groovy - 如何使 configslurper 委托方法调用调用者类

Question

我想创建一个像这样的 groovy 类 og 脚本：

//...
def slurper = new ConfigSlurper().parse(someFile)
//...
//The exact method declaration
def methodCall(def arg){
   //Whatever i want to do
}
//Maybe it is easier with methodMissing
def methodMissing(def args) {
   //Whatever i want to do
}

要 slurp 的文件可能如下所示：

some {
   property = methodCall("with arg")
}

问题是我如何将“methodCall”委托给使用 configslurper 解析的类或脚本？目前它会给你一个methodMissing。

Answer 1

我认为这篇博客文章有一个你想要做什么的例子。它比 a 更复杂，methodMissing但可以完成。

Answer 2

感谢 Sérgio Michels 链接，我找到了解决方案：

public class ScriptWithMethods extends Script {

   String scriptText;

   public ScriptWithMethods(File file) {
      scriptText = file.text
   }

   public void run() {
      GroovyShell shell = new GroovyShell();
      Closure closure = shell.evaluate("{it->$string}");
      closure.resolveStrategy = Closure.DELEGATE_FIRST
      closure.delegate = this
      closure.call()
   }

   def methodCall(def arg){
      //Whatever i want to do
   }
}

//...
def script = new ScriptWithMethods(scriptText:someFile)
def slurper = new ConfigSlurper().parse(script)

当然，您也可以使用“缺少方法”，但这适用于我的用例