@RequestMethod(value = '/sendTcpMessage', method=RequestMethod.POST)
public void sendTcpMessage(@RequestParam(value="ipAddress", required=true) String ipAddress,
@RequestParam(value="message", required=true) String message) {
//send the message contained in the 'message' variable to the IP address located
//at 'ipAddress' - how do I do this?
}
2) 我还需要我的 Spring 后端来监听传递给它的 TCP“消息”,并将它们存储在缓冲区中。我的 Javascript 将每 5 秒左右调用一次路由,并从缓冲区中读取信息。
这是我的 Controller 代码:
@Controller
public class HomeController {
@Resource(name = "userDaoImpl")
private UserDAO userDao;
@Resource(name = "receiveTcp")
private ReceiveTcp tcpMessageReceiver;
@Autowired
SimpleGateway gw;
String tcpBuffer[] = new String[100];
@RequestMapping(value="/")
public String home() {
return "homepage";
}
@RequestMapping(value = "/checkTcpBuffer", method=RequestMethod.POST)
public String[] passTcpBuffer() {
return tcpMessageReceiver.transferBuffer();
}
}
根上下文.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:int="http://www.springframework.org/schema/integration"
xmlns:int-ip="http://www.springframework.org/schema/integration/ip"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://www.springframework.org/schema/integration/ip http://www.springframework.org/schema/integration/ip/spring-integration-ip.xsd
http://www.springframework.org/schema/integration/ http://www.springframework.org/schema/integration/spring-integration.xsd">
<!-- Root Context: defines shared resources visible to all other web components -->
<int:gateway id="gw"
service-interface="net.codejava.spring.interfaces.SimpleGateway"
default-request-channel="input"/>
<bean id="javaSerializer"
class="org.springframework.core.serializer.DefaultSerializer"/>
<bean id="javaDeserializer"
class="org.springframework.core.serializer.DefaultDeserializer"/>
<int-ip:tcp-connection-factory id="server"
type="server"
port="8081"
deserializer="javaDeserializer"
serializer="javaSerializer"
using-nio="true"
single-use="true"/>
<int-ip:tcp-connection-factory id="client"
type="client"
host="localhost"
port="8081"
single-use="true"
so-timeout="10000"
deserializer="javaDeserializer"
serializer="javaSerializer"/>
<int:channel id="input" />
<int:channel id="replies">
<int:queue/>
</int:channel>
<int-ip:tcp-outbound-channel-adapter id="outboundClient"
channel="input"
connection-factory="client"/>
<int-ip:tcp-inbound-channel-adapter id="inboundClient"
channel="replies"
connection-factory="client"/>
<int-ip:tcp-inbound-channel-adapter id="inboundServer"
channel="loop"
connection-factory="server"/>
<int-ip:tcp-outbound-channel-adapter id="outboundServer"
channel="loop"
connection-factory="server"/>
<int:service-activator input-channel="input" ref="receiveTcp" method = "saveValue"/>
</beans>
ReceiveTcp.java
@Component(value = "receiveTcp")
public class ReceiveTcp {
String buf[] = new String[100];
int currentPosition = 0;
@ServiceActivator
public void saveValue(String value){
System.out.println(value);
buf[currentPosition] = value;
currentPosition++;
}
public String[] transferBuffer() {
String tempBuf[] = new String[100];
tempBuf = buf;
buf = new String[100];
return tempBuf;
}
}
我该如何解决这些问题?
最佳答案
您需要使用 ResponseEntity
。看这个answer . tcp 连接是双向连接,因此如果您的方法返回响应而不是 void
,它会自动将其发送回向您发出请求的 ip。您不必手动执行此操作。
关于java - 如何设置 Spring TCP 客户端和服务器模型?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26936721/