android-分配不是表达式,并且在此上下文中仅允许表达式-将Java转换为Kotlin时出错

概述我在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 @T

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))        }    }}

因此,我正在尝试这段代码,但是它显示了编译时错误,例如：

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

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,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))    }  }}

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