Question

我正在使用在端口8080上运行的Glassfish服务器运行我的Web应用程序。对于同一个Web应用程序，我尝试使用node.js集成Stripe API。我的其他Web应用程序在localhost：8080上运行。

那么我如何通过node.js和glassfish监听相同的8080端口，以便我的Web应用程序与Stripe node.js集成。

我应该使用网络套接字吗？

HTML页面：

    var express = require('express');
    var bodyParser = require('body-parser');
    var stripe = require('stripe')('sk_test_bpfjQsY5iK7ZI7W5tJMKpPli');
    var http = require('http');

    var app = express();
    app.use(bodyParser.urlencoded({
          extended: true
    }));
    app.use(bodyParser.json());

      app.post('/acctCreated', function(req, res) { 
        console.log('Inside charge');
        //var stripeToken = req.body.stripeToken;
        var stripeToken = req.body.id;
        var amount = 1000;

        console.log('Calculating charge');
        stripe.charges.create({
            card: stripeToken,
            currency: 'usd', 
            amount: amount
        },
        function(err, charge) {
            if (err) {
                res.send(500, err);
                //res.status(500).send(err);
            } else {
                res.send(204);
                //res.status(204).send(charge);
            }
        });

          var path = "http://localhost:8080/TropoHotelReserv/faces/roomsBooked.xhtml"  ;
          console.log("Get PathName " + path);
          res.writeHead(302, {'Location': path});
          res.end();

        console.log('Complete');
    });

    app.use(express.static(__dirname));
    app.listen(process.env.PORT || 8080);

index.js：

    int pid = fork();
    if(pid == 0)
    {
        close(tcp_socket);
        char buffer[255]={0};
        int bytes_num;
        char serID[] = "0";
        //std::string serID = "0";
       // char serID[7] = "0";
        while(true)
        {
            //recv(int sockfd, void *buf, size_t len, int flags);
            bytes_num = recv(tcp_socket2,buffer,254,0);
            if(bytes_num == 0)
            {
                break;
            }
            if(serID[] == "0")---------------->problem here
            {
                serID[]= buffer[0];
            }
            printf("serverID %s\n", serID);

            if(serID[]=="serverA")---------------->problem here
            {
                 //printf("NOW files!\n");
                strcpy(str_file1[msg_num1],buffer);
                msg_num1++;
            }
            if(serID[]=="serverB")---------------->problem here
            {
                strcpy(str_file2[msg_num2],buffer);
                msg_num2++;
            }
            if(serID[]=="serverC")---------------->problem here
            {
                strcpy(str_file3[msg_num3],buffer);
                msg_num3++;
            }
            if(serID[]=="serverD")---------------->problem here
            {
                strcpy(str_file4[msg_num4],buffer);
                msg_num4++;
            }

            memset(buffer,0,sizeof(buffer));
        }


        if((serID[]=="serverA"))---------------->problem here
        {
            pf_file1 = fopen("rcvd_fileA.txt","w");
            for(int i=0; i< msg_num1;i++)
            {
                 //printf("now we read files!\n");
                fprintf(pf_file1,"%s\n",str_file1[i]);
            }
            fclose(pf_file1);
        }
        if((serID[]=="serverB"))---------------->problem here
        {
            pf_file2 = fopen("rcvd_fileB.txt","w");
            for(int i=0; i< msg_num2;i++)
            {
                fprintf(pf_file2,"%s\n",str_file2[i]);
            }
            fclose(pf_file2);
        }
        if((serID[]=="serverC"))---------------->problem here
        {
            pf_file3 = fopen("rcvd_fileC.txt","w");
            for(int i=0; i< msg_num3;i++)
            {
                fprintf(pf_file3,"%s\n",str_file3[i]);
            }
            fclose(pf_file3);
        }
        if((serID[]=="serverD"))---------------->problem here
        {
            pf_file4 = fopen("rcvd_fileD.txt","w");
            for(int i=0; i< msg_num4;i++)
            {
                fprintf(pf_file4,"%s\n",str_file4[i]);
            }
            fclose(pf_file4);
        }

        printf("The Client receives neighbor information from the Server %s with TCP port number %d and IP address %s \n",serID.c_str(),ntohs(sin.sin_port), ipstr);

        close(tcp_socket2);

        // printf("The Server %c has the following neighbor information: \n",serverID);

        return 0;
    }

谢谢，

Answer 1

你不能直接这样做。你需要一个负载平衡器/路由器在前面听8080。

在端口8081上运行Glassfish，在8082上运行Node.js，然后在前面使用负载均衡器（例如stud，haproxy，apache httpd或varnish）并设置localhost：8081和localhost：8082作为相应的后端网址路径。

以下是以这种方式使用HAProxy的示例

您可以使用单个HAProxy服务器根据URL和负载平衡隔离请求。您的配置将具有以下内容：

frontend http
acl app1 path_end -i /app1/123 #matches path ending with "/app/123"
acl app2 path_end -i /app2/123 
acl app3 path_end -i /app3/123 


use_backend srvs_app1    if app1
use_backend srvs_app2    if app2
use_backend srvs_app3    if app3

backend srvs_app1 #backend that lists your servers. Use a balancing algorithm as per your need.
   balance roundrobin 
   server host1 REGION1_HOST_FOR_APP1:PORT 
   server host2 REGION2_HOST_FOR_APP1:PORT

backend srvs_app2
   balance roundrobin
   server host1 REGION1_HOST_FOR_APP2:PORT 
   server host2 REGION2_HOST_FOR_APP2:PORT

backend srvs_app3
   balance roundrobin
   server host1 REGION1_HOST_FOR_APP3:PORT 
   server host2 REGION2_HOST_FOR_APP3:PORT

更多信息可以在[主页] [1]上找到。

[1]：http://haproxy.1wt.eu/download/1.5/doc/configuration.txt

来源：https://stackoverflow.com/a/20640578

在Windows上，您可以在Apache httpd：

中使用mod_proxy

ProxyPass /nodejspath http://localhost:8081/nodejspath
ProxyPass /glassfishpath http://localhost:8081/glassfishpath

更多信息：http://httpd.apache.org/docs/2.2/mod/mod_proxy.html

如何让Node.js和Glassfish服务器监听同一个端口？

1 个答案: