我在Java中管理良好的代码和项目.但是我需要在Kotlin中开发另一个项目.因此,我在Kotlin中转换了所有代码.但是有ZipFileManager.kt的代码,用于zip / unzip文件.
这是代码(科特琳):
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,BUFFER_SIZE)) != -1) {
out.write(data,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,read)
}
zin.closeEntry()
}
}
}
}
} catch (e: Exception) {
e.printStackTrace()
Log.e("UnZipException",Log.getStackTraceString(e))
}
}
}
赋值不是表达式,在此上下文中,仅当((count = origin.read(data,BUFFER_SIZE))!= -1)处的乐趣zip中才允许使用表达式.
并在while((ze = zin.nextEntry)!= null)行和while((read = zin.read(buffer))!= -1)行给出另一个相同的编译时错误.
因此,我最大的问题是在Kotlin中使用此代码.那么,任何人都可以帮助了解Kotlin的人,如何在Kotlin中使用这种类型的循环结构?
如果有人想看的话,我也有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,BUFFER_SIZE)) != -1) {
out.write(data,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,read);
}
zin.closeEntry();
}
}
}
}
} catch (Exception e) {
e.printStackTrace();
Log.e("UnZipException",Log.getStackTraceString(e));
}
}
}
我也尝试管理像这样的循环:
do {
ze = zin.nextEntry
} while (ze != null)
但是,文件无法正确解压缩或损坏.因此,如果有人有管理这种循环的想法,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
分配不是表达式,并且在此上下文中仅允许表达式
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,BUFFER_SIZE);
while (count != -1) {
out.write(data,count)
count = origin.read(data,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,read)
read = zin.read(buffer)
}
zin.closeEntry()
}
}
ze = zin.nextEntry
}
}
} catch (e: Exception) {
e.printStackTrace()
Log.e("UnZipException",Log.getStackTraceString(e))
}
}
}