下面你可以看到我的代码。它读取sort.txt其中存在由lenghts第一格式化然后像alhabeticly这样的话:如何在java中跳过txt文件中的行
- 一个
- b
- AB
- 交流
- ABC
- ACD
- AAAA
每个字中的一个字线。下面的程序是什么增加了单词列表1中的单词长度1,单词长度2数组列表2等。它工作正常,但是当它读取让我们说arraylist3它读取整个sort.txt。我怎样才能做到这一点,所以它跳过单词长度1和2直接到单词长度3并将它们添加到列表3。然后当第一个单词长度为4时发现它停止。所以我不必一遍又一遍读整个文件?
package test;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.InputMismatchException;
import java.util.Scanner;
public class swithc {
public static void main(String[] args) {
String izbira;
int dolzina=0;
Scanner in = new Scanner(System.in);
String vnos;
Scanner input = new Scanner(System.in);
ArrayList list1 = new ArrayList();
ArrayList list2 = new ArrayList();
ArrayList list3 = new ArrayList();
ArrayList list4 = new ArrayList();
ArrayList list5 = new ArrayList();
ArrayList list6 = new ArrayList();
ArrayList list7 = new ArrayList();
ArrayList list8 = new ArrayList();
ArrayList list9 = new ArrayList();
ArrayList list10plus = new ArrayList();
try {
File file = new File("sort.txt");
FileReader fileReader = new FileReader(file);
BufferedReader bufferedReader = new BufferedReader(fileReader);
String vrstica;
while ((vrstica = bufferedReader.readLine()) != null) {
if (vrstica.length() == 1) {
list1.add(vrstica);
}
if (vrstica.length() == 2) {
list2.add(vrstica);
}
if (vrstica.length() == 3) {
list3.add(vrstica);
}
if (vrstica.length() == 4) {
list4.add(vrstica);
}
if (vrstica.length() == 5) {
list5.add(vrstica);
}
if (vrstica.length() == 6) {
list6.add(vrstica);
}
if (vrstica.length() == 7) {
list7.add(vrstica);
}
if (vrstica.length() == 8) {
list8.add(vrstica);
}
if (vrstica.length() == 9) {
list9.add(vrstica);
}
if (vrstica.length() > 9) {
list10plus.add(vrstica);
}
}
do{
do {
System.out.println("Vnesi dožino besede, ki jo iščeš:");
if (in.hasNextInt()) {
dolzina = in.nextInt();
} else if (in.hasNextLine()) {
System.out.printf("Napačen vnos! Poskusi ponovno:%n ",
in.nextLine());
}
} while (dolzina <= 0);
System.out.println("Vnesi besedo za neznano črko vpiši * :");
vnos = input.nextLine();
vnos = vnos.replace("*", ".");
if (dolzina == 1) {
for (int i = 0; i < list1.size(); i++) {
String s = (String) list1.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 2) {
for (int i = 0; i < list2.size(); i++) {
String s = (String) list2.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 3) {
for (int i = 0; i < list3.size(); i++) {
String s = (String) list3.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 4) {
for (int i = 0; i < list4.size(); i++) {
String s = (String) list4.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 5) {
for (int i = 0; i < list5.size(); i++) {
String s = (String) list5.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 6) {
for (int i = 0; i < list6.size(); i++) {
String s = (String) list6.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 7) {
for (int i = 0; i < list7.size(); i++) {
String s = (String) list7.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 8) {
for (int i = 0; i < list8.size(); i++) {
String s = (String) list8.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina == 9) {
for (int i = 0; i < list9.size(); i++) {
String s = (String) list9.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
if (dolzina > 9) {
for (int i = 0; i < list10plus.size(); i++) {
String s = (String) list10plus.get(i);
if (s.matches(vnos))
System.out.println(s);
}
}
dolzina=-1;
System.out.println("Ponovni vnos (da/ne):");
Scanner inn= new Scanner (System.in);
izbira = inn.next();
}while (izbira.equalsIgnoreCase("da"));
bufferedReader.close();
} catch (IOException e) {
e.printStackTrace();
}
}}
点,而不是列表1,列表2,项目list3,list4等,有一个列表 - 列表ArrayList lists = new ArrayList ()'然后简单地抓住lists.get(...)来添加你的项。而不是10个块都完全相同,现在你有1个块,这是更容易分析和修复。 –
2014-11-03 19:28:14
@ Mike'Pomax'Kamermans奇怪的是,根据他自己的要求,他似乎并不需要多个列表。 – furkle 2014-11-03 19:33:06