kotlin - 从 Kotlin Psi API 检索继承类全名

Question

我正在尝试开发一个 codegen IDEA-Plugin。这个插件应该分析KtClassInheritance 并获取所有继承类的全名（如 com.example.config.TestConfig）

我试图通过查看 PsiViewer 找到任何有用的信息。我发现 KtClass 的所有继承信息都存储在 中KtSuperTypeEntry，我尽力获取继承类的全名。

对于类Dest：

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

1. superTypeListEntry.typeAsUserType.referenceExpression.getReferencedName() -return->"Config"
2. superTypeListEntry.importReceiverMembers() -return->null

看似 SuperTypeListEntry 只包含继承类简单名称信息。

我也尝试通过 KtFile 查找继承类全名，但不知道何时将继承类作为通配符导入到此 KtFile 中：

fun KtSuperTypeListEntry.getType(ktFile: KtFile): String {
    val simpleName = superTypeListEntry.text

    // try to find by declared KtClass ...
    ktFile.children.filterIsInstance<KtClass>().filter { it.name == simpleName }

    // try to find by import ...
    ktFile.importDirectives.filter { it.importPath.toString().contains(simpleName) }

    
    // try to find by import wildcards ...
    ktFile.importDirectives.filter { it.importPath.toString().endWith("*") }.forEach {
        val split = it.importPath.split(".")
        split.set(split.size - 1, simpleName)
        val maybeFullName = split.joinToString(",") { it }
        // confused on how to detect "maybeFullName" is correct ...
    }
}

• 问题

如何从 Kotlin Psi API 检索所有继承类全名？谢谢！

Answer 1

经过千方百计的排查和调试，我发现通过BindingContext可以找到一个类的继承类。BindingContext 可以分析 TypeReference 并找到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 获取类的超类型全名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
}