Question

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

Answer 1

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“加载平衡规则”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“探头”字段中，选择之前创建的探头。

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

Azure Service结构，通过域名访问应用程序

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