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Ob'ekt
File
hali mavjud bo'lmagan faylga mos kelishi mumkinmi?Ha, agar siz katalog qiymatini konstruktorga o'tkazsangiz.
String dirPath = "/"; File f = new File(dirPath); File[] files = f.listFiles();
Bu, masalan, fayllar qatorini olish uchun amalga oshiriladi.
public class MyClass { public static void main(String[] args) { boolean isObject = false; File file = new File("/"); if (file instanceof Object){ isObject = true; } boolean isFile = file.isFile(); // Tests whether the file denoted by this abstract pathname is a normal file. Это из documentации System.out.println(isObject + " "+ isFile); } }
Chiqish:
haqiqiy yolg'onFile
dan meros oladiobject
. Javob: ha! Fikrlaringizni kutaman. -
Ob'ektni
File
turga qanday aylantirish mumkinPath
?Usul
toPath();
toPath(); //Returns a java.nio.file.Path object constructed from the this abstract path.
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Files klassi nima uchun kerak?
Biz sinfni asos qilib oldik
File
, unga ozgina yangi narsa qo'shdik, usullarni o'zgartirdik va oxirida uni ikkiga bo'ldik. Shunday qilib, endi ikkita yangi sinf mavjud -Path
vaFiles
.Path
- bu, aslida, sinfning yangi analogidirFile
vaFiles
- bu yordamchi sinf (Arrays
& classes bilan o'xshashlik bo'yichaCollections
), sinfning barcha statik usullari unga o'tkazildiFile
. Bu OOP.M nuqtai nazaridan "to'g'riroq"Hujjatlardan ba'zilari:
public final class Files
extends Object
Bu sinf faqat fayllar, kataloglar yoki boshqa turdagi fayllarda ishlaydigan statik usullardan iborat.
Ko'pgina hollarda, bu erda belgilangan usullar fayl operatsiyalarini bajarish uchun tegishli fayl tizimi provayderiga vakolat beradi. -
Qanday arxivlash darslarini bilasiz?
Ushbu mavzu bo'yicha yaxshi maqola va undan parcha: Java'da arxivlash
Java spetsifikatsiyasida arxivlar bilan ishlash uchun ikkita paket mavjud -
java.util.zip
mosjava.util.jar
ravishda zip va jar arxivlari uchun. Jar va zip formatlari o'rtasidagi farq faqat zip arxivining kengaytmasida. Paket konstruktorlarni amalga oshirish va sinf usulidan tashqarijava.util.jar
paketga o'xshaydi . Faqat paketi quyida muhokama qilinadi . Barcha misollarni zip arxividan foydalanishga aylantirish uchun kodning hamma joyida Jar ni Zip bilan almashtiring.java.util.zip
voidputNextEntry(ZipEntry e)
JarOutputStream
java.util.jar
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Arxivga katalogni qanday qo'shish mumkin?
O'zim uchun bu savolni tugallangan arxivga bo'sh katalog qo'shish deb tushundim. Men hech qanday ishlaydigan misol topmadim. Mana kod: (Bu arxivga istalgan faylni qo'yish mumkinligini aniq ko'rsatib turibdi, lekin bo'sh katalog bilan... Men qanday javob berishni bilmayman, StackOverFlow-ga yozmadim, bunday savol albatta bo'ladi. pastga ovoz berilgan) Agar kimdir biron bir taklif bo'lsa, yozing.
public class Main { public static void main(String[] args) { String[] myFiles = {"D:\\forJava\\MyArtifactName\\packForTest\\res2.txt", "D:\\forJava\\MyArtifactName\\packForTest\\res.txt", "D:\\forJava\\MyArtifactName\\packForTest\\res4.txt", "D:\\forJava\\MyArtifactName\\packForTest\\testDir\\" }; String zipFile = "D:\\forJava\\MyArtifactName\\packForTest\\res.zip"; ZipUtility zipUtil = new ZipUtility(); try { zipUtil.zip(myFiles, zipFile); } catch (Exception ex) { // some errors occurred ex.printStackTrace(); } } }
Savol oxirgi testDir haqida, JVM uni olingan arxivga qo'ymaydi, boshqa barcha txt fayllari bilan u yaxshi ishlaydi.
