konvertieren Konfigurationsdatei application.yml zu application.groovy in Grails 3.x

Question

Ich versuche, ein einfaches Grails 3 Projekt zu erstellen und blieb mit etwas wirklich einfach stecken. Ich möchte, dass meine Datenquelleneigenschaften von VM-Optionen stammen, die ich in meiner IntelliJ-IDE festgelegt habe. Bevor in Grails 2.x, habe ich nur so etwas wie:

environments { 
    development{ 
     //Database connection properties 
     def dbserver = System.properties.getProperty('dbserver') 
     def dbport = System.properties.getProperty('dbport') 
     ............ 
     dataSource { 
      url: "jdbc:sqlserver://${dbserver}:${dbport};databaseName=${dbname} 
     } 
    } 

Nun, da ich habe application.yml, wie greife ich auf „System.properties“ und binde sie in yml? Ich habe gelesen, dass wir stattdessen application.groovy verwenden können, wenn YML es nicht unterstützt, in diesem Fall das ist, was die application.groovy wie folgt aussehen:

grails { 
    profile = 'web' 
    codegen { 
     defaultPackage = 'defPack' 
    } 
} 

info { 
    app { 
     name = '@info.app.name@' 
     version = '@info.app.version@' 
     grailsVersion = '@info.app.grailsVersion@' 
    } 
} 

spring { 
    groovy { 
     template['check-template-location'] = false 
    } 
} 

hibernate { 
    naming_strategy = 'org.hibernate.cfg.DefaultNamingStrategy' 
    cache { 
     queries = false 
    } 
} 

grails { 
    mime { 
     disable { 
      accept { 
       header { 
        userAgents = ['Gecko', 'WebKit', 'Presto', 'Trident'] 
       } 
      } 
     } 

     types { 
      all = '*/*' 
      atom = 'application/atom+xml' 
      css = 'text/css' 
      csv = 'text/csv' 
      form = 'application/x-www-form-urlencoded' 
      html = ['text/html', 'application/xhtml+xml'] 
      js = 'text/javascript' 
      json = ['application/json', 'text/json'] 
      multipartForm = 'multipart/form-data' 
      rss = 'application/rss+xml' 
      text = 'text/plain' 
      hal = ['application/hal+json', 'application/hal+xml'] 
      xml = ['text/xml', 'application/xml'] 
     } 
    } 
    urlmapping { 
     cache { 
      maxsize = 1000 
     } 
    } 
    controllers { 
     defaultScope = 'singleton' 
    } 
    converters { 
     encoding = 'UTF-8' 
    } 
    views { 
     default { codec = 'html' } 
     gsp { 
      encoding = 'UTF-8' 
      htmlcodec = 'xml' 
      codecs { 
       expression = 'html' 
       scriptlets = 'html' 
       taglib = 'none' 
       staticparts = 'none' 
      } 
     } 
    } 
} 
dataSource { 
    pooled = true 
    jmxExport = true 
    driverClassName = 'com.microsoft.sqlserver.jdbc.SQLServerDriver' 
    dbCreate = '' 
    username = 'someUsername' 
    password = 'somePass' 
} 
environments { 
    development { 
     dataSource { 
      url = 'jdbc:sqlserver://localhost:1234;databaseName=someDbName;' 
     } 
    } 
} 

Dank.

UPDATE:

application.groovy ist standardmäßig eingelassenes nicht werden, auch wenn ich application.yml

Answer 1

stellte sich heraus, entfernte ich ein Problem hatte, musste ich ‚default‘ Schlüsselwort setzen innerhalb Zitate. Wie:

grails { 
    profile = 'web' 
    codegen { 
     defaultPackage = 'defPack' 
    } 
} 

info { 
    app { 
     name = '@info.app.name@' 
     version = '@info.app.version@' 
     grailsVersion = '@info.app.grailsVersion@' 
    } 
} 

spring { 
    groovy { 
     template['check-template-location'] = false 
    } 
} 

hibernate { 
    naming_strategy = 'org.hibernate.cfg.DefaultNamingStrategy' 
    cache { 
     queries = false 
    } 
} 

grails { 
    mime { 
     disable { 
      accept { 
       header { 
        userAgents = ['Gecko', 'WebKit', 'Presto', 'Trident'] 
       } 
      } 
     } 

     types { 
      all = '*/*' 
      atom = 'application/atom+xml' 
      css = 'text/css' 
      csv = 'text/csv' 
      form = 'application/x-www-form-urlencoded' 
      html = ['text/html', 'application/xhtml+xml'] 
      js = 'text/javascript' 
      json = ['application/json', 'text/json'] 
      multipartForm = 'multipart/form-data' 
      rss = 'application/rss+xml' 
      text = 'text/plain' 
      hal = ['application/hal+json', 'application/hal+xml'] 
      xml = ['text/xml', 'application/xml'] 
     } 
    } 
    urlmapping { 
     cache { 
      maxsize = 1000 
     } 
    } 
    controllers { 
     defaultScope = 'singleton' 
    } 
    converters { 
     encoding = 'UTF-8' 
    } 
    views { 
     'default' { codec = 'html' }//THIS WAS THE SOURCE OF ERROR 
     gsp { 
      encoding = 'UTF-8' 
      htmlcodec = 'xml' 
      codecs { 
       expression = 'html' 
       scriptlets = 'html' 
       taglib = 'none' 
       staticparts = 'none' 
      } 
     } 
    } 
} 

def dbserver = System.properties.getProperty('dbserver') 
def dbport = System.properties.getProperty('dbport') 
def dbusername = System.properties.getProperty('dbusername') 
def dbpassword = System.properties.getProperty('dbpassword') 
def dbname = System.properties.getProperty('dbname') 

dataSource { 
    pooled = true 
    jmxExport = true 
    driverClassName = 'com.microsoft.sqlserver.jdbc.SQLServerDriver' 
    dbCreate = '' 
    username = dbusername 
    password = dbpassword 
} 
environments { 
    development { 
     dataSource { 
      url = 'jdbc:sqlserver://${dbserver}:${dbport};databaseName=${dbname}' 
     } 
    } 
} 

Answer 2

Kann keine direkte Antwort auf Ihre Frage sein. Sie können Systemeigenschaften in application.yml zugreifen, indem Sie unter Zugabe von

tasks.withType(org.springframework.boot.gradle.run.BootRunTask) { 
systemProperties = System.properties 

Verweis auf build.gradle: https://github.com/grails/grails-core/issues/9086