英文:
Java: BufferedReader not printing out File content on TXT file
问题
我在使用BufferedReader读取txt文件夹中的内容时遇到问题，该文件夹通过方法`showEditFile()`调用，方法中使用了来自方法`pideNumero.preguntaUno();`的用户输入数组来调用，该方法接受一个整数来迭代数组位置：
遍历文件夹"Archivos"的数组。
```java
public static String[] testFiles() {
String endPath = System.getProperty("user.dir");
String separator = File.separator;
String folderPath = endPath + separator + "Archivos";
File carpeta = new File(folderPath);
String[] lista = carpeta.list();
return lista;
}
以下是读取内容的方法，应该读取内容的第一行为Hellowwwww：
public static void showEditFile() throws IOException {
System.out.println("Por Favor, elige un archivo con su numero para mostrar su contenido ");
System.out.println("Los archivos dentro la carpeta Archivos son: ");
Menu.listFiles.nomFiles();
String[] archivos = Menu.listFiles.testFiles();
int menu = Menu.pideNumero.preguntaUno();
File document = new File(archivos[menu - 1]);
try {
FileReader fr = new FileReader(document);
BufferedReader br = new BufferedReader(fr);
String line;
line = br.readLine();
System.out.println(line);
} catch (FileNotFoundException e) {
System.out.println("File not found." + document.toString());
} catch (IOException e) {
System.out.println("Unable to read file: " + document.toString());
}
}
我尝试在调试模式下检查，并发现在FileReader fr = new FileReader(document);这一行，它会直接跳转到FileNotFoundException，FilePath为null，我认为问题是从这里引起的。
似乎它不知道"Archivos"之后的路径。
路径：
Root\Archivos\kiki.txt
我已经卡在这里整整一天了，有人可以帮忙吗！
<details>
<summary>英文:</summary>
I'm having issues with BufferedReader reading out the content of a txt file within a folder which is called via a method `showEditFile()` using an array with input of the user from method `pideNumero.preguntaUno();` which takes an int to iterate through the array positions :
Array that loops through the Folder "Archivos".
public static String[] testFiles() {
String endPath = System.getProperty("user.dir");
String separator = File.separator;
String folderPath = endPath + separator + "Archivos";
File carpeta = new File(folderPath);
String[] lista = carpeta.list();
return lista;
}
Method which would read the first line of the content which should be **Hellowwwww**:
public static void showEditFile() throws IOException {
System.out.println("Por Favor, elige un archivo con su numero para mostrar su contenido ");
System.out.println("Los archivos dentro la carpeta Archivos son: ");
Menu.listFiles.nomFiles();
String[] archivos = Menu.listFiles.testFiles();
int menu = Menu.pideNumero.preguntaUno();
File document = new File(archivos[menu - 1]);
try {
FileReader fr = new FileReader(document);
BufferedReader br = new BufferedReader(fr);
String line;
line = br.readLine();
System.out.println(line);
} catch (FileNotFoundException e) {
System.out.println("File not found." + document.toString());
} catch (IOException e) {
System.out.println("Unable to read file: " + document.toString());
}
}
I tried to check in Debug mode an see that on line `FileReader fr = new FileReader(document); ` it would jump straight into the FileNotFoundException with the FilePath == **null** which I think comes the problem from.
It seems it doesn't know the path after "Archivos"
Path:
Root\Archivos\kiki.txt
**I've been stuck on this for a full day now can someone please help!**
</details>
# 答案1
**得分**: 1
```carpeta.list()```不提供完全限定的路径，它只提供文件名。因此，在接下来的调用中，```new File(archivos[menu - 1])```将失败。在```new File(archivos[menu - 1])```中，您需要提供完整的路径，这样您就不会收到异常。请参考 https://docs.oracle.com/javase/7/docs/api/java/io/File.html#list()。
<details>
<summary>英文:</summary>
```carpeta.list()``` does not give fully qualified path. It only gives you file name. So next call ```new File(archivos[menu - 1])``` will fail. In ```new File(archivos[menu - 1])``` you will need to provide full and then you will not get Exception. Refer to https://docs.oracle.com/javase/7/docs/api/java/io/File.html#list()
</details>
# 答案2
**得分**: 0
感谢@Jags，我解决了这个问题。以下是解决方案：
在我的数组中，我返回了一个String[]，它基本上保存了字符串... 因此，当我尝试从另一个方法调用它以读取内容时，它会显示文件的名称（它是一个字符串，但是不会分配文件路径，因为它只是一个字符串）。
我在这里做出的更改：
```java
public static File[] testFiles() {
String endPath = System.getProperty("user.dir");
String separator = File.separator;
String folderPath = endPath + separator + "Archivos";
File carpeta = new File(folderPath);
// 正如你所看到的，我在这里使用了listFiles()方法来列出所有文件并将它们保存到File[]数组中
File[] lista = carpeta.listFiles();
return lista;
}
现在在调用方法showEditFiles()时：
public static void showEditFile() throws IOException {
System.out.println("Por Favor, elige un archivo con su numero para mostrar su contenido ");
System.out.println("Los archivos dentro la carpeta Archivos son: ");
Menu.listFiles.nomFiles();
File[] archivos = Menu.listFiles.testFiles();
int menu = Menu.pideNumero.preguntaUno();
File document = new File(archivos[menu - 1]);
try {
FileReader fr = new FileReader(document);
BufferedReader br = new BufferedReader(fr);
String line;
line = br.readLine();
System.out.println(line);
} catch (FileNotFoundException e) {
System.out.println("File not found." + document.toString());
} catch (IOException e) {
System.out.println("Unable to read file: " + document.toString());
}
}
现在它会打印出文件的第一行（在这种情况下是"hello"）。
英文:
So thanks to @Jags I figured the issue out. Solution below:
In my array I was returning a String[] which saves basically strings... So when I try to call it from my other method to read out the content it would show me the name of the file (which is a string but wouldn't have a file path assigned to it because it's just a string).
Changes I made here:
public static File[] testFiles() {
String endPath = System.getProperty("user.dir");
String separator = File.separator;
String folderPath = endPath + separator + "Archivos";
File carpeta = new File(folderPath);
// here as you can see i used the listFiles() method to list all the files and
// and save them into the File[] array
File[] lista = carpeta.listFiles();
return lista;
}
Now when calling the method in my other method showEditFiles() :
public static void showEditFile() throws IOException {
System.out.println("Por Favor, elige un archivo con su numero para mostrar su contenido ");
System.out.println("Los archivos dentro la carpeta Archivos son: ");
Menu.listFiles.nomFiles();
File[] archivos = Menu.listFiles.testFiles();
int menu = Menu.pideNumero.preguntaUno();
File document = new File(archivos[menu - 1]);
try {
FileReader fr = new FileReader(document);
BufferedReader br = new BufferedReader(fr);
String line;
line = br.readLine();
System.out.println(line);
} catch (FileNotFoundException e) {
System.out.println("File not found." + document.toString());
} catch (IOException e) {
System.out.println("Unable to read file: " + document.toString());
}
}
Now it prints out the first line of the file (which in this case was hellow).
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