Can';Grails2.4.4文档中的示例不起作用。(扩展RestfulController超类)

Can';Grails2.4.4文档中的示例不起作用。(扩展RestfulController超类),grails,groovy,Grails,Groovy,我一直试图在Grails2.4.4文档中实现这个示例,但似乎无法编译源代码 我遇到问题的代码是: @Override protected Score queryForResource(Serializable id) { Score.where { id == id && Game.id = params.GameId //Error }.find() } IDE在上面代码中标记为error的行发出错误警告。IDE错误为: ((id == id

我一直试图在Grails2.4.4文档中实现这个示例,但似乎无法编译源代码

我遇到问题的代码是:

@Override
protected Score queryForResource(Serializable id) {
    Score.where {
        id == id && Game.id = params.GameId  //Error
    }.find()
}
IDE在上面代码中标记为error的行发出错误警告。IDE错误为:

((id == id) && Game.id) is a binary expression, but it should be a variable expression at line: 16 column: 33.
  [groovyc] org.codehaus.groovy.control.MultipleCompilationErrorsException: startup failed:
  [groovyc] C:\Users\PC_2\Documents\Grails\WS\grails-app\controllers\ws\ScoreController.groovy: 16: 
  [groovyc] ((id == id) && Game.id) is a binary expression, but it should be a variable expression at line: 16 column: 33. File: C:\Users\PC_2\Documents\Grails\WS\grails-app\controllers\ws\ScoreController.groovy @ line 16, column 33.
  [groovyc]                id == id && Game.id = params.authorId
  [groovyc]                                    ^
  [groovyc] 
  [groovyc] 1 error
编译器错误为:

((id == id) && Game.id) is a binary expression, but it should be a variable expression at line: 16 column: 33.
  [groovyc] org.codehaus.groovy.control.MultipleCompilationErrorsException: startup failed:
  [groovyc] C:\Users\PC_2\Documents\Grails\WS\grails-app\controllers\ws\ScoreController.groovy: 16: 
  [groovyc] ((id == id) && Game.id) is a binary expression, but it should be a variable expression at line: 16 column: 33. File: C:\Users\PC_2\Documents\Grails\WS\grails-app\controllers\ws\ScoreController.groovy @ line 16, column 33.
  [groovyc]                id == id && Game.id = params.authorId
  [groovyc]                                    ^
  [groovyc] 
  [groovyc] 1 error

如果需要其他信息来解决问题,我会尽力提供。谢谢

Game.id=params.GameId
应该是
Game.id==params.GameId
谢谢您的回复。这修复了语法错误,但我现在有另一个问题。你知道我为什么会这样吗?文档中的示例中似乎没有定义它。您不能在
where
query中引用
params
。将其分配到where query外部的另一个变量,并使用该变量。感谢您的帮助,我使应用程序正常工作。