从 Kotlin Psi API 检索继承类全名

Question

我\xe2\x80\x99m 尝试开发一个codegen IDEA-Plugin。这个插件应该分析KtClass继承并获取所有继承类的全名（例如 com.example.config.TestConfig）

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我尝试通过查看 PsiViewer 来查找任何有用的信息。我发现KtClass的所有继承信息都存储在 中KtSuperTypeEntry，并且我尽力获取继承类的全名。

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上课Dest：

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data class Dest(\n    val stringValue: String = "123",\n    override val stringConfig: String = "123",\n    override val iConfigStr: String = "123",\n    val b: B = B(),\n    val c: List<List<Set<Map<String, out Any?>>>> = listOf(),\n    val config: Config? = Config()\n) : Config()\n

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1. superTypeListEntry.typeAsUserType.referenceExpression.getReferencedName() -return->"Config"
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2. superTypeListEntry.importReceiverMembers() -return->null
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看起来SuperTypeListEntry只包含继承类的简单名称信息。

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我还尝试通过 KtFile 查找继承类全名，但不知道继承类何时作为通配符在此 KtFile 中导入：

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fun KtSuperTypeListEntry.getType(ktFile: KtFile): String {\n    val simpleName = superTypeListEntry.text\n\n    // try to find by declared KtClass ...\n    ktFile.children.filterIsInstance<KtClass>().filter { it.name == simpleName }\n\n    // try to find by import ...\n    ktFile.importDirectives.filter { it.importPath.toString().contains(simpleName) }\n\n    \n    // try to find by import wildcards ...\n    ktFile.importDirectives.filter { it.importPath.toString().endWith("*") }.forEach {\n        val split = it.importPath.split(".")\n        split.set(split.size - 1, simpleName)\n        val maybeFullName = split.joinToString(",") { it }\n        // confused on how to detect "maybeFullName" is correct ...\n    }\n}\n

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• 问题
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如何从 Kotlin Psi API 检索所有继承类的全名？谢谢你！

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Answer 1

经过数千次的调查和调试，我发现通过BindingContext找到一个类的继承类是可能的。BindingContext 可以分析 TypeReference 并找到 的引用KotlinType。代码可能是这样的：

ktClass.superTypeListEntries.map { superTypeEntry ->
    val typeReference = superTypeEntry.typeReference
    val bindingContext = typeReference.analyze()
    bindingContext.get(BindingContext.TYPE, typeReference)
}.forEach { kotlinType ->
    val classId = kotlinType.constructor.declarationDescriptor.classId
    val packageName = classId.packageFqName
    val simpleName = classId.relativeClassName
    // It can also get the generics of this class by KotlinType.arguments
    val generics = kotlinType.arguments
}

另外，你可以通过 获取超类型的类全名KtLightClass，代码可能是这样的：

val ktLightClass = ktClass.toLightClass()
val superTypesFullName = ktLightClass?.supers?.forEach { superType ->
    val packageName = superType.qualifiedName
    val simpleName = superType.name
    // you can get as KtClass by this, which can help you unify design of interface.
    val ktClass = superType.kotlinOrigin
}