ZipUtility.java
:import java.io.BufferedInputStream; import java.io.File; import java.io.FileInputStream; import java.io.FileOutputStream; import java.io.IOException; import java.util.ArrayList; import java.util.List; import java.util.zip.ZipEntry; import java.util.zip.ZipOutputStream; public class ZipUtility { private static final int BUFFER_SIZE = 4096; public void zip(List<File> listFiles, String destZipFile) throws IOException { ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(destZipFile)); for (File file : listFiles) { if (file.isDirectory()) { zipDirectory(file, file.getName(), zos); } else { zipFile(file, zos); } } zos.flush(); zos.close(); } public void zip(String[] files, String destZipFile) throws IOException { List<File> listFiles = new ArrayList<File>(); for (int i = 0; i < files.length; i++) { listFiles.add(new File(files[i])); } zip(listFiles, destZipFile); } private void zipDirectory(File folder, String parentFolder, ZipOutputStream zos) throws IOException { for (File file : folder.listFiles()) { if (file.isDirectory()) { zipDirectory(file, parentFolder + "/" + file.getName(), zos); continue; } zos.putNextEntry(new ZipEntry(parentFolder + "/" + file.getName())); BufferedInputStream bis = new BufferedInputStream( new FileInputStream(file)); long bytesRead = 0; byte[] bytesIn = new byte[BUFFER_SIZE]; int read = 0; while ((read = bis.read(bytesIn)) != -1) { zos.write(bytesIn, 0, read); bytesRead += read; } zos.closeEntry(); } } private void zipFile(File file, ZipOutputStream zos) throws IOException { zos.putNextEntry(new ZipEntry(file.getName())); BufferedInputStream bis = new BufferedInputStream(new FileInputStream( file)); long bytesRead = 0; byte[] bytesIn = new byte[BUFFER_SIZE]; int read = 0; while ((read = bis.read(bytesIn)) != -1) { zos.write(bytesIn, 0, read); bytesRead += read; } zos.closeEntry(); } }
Bu yerdan kod
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Nima uchun ular kerak
Properties
?Properties
xususiyatlar faylidir. Uning tuzilishi: kalit - qiymat. Bunday fayllar bilan ishlash uchun Java-da sinf mavjudProperties
, u meros qilib olinganHashTable<Object, Object>
Uni manipulyatsiya qilish haqida maqola bor - Java Properties fayl namunalari
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Faylda ma'lumotlar qanday shaklda saqlanadi
.properties
?Kalit - bu ma'no.
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Properties
Ob'ektdagi ma'lumotlarni fayldan yuklagandan keyin o'zgartirish mumkinmi ?Agar u dan meros bo'lib o'tgan bo'lsa
HashMap
, unda siz mumkin, faqat keyin siz ushbu faylga kiritilgan o'zgarishlarni yozishni bekor qilishingiz kerak bo'ladi. Buning uchun bir usul mavjudsetProperty
.Mana kod:
import java.io.FileInputStream; import java.io.IOException; import java.io.InputStream; import java.util.Properties; /** * Created by Роман on 12.09.2016. */ public class LoadAndSetProperties { public static void main(String[] args) { Properties prop = new Properties(); InputStream input = null; try { input = new FileInputStream("D:\\forJava\\MyArtifactName\\packForTest\\config.properties"); // load a properties file prop.load(input); // get the property value and print it out prop.setProperty("database", "ddfdfdfdfdf"); System.out.print(prop.getProperty("database")); } catch (IOException ex) { ex.printStackTrace(); } finally { if (input != null) { try { input.close(); } catch (IOException e) { e.printStackTrace(); } } } } }
Вывод:
ddfdfdfdfdf -
Зачем нужен класс
FileReader
?Java Docs:
public class FileReader
extends InputStreamReader
Convenience class for reading character files. The constructors of this class assume that the default character encoding and the default byte-buffer size are appropriate. To specify these values yourself, construct an InputStreamReader on a FileInputStream.
FileReader is meant for reading streams of characters.
Класс для чтения символов файлов. Конструкторы этого класса предполагают, что codeировка символов дефолтная и дефолтный размер буфера являются подходящими. Whatбы задать эти значения самостоятельно, следует построить
InputStreamReader
надFileInputStream
.FileReader
предназначен для считывания потоков символов. -
Зачем нужен класс
FileWriter
?public class FileWriter
extends OutputStreamWriter
Convenience class for writing character files. The constructors of this class assume that the default character encoding and the default byte-buffer size are acceptable. To specify these values yourself, construct an OutputStreamWriter on a FileOutputStream.
Whether or not a file is available or may be created depends upon the underlying platform. Some platforms, in particular, allow a file to be opened for writing by only one FileWriter (or other file-writing object) at a time. In such situations the constructors in this class will fail if the file involved is already open. FileWriter is meant for writing streams of characters.
Класс для записи символов файлов. Конструкторы этого класса предполагают, что codeировка символов дефолтная и дефолтный размер буфера являются приемлемым. Whatбы задать эти значения самостоятельно, следует построить
OutputStreamWriter
надFileOutputStream
. Является ли файл доступен для записи, зависит от используемой платформы. Некоторые платформы разрешают держать файл для записи только однимFileWriter
(or другого an object записи file), в одно время.FileWriter
предназначен для записи потоков символов. Для написания потоков необработанных byteов, используйтеFileOutputStream
.Эти классы (
FileReader
иFileWriter
) специально ориентированы для работы с текстом и строками.
DefNeo
Daraja
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