Kotlin中的错误"不能为空"

Question

.zip文件中有多个文件,我正在尝试获取.尝试解压缩文件提供了java.lang.IllegalStateException:zis.nextEntry不能为null.怎么做正确的方法？

@Throws(IOException::class)
    fun unzip(zipFile: File, targetDirectory: File) {
        val zis = ZipInputStream(
                BufferedInputStream(FileInputStream(zipFile)))
        try {
            var ze: ZipEntry
            var count: Int
            val buffer = ByteArray(8192)
            ze = zis.nextEntry
            while (ze != null) {
                val file = File(targetDirectory, ze.name)
                val dir = if (ze.isDirectory) file else file.parentFile
                if (!dir.isDirectory && !dir.mkdirs())
                    throw FileNotFoundException("Failed to ensure directory: " + dir.absolutePath)
                if (ze.isDirectory)
                    continue
                val fout = FileOutputStream(file)
                try {
                    count = zis.read(buffer)
                    while (count != -1) {
                        fout.write(buffer, 0, count)
                        count = zis.read(buffer)
                    }
                } finally {
                    fout.close()
                    zis.closeEntry()
                    ze = zis.nextEntry
                }
            }
        } finally {
            zis.closeEntry()
            zis.close()
        }
    }

Answer 1

在ZipEntry你从流中读取会null当你达到了文件的末尾,所以你必须让你将其存储在可空的变量:

var ze: ZipEntry?

您可以将读取的值分配给不可为空的变量,因为它们具有平台类型ZipEntry!,因为它是Java API - 在这种情况下,您必须确定它是否可以null.有关更多信息,请参阅有关平台类型的文档.