英文:
Getting a file from user input and store integers into array
问题
以下是您提供的代码翻译后的内容：
我正试图在Java中从用户输入中打开一个文件，然后读取该文件，仅获取每行中的整数，然后将其放入一个数组中并返回该数组。在从文件中获取项目方面，我对Python比Java更熟悉。
文件的示例内容（一行）：
34 a 55 18 47 89 b 45 67 59 abbbb 88 37 20 27 10 78 39 21 n m ghff
我的代码：
private static int[] getArray(){
List<Integer> temp = new ArrayList<Integer>();
System.out.print("请输入要打开的文件名：");
try{
String filename = in.next();
File file = new File(filename);
Scanner inputFile = new Scanner(file);
while (inputFile.hasNext()){
if (inputFile.hasNextInt()){
temp.add(inputFile.nextInt());
}
else{
inputFile.next();
}
}
} catch (FileNotFoundException e){
System.out.println("---文件未找到！退出程序！---");
System.exit(0);
}
int[] array = new int[temp.size()];
for (int i = 0; i < array.length; i++){
array[i] = temp.get(i);
}
return array;
}
编辑：
我弄明白了，我的 while 循环写错了。应该像这样：
while (inputFile.hasNext()){
if (inputFile.hasNextInt()){
temp.add(inputFile.nextInt());
}
else{
inputFile.next();
}
}
英文:
I am trying to open a file from user input in java and then read that file and grab only the integers in each line then place it into an array and return the array. I'm more familiar with python in grabbing items in a file than java.
Sample contents of the file in one line:
34 a 55 18 47 89 b 45 67 59 abbbb 88 37 20 27 10 78 39 21 n m ghff
My code:
private static int[] getArray(){
List<Integer> temp = new ArrayList<Integer>();
System.out.print("Please input the name of the file to be opened: ");
try{
String filename = in.next();
File file = new File(filename);
Scanner inputFile = new Scanner(file);
while (inputFile.hasNextInt()){
temp.add(inputFile.nextInt());
}
} catch (FileNotFoundException e){
System.out.println("---File Not Found! Exit program!---");
System.exit(0);
}
int[] array = new int[temp.size()];
for (int i = 0; i < array.length; i++){
array[i] = temp.get(i);
}
return array;
}
Edit:
I figured it out my while loop was wrong. It should be like this:
while (inputFile.hasNext()){
if (inputFile.hasNextInt()){
temp.add(inputFile.nextInt());
}
else{
inputFile.next();
}
}
答案1
得分: 2
在您的情况下，我不会使用Scanner对象来获取int值，因为当您使用scanner时，只有在存在EOF（文件结尾）字符时才会为`hasNext()`返回`false`。
因此，一旦检索到`filename`，请使用下面的代码，它将消除给定字符串中的所有字符，并用一个单独的`whitespace`替换它。
String stringWithSingleWS = filename.trim().replaceAll("([a-zA-Z])"," ").replaceAll("\\s{2,}"," ");
然后您可以将其解析为数组，并直接返回结果，甚至无需转换为List对象变量。
int[] values = java.util.Arrays.stream(stringWithSingleWS.split(" "))
.mapToInt(Integer::parseInt)
.toArray();
return values;
英文:
In your case, I wouldn't use Scanner object to get the int values, because when you use scanner, you only get false for hasNext() when you have a EOF(end of file) character.
So once you retrieve the filename use the code below, which will eliminate all of the characters from given string and replace it with a single whitespace
String stringWithSingleWS = filename.trim().replaceAll("([a-zA-Z])"," ").replaceAll("\\s{2,}"," ");
Then you can parse it into array and directly return the result without even casting into a List object variable.
int[] values = java.util.Arrays.stream(stringWithSingleWS.split(" "))
.mapToInt(Integer::parseInt)
.toArray();
return values;
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