我在分割从输入文件读入的字符串时遇到一些问题，确保它有效，然后将其保存到变量。
让我们说这是第一个字符串： 12345 5 59.28
我想分开12345,5和59.28。 在验证它们是正确的格式（00000-99999,0-5,000.00 0 100.00）之后，我会将其分配给变量。
我的主要两个障碍是我不能在这个程序中使用数组，因此我不确定如何分割字符串。我试过把每个部分作为一个整数来拉，但这似乎没有用。
我的另一个问题是，我不确定如何验证它。我会使用这样的东西：
conm fyear dvpayout industry firmycount ipodate dv age
46078 CAESARS ... 2003 0.226813 Services 22 19891213.0 Q2 27
46079 CAESARS ... 2004 0.226813 Services 22 19891213.0 Q2 27
46080 CAESARS ... 2005 0.226813 Services 22 19891213.0 Q2 27
46091 CAESARS ... 2016 0.226813 Services 22 19891213.0 Q2 27
114620 CAESARSTONE LTD 2010 0.487543 Manufacturing 10 20120322.0 Q3 4
114621 CAESARSTONE LTD 2011 0.487543 Manufacturing 10 20120322.0 Q3 4
114622 CAESARSTONE LTD 2012 0.487543 Manufacturing 10 20120322.0 Q3 4
114623 CAESARSTONE LTD 2013 0.487543 Manufacturing 10 20120322.0 Q3 4
114624 CAESARSTONE LTD 2014 0.487543 Manufacturing 10 20120322.0 Q3 4
114625 CAESARSTONE LTD 2015 0.487543 Manufacturing 10 20120322.0 Q3 4
114626 CAESARSTONE LTD 2016 0.487543 Manufacturing 10 20120322.0 Q3 4
132524 CAFEPRESS INC 2010 0.000000 Retail Trade 7 20120329.0 Q1 4
132525 CAFEPRESS INC 2011 0.000000 Retail Trade 7 20120329.0 Q1 4
132526 CAFEPRESS INC 2012 -0.000000 Retail Trade 7 20120329.0 Q1 4
132527 CAFEPRESS INC 2013 -0.000000 Retail Trade 7 20120329.0 Q1 4
132528 CAFEPRESS INC 2014 -0.000000 Retail Trade 7 20120329.0 Q1 4
132529 CAFEPRESS INC 2015 -0.000000 Retail Trade 7 20120329.0 Q1 4
132530 CAFEPRESS INC 2016 -0.000000 Retail Trade 7 20120329.0 Q1 4
120049 CAI INTERNATIONAL INC 2006 0.000000 Services 12 20070516.0 Q1 0
120050 CAI INTERNATIONAL INC 2007 0.000000 Services 12 20070516.0 Q1 0
3896 CALAMP CORP 1999 -0.000000 Manufacturing 23 NaN Q1 Nan
3897 CALAMP CORP 2000 0.000000 Manufacturing 23 NaN Q1 Nan
3898 CALAMP CORP 2001 0.000000 Manufacturing 23 NaN Q1 Nan
3899 CALAMP CORP 2002 0.000000 Manufacturing 23 NaN Q1 Nan
21120 CALATLANTIC GROUP INC 1995 -0.133648 Construction 22 NaN Q1 Nan
21121 CALATLANTIC GROUP INC 1996 -0.133648 Construction 22 NaN Q1 Nan
21122 CALATLANTIC GROUP INC 1997 -0.133648 Construction 22 NaN Q1 Nan
21123 CALATLANTIC GROUP INC 1998 -0.133648 Construction 22 NaN Q1 Nan
21124 CALATLANTIC GROUP INC 1999 -0.133648 Construction 22 NaN Q1 Nan
21125 CALATLANTIC GROUP INC 2000 -0.133648 Construction 22 NaN Q1 Nan
21126 CALATLANTIC GROUP INC 2001 -0.133648 Construction 22 NaN Q1 Nan
21127 CALATLANTIC GROUP INC 2002 -0.133648 Construction 22 NaN Q1 Nan
21128 CALATLANTIC GROUP INC 2003 -0.133648 Construction 22 NaN Q1 Nan
我是Java的初学者，请原谅我的困惑。
答案 0 :(得分：0)
试试这个。无效格式将在next方法调用期间抛出异常。
import java.util.Scanner;
class Main {
public static void main(String[] args) {
Scanner in = new Scanner("12345 5 59.28");
in.useDelimiter(" "); // reads per space
