使用 JAVA 控制台应用程序从桌面访问文本文件
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Accessing a text file from desktop using JAVA console application
提问by Vipin Nair
I have a text file in my desktop named "MyText.txt" and it have some contents in it.
我的桌面上有一个名为“MyText.txt”的文本文件，其中包含一些内容。
Since i am learning JAVA i thought of simply creating an console application which will read the contents from the txt file and then show it in console.
由于我正在学习 JAVA，我想简单地创建一个控制台应用程序，它将从 txt 文件中读取内容，然后在控制台中显示它。
But since i am new to this ,so i am unable to figure out the way it works.
但由于我是新手，所以我无法弄清楚它的工作方式。
Can anyone get me into a right direction.
任何人都可以让我进入正确的方向。
All help will be appreciated.
所有帮助将不胜感激。
采纳答案by Nandeshwar
You can try the following example to read file character basis.
您可以尝试以下示例来读取文件字符基础。
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
package io;
import java.io.*;
/**
*
* @author Nandeshwar
*/
public class FileExample {
public static void main(String[] args) {
File inFile = new File("c:/users/toshiba/desktop/MyText.txt");
FileReader ins = null;
try {
ins = new FileReader(inFile);
int ch;
while ((ch = ins.read()) != -1) {
System.out.println((char) ch);
}
} catch (Exception e) {
System.out.println(e);
} finally {
try {
ins.close();
} catch (Exception e) {
}
}
}
}
回答by pinkpanther
There are many alternatives for doing this. A higher level thing could be using a Scannerclass. I assume, since you are learning java, you might have come across Scannerclass for reading the input from console. You can use the Scannersame way to read the file also.
有很多替代方法可以做到这一点。更高层次的事情可能是使用一个Scanner类。我假设，由于您正在学习 Java，因此您可能遇到过从Scanner控制台读取输入的类。您也可以使用Scanner相同的方式读取文件。
You can use Scanner#nextInt(), Scanner#next(), etc... methods for reading the input. You can use args[]array for taking the command line arguments.
您可以使用Scanner#nextInt()、Scanner#next()等... 方法来读取输入。您可以使用args[]数组来获取命令行参数。
Since, you haven't mention exactly what kind of way your data is stored in file, it's hard to give a working example.
因为，您还没有确切地提到您的数据以何种方式存储在文件中，所以很难给出一个有效的例子。
import java.util.Scanner;
public class Tester {
public static void main(String args[]) {
if (args.length > 0) {
Scanner sc = new Scanner(new File(args[0]));
//use here the functions such as sc.nextInt() and so on
}
}
}
Link to the docs: http://docs.oracle.com/javase/1.5.0/docs/api/java/util/Scanner.html
链接到文档：http: //docs.oracle.com/javase/1.5.0/docs/api/java/util/Scanner.html
If you are reading big file in java (see my answer to other question): file size too big for java:
如果您正在使用 java 读取大文件（请参阅我对其他问题的回答）：文件大小对于 java 来说太大了：
Working Example:
工作示例：
This program prints the file contents to console.
该程序将文件内容打印到控制台。
Execute this program using java Tester yourFileNameafter compilation.
java Tester yourFileName编译后使用执行此程序。
import java.util.*;
import java.io.*;
public class Tester {
public static void main(String args[]) {
try {
if (args.length > 0) {
Scanner sc = new Scanner(new File(args[0]));
while (sc.hasNext()) {
System.out.println(sc.next());
}
} else {
System.out.println("No file name given");
}
}
catch(FileNotFoundException e) {
e.printStackTrace();
}
}
}
回答by zerocool
This should help. Create a BufferedReader object like this:
这应该有帮助。像这样创建一个 BufferedReader 对象：
new BufferedReader(new FileReader("Location of the file"));
new BufferedReader(new FileReader("文件位置"));
Read this doc, for reference:http://docs.oracle.com/javase/6/docs/api/java/io/BufferedReader.html
阅读此文档，以供参考：http: //docs.oracle.com/javase/6/docs/api/java/io/BufferedReader.html
