为多个配置文件设置弹簧配置

Question

我的应用程序配置文件中有多个配置文件application.yml，如下例所示：

spring:
  application:
    name: my-super-app
database:
  secret: "default secret"
this:
  that: "default value..."

---

spring:
  profiles: staging


---

spring:
  profiles: qa
database:
  secret: "foo bar"

---

spring:
  profiles: playground
database:
  secret: "foo bar"

---

spring:
  profiles: production
database:
  secret: "foo bar"

很明显，我为qa、playground 和production 配置文件冗余设置了database.secret 配置，staging 除外。有没有办法为这三个配置文件设置一次，对配置文件进行分组或从基本配置文件继承？

Answer 1

您可以在逗号分隔的列表 (qa,playground,production) 中组合配置文件，如下所示：

spring:
  application:
    name: my-super-app
common-secret: "foo bar"
database:
  secret: "default secret"
this:
  that: "default value..."

---

spring:
  profiles: staging


---

spring:
  profiles: qa

---

spring:
  profiles: playground

---

spring:
  profiles: production

---

spring:
  profiles: qa,playground,production
database:
    secret: "foo bar"

您也可以设置一个“共享变量”，如下所示：

spring:
  application:
    name: my-super-app
common-secret: "foo bar"
database:
  secret: "default secret"
this:
  that: "default value..."

---

spring:
  profiles: staging


---

spring:
  profiles: qa
database:
  secret: ${common-secret}

---

spring:
  profiles: playground
database:
  secret: ${common-secret}

---

spring:
  profiles: production
database:
    secret: ${common-secret}