The information comes from the internet. Please forget it. application scenario: assume that there are three devices A, B, and C, AB can access each other, BC can access each other, A and C cannot access each other directly, and Oracle database is installed on C, how can A access C databases? Specific environment: A (192.168.8.253), windowvistaB (192.168.8.150), and linuxC application scenarios:
Assume that there are three devices A, B, and C. A <=> B can access each other, B <=> C can access each other, and A and C cannot access each other directly, oracle Database is installed on C. How can A access C databases?
Specific environment:
A (192.168.8.253), window vista
B (192.168.8.150), linux CentOS 5.5
C (192.168.8.49), linux Redhat AS4, Oracle (SID: ora9i, Port: 1521)
Procedure:
1. B (192.168.8.150) allows IP packet forwarding. modify net. ipv4.ip _ forward = 1 in the configuration file/etc/sysctl. conf (1: allows forwarding. the default value is 0)
PS: There may be differences between linux versions. this article passes the test under CentOS5.5
2. modify B (192.168.8.150)
IptablesRules in:
Shell script:
$ Iptables-t nat-a prerouting-p tcp-m tcp -- dport 11521-jDNAT -- to-destination 192.168.8.49: 1521
$ Iptables-t nat-a postrouting-p tcp-m tcp -- dport 1521-jSNAT -- to-source 192.168.8.150
$ Service iptables save
$ Service iptables restart
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