我在 Java 中管理良好的代码和项目。但我需要用它在 Kotlin 中开发另一个项目。所以，我在 Kotlin 中尽可能地转换了所有代码。但是有 ZipFileManager.kt 的代码，用于压缩/解压缩文件。
这是代码(Kotlin):
object ZipFileManager {
private val BUFFER_SIZE = 6 * 1024
@Throws(IOException::class)
fun zip(files: Array<String>, zipFile: String) {
var origin: BufferedInputStream? = null
val out = ZipOutputStream(BufferedOutputStream(FileOutputStream(zipFile)))
try {
val data = ByteArray(BUFFER_SIZE)
for (file in files) {
val fi = FileInputStream(file)
origin = BufferedInputStream(fi, BUFFER_SIZE)
try {
val entry = ZipEntry(file.substring(file.lastIndexOf("/") + 1))
out.putNextEntry(entry)
var count: Int
while ((count = origin.read(data, 0, BUFFER_SIZE)) != -1) {
out.write(data, 0, count)
}
} finally {
origin.close()
}
}
} finally {
out.close()
}
}
fun unzip(zipFileUrl: String, fileLocation: String) {
try {
val f = File(fileLocation)
if (!f.isDirectory) {
f.mkdirs()
}
ZipInputStream(FileInputStream(zipFileUrl)).use { zin ->
var ze: ZipEntry? = null
while ((ze = zin.nextEntry) != null) {
// Log.e("UnZipFILE", "Unzipping....");
val path = fileLocation + ze!!.name
if (ze.isDirectory) {
val unzipFile = File(path)
if (!unzipFile.isDirectory) {
unzipFile.mkdirs()
}
} else {
FileOutputStream(path, false).use { fout ->
val buffer = ByteArray(1024)
var read: Int
while ((read = zin.read(buffer)) != -1) {
fout.write(buffer, 0, read)
}
zin.closeEntry()
}
}
}
}
} catch (e: Exception) {
e.printStackTrace()
Log.e("UnZipException", Log.getStackTraceString(e))
}
}
}
所以，我正在尝试通过这段代码，但它显示编译时错误，如:
赋值不是表达式，在此上下文中只允许使用表达式 fun zip 中的行 while ((count = origin.read(data , 0, BUFFER_SIZE)) != -1)
并在 while ((ze = zin.nextEntry) != null) 行和 while ((read = zin.read(buffer) ) != -1).
所以，我的大问题是在 Kotlin 中使用这段代码。因此，任何了解 Kotlin 的人都可以提供帮助，以及如何在 Kotlin 中使用这种类型的循环结构？
如果有人想看的话，我还有Java代码:
public class ZipFileManager {
private static int BUFFER_SIZE = 6 * 1024;
public static void zip(String[] files, String zipFile) throws IOException {
BufferedInputStream origin = null;
ZipOutputStream out = new ZipOutputStream(new BufferedOutputStream(new FileOutputStream(zipFile)));
try {
byte data[] = new byte[BUFFER_SIZE];
for (String file : files) {
FileInputStream fi = new FileInputStream(file);
origin = new BufferedInputStream(fi, BUFFER_SIZE);
try {
ZipEntry entry = new ZipEntry(file.substring(file.lastIndexOf("/") + 1));
out.putNextEntry(entry);
int count;
while ((count = origin.read(data, 0, BUFFER_SIZE)) != -1) {
out.write(data, 0, count);
}
} finally {
origin.close();
}
}
} finally {
out.close();
}
}
public static void unzip(String zipFileUrl, String fileLocation) {
try {
File f = new File(fileLocation);
if (!f.isDirectory()) {
f.mkdirs();
}
try (ZipInputStream zin = new ZipInputStream(new FileInputStream(zipFileUrl))) {
ZipEntry ze = null;
while ((ze = zin.getNextEntry()) != null) {
// Log.e("UnZipFILE", "Unzipping....");
String path = fileLocation + ze.getName();
if (ze.isDirectory()) {
File unzipFile = new File(path);
if (!unzipFile.isDirectory()) {
unzipFile.mkdirs();
}
} else {
try (FileOutputStream fout = new FileOutputStream(path, false)) {
byte[] buffer = new byte[1024];
int read;
while ((read = zin.read(buffer)) != -1) {
fout.write(buffer, 0, read);
}
zin.closeEntry();
}
}
}
}
} catch (Exception e) {
e.printStackTrace();
Log.e("UnZipException", Log.getStackTraceString(e));
}
}
}
我也尝试像这样管理循环:
do {
ze = zin.nextEntry
} while (ze != null)
但文件未正确解压缩或已损坏。因此，如果任何人有管理此类循环的想法，那将非常有帮助。
最佳答案
我正在将您的 Java 代码转换为 Kotlin
我之前遇到过这个问题
赋值不是表达式，在此上下文中只允许使用表达式
在此处使用此代码是您的解决方案
object ZipFileManager {
private val BUFFER_SIZE = 6 * 1024
@Throws(IOException::class)
fun zip(files: Array<String>, zipFile: String) {
var origin: BufferedInputStream? = null
val out = ZipOutputStream(BufferedOutputStream(FileOutputStream(zipFile)))
try {
val data = ByteArray(BUFFER_SIZE)
for (file in files) {
val fi = FileInputStream(file)
origin = BufferedInputStream(fi, BUFFER_SIZE)
try {
val entry = ZipEntry(file.substring(file.lastIndexOf("/") + 1))
out.putNextEntry(entry)
var count: Int= origin.read(data, 0, BUFFER_SIZE);
while (count != -1) {
out.write(data, 0, count)
count = origin.read(data, 0, BUFFER_SIZE)
}
} finally {
origin.close()
}
}
} finally {
out.close()
}
}
fun unzip(zipFileUrl: String, fileLocation: String) {
try {
val f = File(fileLocation)
if (!f.isDirectory) {
f.mkdirs()
}
ZipInputStream(FileInputStream(zipFileUrl)).use { zin ->
var ze: ZipEntry? = null
ze = zin.nextEntry
while (ze != null) {
// Log.e("UnZipFILE", "Unzipping....");
val path = fileLocation + ze!!.name
if (ze.isDirectory) {
val unzipFile = File(path)
if (!unzipFile.isDirectory) {
unzipFile.mkdirs()
}
} else {
FileOutputStream(path, false).use { fout ->
val buffer = ByteArray(1024)
var read: Int= zin.read(buffer)
while (read != -1) {
fout.write(buffer, 0, read)
read = zin.read(buffer)
}
zin.closeEntry()
}
}
ze = zin.nextEntry
}
}
} catch (e: Exception) {
e.printStackTrace()
Log.e("UnZipException", Log.getStackTraceString(e))
}
}
}
关于android - Assignments are not expressions, and only expressions are allowed in this context - 将 Java 转换为 Kotlin 时出错，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56488871/