我有一个程序来拆分.txt文件,将其分为101个文件。必须分割的文件File.txt包含以\n
分隔的URL。事实是该程序将文件分成相等的部分,当文件达到最大大小时,它将剪切url并启动一个新文件。如何将其拆分为不超过1Mb的大小,并且包含经过良好拆分的网址?
import java.io.*;
import java.util.Scanner;
public class readfile {
public static int SubfileName;
public static int[] Murl = new int[2000000];
public static int x = 0;
public static long usemem = 0;
public static long Numberofmailto = 0;
static byte[] subfich; //Subfile data (global var)
static long NumberUrl;
static int[] indURL; //Indices de las URLs en "subfich"
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter the file name like url.txt to read but it should be in E:\\url\\");
String name = in.nextLine();
readfile(name);
try {
//now create 100 subfile
GeneraFicheros();
} catch (Exception e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
// not used
public static void readfile(String filename) {
try {
// file path
leeSubfichero("E:\\url\\" + filename);
creaIndices();
} catch (Exception e) {
}
}
//read danger file
static void leeSubfichero(String nomfich) throws IOException { // read file
File fich = new File(nomfich);
int tam = (int) fich.length(); //Tamaño bytes // size byte
subfich = null;
subfich = new byte[tam];
try (FileInputStream fis = new FileInputStream(fich)) {
NumberUrl = fis.read(subfich);
// find the mailto urls
}
}
static void creaIndices() {
// 1. Count the number of URLs
int n = 0;
int x = 0;
boolean dangerurl = false;
for (int i = 0; i < subfich.length; i++) {
if (subfich[i] == 10) {
n++;
}
}
//2. Store separators position
indURL = null;
indURL = new int[n];
//Murl = new int[n];
int k = 0;
for (int i = 0; i < subfich.length; i++) {
if (subfich[i] == 10) {
indURL[k++] = i;
}
}
}
// create 100 files
public static void GeneraFicheros() throws Exception {
String zero = "00";
RandomAccessFile raf = new RandomAccessFile("E:\\url\\danger.txt", "r");
long numSplits = 100; //divid in 100 subfiles
long sourceSize = raf.length(); // danger.txt file size
long bytesPerSplit = sourceSize / numSplits; // number of bytes each file will have
long remainingBytes = sourceSize % numSplits;
int maxReadBufferSize = 8 * 1024; //8KB
for (int destIx = 1; destIx <= numSplits; destIx++) {
// each literation create a new file like 000
System.out.println("Escrito Subfichero " + zero + destIx + ".txt");
runtime();
if (destIx > 9) {
zero = "0";
}
// write the file with name like 000.txt
BufferedOutputStream bw = new BufferedOutputStream(new FileOutputStream("E:\\url\\" + zero + destIx + ".txt"));
if (bytesPerSplit > maxReadBufferSize) {
// total number of bytes to read
long numReads = bytesPerSplit / maxReadBufferSize;
// total number of bytes remaining for other files
long numRemainingRead = bytesPerSplit % maxReadBufferSize;
for (int i = 0; i < numReads; i++) {
readWrite(raf, bw, maxReadBufferSize);
}
// if bytes are remaining write the file
if (numRemainingRead > 0) {
readWrite(raf, bw, numRemainingRead);
}
} else {
readWrite(raf, bw, bytesPerSplit);
}
bw.close();
}
// if dividion didn't work extra store here
if (remainingBytes > 0) {
BufferedOutputStream bw = new BufferedOutputStream(new FileOutputStream("split." + (numSplits + 1) + ".txt"));
readWrite(raf, bw, remainingBytes);
bw.close();
}
raf.close();
}
// write 8kb each time in the file
static void readWrite(RandomAccessFile raf, BufferedOutputStream bw, long numBytes) throws IOException {
byte[] buf = new byte[(int) numBytes];
int val = raf.read(buf);
if (val != -1) {
bw.write(buf);
}
}
static long startTime = System.nanoTime();
public static void runtime() {
long endTime = System.nanoTime();
long totalTime = endTime - startTime;
double seconds = (double) totalTime / 1000000000.0;
System.out.println("Toatl seconds" + seconds);
}
}
答案 0 :(得分:4)
无法完成这项工作,就像您设置代码的方式一样。您的代码基本上是基于字节的(这使得不可能读取诸如换行符之类的字符,这是一个字符,而不是字节),并且它“预先计算”了要读取的数量,这也使得它不可能。想象一下,输入文件中的URL确实很长。几乎与所需的尺寸一样长。
最后,没有可行的说法:“将这个文件精确地分成100个部分,但是这样做的方式是每个文件都小于totalsize / 100 ...并且也不将任何行切成两半。 ” –可能会有101个文件,甚至可能有250个文件。如果整个文件只是一个非常长的URL,则可能也只有1个文件。
在编写程序之前,您必须准确指定程序要执行的操作。
一些提示:
Files.newBufferedReader
,传入字符集。您需要离开InputStream
,因为那是字节,而不是字符。