Azure Service结构,通过域名访问应用程序

我的要求是基于租户域访问Fabric应用程序

eg:  i  had  3  services  

1)CustomerApps
2)CompanyApps
3)SignInApps

All  the   above  apps  deployed  on  azure   service  fabric  Cluster

我已经在Azure Active目录中创建了自定义域并使用Register验证该域(例如:godady)

So,  now   i  want  to  display   each  fabric  Application  based  on  Tenant/Domain  name like, assumed  i  had  custom  domain  dev-tennat.xyz.com

so,application   url  will  be  
dev-tennat.xyz.com/SignInApps
dev-tennat.xyz.com/CompanyApps
dev-tennat.xyz.com/CustomerApps

最佳答案 1)您的域名例如“your-domain.xyz.com”应为A类型,并且必须指向您的负载均衡器公共IP.您可以在负载均衡器的概述下找到IP.

lbip

2)您需要为负载余额配置自定义探针.

go to “Probes” under “Settings” category.
Select HTTP protocol
i’ve set port to 80 as i wanted something to be visible by default
In “Path” field you specify endpoint you want to expose it can be “/” or “/api/customapi”
Interval – set value in seconds. Every ‘n’ seconds your “Path” will be triggered to check if app is alive

3)您需要配置自定义规则

转到“设置”类别下的“负载平衡规则”并创建新规则.创建新规则时:

“Port” is your publicly exposed port (80 in my case),
“Backend port” is internal port of your app (you can find it in service manifest in <EndPoint /> element. In my case it’s 8164)

在“探测”(Probe)字段中,选择之前创建的探测器.

如果需要,有一些屏幕截图:
https://eduardlos.wordpress.com/2016/10/10/how-to-access-your-app-when-deployed-to-azure-service-fabric-cluster/

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