我正在写一个酒店控制台程序，我现在遇到的问题是当用户按下从文件加载的选项时，将保存的文件加载回String[]。
文本文件包含先前保存的访客名称。
这是文件数据
tom
mo
jo
john
meg
bob
jack
veronica
jessica
angelica
这是我的所有代码
是的，谢谢，我知道数组是0索引。 for循环从1开始，因为 我希望有 Room1代替Room0作为第一个
谢谢你解决了问题
public class FileLoad {
public String[] readLines(String filename) throws IOException {
FileReader fileReader = new FileReader(filename);
BufferedReader bufferedReader = new BufferedReader(fileReader);
List<String> lines = new ArrayList<String>();
String line = null;
while ((line = bufferedReader.readLine()) != null) {
lines.add(line);
}
bufferedReader.close();
return lines.toArray(new String[lines.size()]);
}
public class Hotel_array {
if (Menu.equalsIgnoreCase("S")) {
save(hotel);
}
if (Menu.equalsIgnoreCase("L")) {
load(hotel);
}
}
}
private static void save(String hotel[]) {
try {
PrintWriter pr = new PrintWriter("data.txt");
for (int i = 1; i < 11; i++) {
pr.println(hotel[i]);
}
pr.close();
} catch (Exception e) {
e.printStackTrace();
System.out.println("No such file exists.");
}
}
public static void load(String[] args) {
FileLoad rf = new FileLoad();
String file = "data.txt";
try {
String[] hotel = rf.readLines(file);
for (String line : hotel) {
System.out.println(line); // IT PRINTS FILE NOT LOADS TO ARRAY
}
} catch (IOException e) {
System.out.println("Unable to create " + file + ": " + e.getMessage());
}
}
}
答案 0 :(得分：1)
您可以更改FileLoad类并添加另一种方法将数组写入文件，只是为了将所有文件IO保留在一个类中。
public class FileLoad {
public static String[] readHotelArray(String filename) throws IOException {
BufferedReader bufferedReader = new BufferedReader(new FileReader(filename));
List<String> lines = new ArrayList<String>();
String line = null;
while ((line = bufferedReader.readLine()) != null) {
lines.add(line);
}
bufferedReader.close();
return lines.toArray(new String[lines.size()]);
}
public static void writeHotelArray(String filename, String[] hotel) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(filename, false));
//Write each string from the array to the file as a new line
for (String s : hotel)
bufferedWriter.write(s + "\n");
bufferedWriter.flush();
bufferedWriter.close();
}
}
注意：两种方法都是静态的，因此您不必实例化新对象，因为该对象上始终只有一个方法调用
现在您必须更改Hotel_array课程中保存和加载数组的方式。你可以使用这样的东西：
//...
private static void save(String[] hotel) {
try {
FileLoad.writeHotelArray("data.txt", hotel);
} catch (Exception e) {
e.printStackTrace();
System.out.println("No such file exists.");
}
}
public static String[] load() {
String file = "data.txt";
String[] hotelArray = null;
try {
hotelArray = FileLoad.readHotelArray(file);
} catch (IOException e) {
System.out.println("Unable to create " + file + ": " + e.getMessage());
}
return hotelArray;
}
//...
由于java中的参数始终是按值传递（更多关于here），因此您需要在String方法中返回load()数组。因此，您还必须在main方法中更改一小段代码
从：
//...
if (Menu.equalsIgnoreCase("L")) {
load(hotel);
}
//...
要：
//...
if (Menu.equalsIgnoreCase("L")) {
hotel = load();
}
//...
希望有所帮助（：
答案 1 :(得分：1)
tomaszsvd，我将在此处留下您的评论...我认为它可能有助于您的Java学习曲线。我鼓励您将下面的load（）方法与原始代码进行比较。还要研究示例输出，看看有什么不同。
fwiw，我喜欢scsere的答案，这是一个干净的设计。你应该追求并将其标记为答案。
让我们专注于Hotel_array.load（String [] args）的代码。
一旦Hotel_array.load（）调用rf.readLines（），你就会在内存中有2个数组。
1st array: Hotel_array's main()'s local variable "hotel".
2nd array: load()'s local variable "hotel", which is a temporary variable.
在Hotel_array.load（）内部，请记住args参数绑定到main（）＆＃34; hotel＆＃34; hotel＆＃34;变量
所以加载（）的本地变量＆＃34; hotel＆＃34;没有没有与main（）＆＃34; s＆＃34; hotel＆＃34;变量
为了使这一点更加清晰，我将调整你的load（）方法：
$ javac *.java
$ cat data.txt
alpha
beta
gamma
delta
$ java Hotel_array
WELCOME TO THE HOTEL BOOKING
Hotel Booking Options
A: To Add customer to a room
V: To View all rooms
E: To Display empty rooms
D: To Delete customer from a room
F: Find room from customer name
O: View rooms alphabetically by name
S: Save to file
L: Load from file
L
Loaded 4 lines from filedata.txt
args[1]=empty, will assign line=alpha
args[2]=empty, will assign line=beta
args[3]=empty, will assign line=gamma
args[4]=empty, will assign line=delta
Hotel Booking Options
A: To Add customer to a room
V: To View all rooms
E: To Display empty rooms
D: To Delete customer from a room
F: Find room from customer name
O: View rooms alphabetically by name
S: Save to file
L: Load from file
V
room 1 is occupied by alpha
room 2 is occupied by beta
room 3 is occupied by gamma
room 4 is occupied by delta
room 5 is empty
room 6 is empty
room 7 is empty
room 8 is empty
room 9 is empty
room 10 is empty
Hotel Booking Options
A: To Add customer to a room
V: To View all rooms
E: To Display empty rooms
D: To Delete customer from a room
F: Find room from customer name
O: View rooms alphabetically by name
S: Save to file
L: Load from file
^C$
public static void load(String[] args) {
FileLoad rf = new FileLoad();
String file = "data.txt";
try {
// ORIGINAL String[] hotel = rf.readLines(file);
String[] tempHotelData = rf.readLines(file); // Note the different var name.
System.out.println("Loaded "+tempHotelData.length+" lines from file"+file);
int i = 1; // Following your convetion of staring from index #1.
for (String line : tempHotelData ) {
// ORIGINAL: System.out.println(line); // IT PRINTS FILE NOT LOADS TO ARRAY
// NEW...
// Let's print out what is oging on...
// So let's assign "line" to the "args" array.
// Remember that "args" ties back to main()'s "hotel" variable.
System.out.println("args["+i+"]="+args[i]+", will assign line="+line);
args[i] = line;
i = i + 1;
}
} catch (IOException e) {
// probably should say "Unable to LOAD" vs "Unable to CREATE"...
System.out.println("Unable to create " + file + ": " + e.getMessage());
}
}
您需要考虑一些额外的事情......
1) Do you want to assign a line from a file if somebody is already in a room?
(e.g. it isn't empty).
2) What happens if "data.txt" has more lines than you have rooms?