String next = in.next("\\d{5}"); // reads next 5 digits
int numbers = Integer.valueOf(next);
System.out.println(numbers);
next = in.next("\\d{1}"); // reads next 1 digit
int studentId = Integer.valueOf(next);
System.out.println(studentId);
next = in.next("\\d{2}\\.\\d{2}"); // reads next a decimal with two digits before and after point
float floatingNumbers = Float.valueOf(next);
System.out.println(floatingNumbers);
}
}
<script src="//repl.it/embed/IWzC/0.js"></script>
答案 1 :(得分：0)
如果你知道文件的结构是什么，例如它的格式如下：
int int double
然后，您只需致电nextInt()，nextInt()，然后nextDouble()即可通过该方式解析数据。
也许是这样的
do
{
num1 = scanner.nextInt();
num2 = scanner.nextInt();
num3 = scanner.nextDouble();
} while (scanner.hasNextInt());
这样做是为了收集您的所有数据，但如果您有大量的数据需要阅读，那么您可能需要很多变量
或者，如果有错误的数据有时会在它之后立即生成正确的数据，那么就可以跳过错误的数据，即使它不是很漂亮
do
{
if (scanner.hasNextInt())
{
num1 = scanner.nextInt();
}
else
{
scanner.next() // move past whatever bad data there was
num1 = scanner.nextInt();
}
if (scanner.hasNextInt())
{
num2 = scanner.nextInt();
}
else
{
scanner.next() // move past whatever bad data there was
num2 = scanner.nextInt();
}
if (scanner.hasNextDouble())
{
num3 = scanner.nextDouble();
}
else
{
scanner.next() // move past whatever bad data there was
num3 = scanner.nextDouble();
}
} while (scanner.hasNext());
答案 2 :(得分：0)
我认为你的老师会给你这个作业来练习你的if-else条件或转换语句以及循环（基本）技能。
在这里我做了什么，这可能与你的作业问题不完全匹配，但使用它你可以得到完整的想法，并想出一种方法来减少这一点。嘿!因为我们不在这里做你的任务。你必须解决你的问题并熟悉它们。
尝试理解这些，做出改变看看会发生什么：
public static void main(String[] args) {
Scanner fileInput = new Scanner(System.in);
//Declare variables
String numbers = "";
String firstNum = "";
String secondNum = "";
String thirdNum = "";
int studentID = 0;
int secondDigit = 0;
double thirdDigit = 0;
System.out.print("Input: ");
numbers = fileInput.nextLine();
int firstIndex = 0;
int secondIndex = 0;
int thirdIndex = 0;
firstIndex = numbers.indexOf(" ");
if(firstIndex <= 4){
System.out.println("Number should be 5");
}else{
firstNum = numbers.substring(0, firstIndex);
numbers = numbers.substring(firstIndex+1);
studentID = Integer.parseInt(firstNum);
if(studentID > 0 && studentID < 99999){
System.out.println("First num: " +firstNum);
}else{
System.out.println("first digits not in a range ");
}
}
secondIndex = numbers.indexOf(" ");
if(secondIndex == 0){
System.out.println("no number");
}else{
secondNum = numbers.substring(0, secondIndex);
numbers = numbers.substring(secondIndex+1);
secondDigit = Integer.parseInt(secondNum);
if(secondDigit >= 0 && secondDigit <= 5){
System.out.println("Second num: " +secondNum);
}else{
System.out.println("second digit not in a range ");
}
}
thirdIndex = numbers.length();
if(thirdIndex < 3){
System.out.println("3 numbers should be there");
}else{
thirdNum = numbers.substring(0, thirdIndex);
thirdDigit = Double.parseDouble(thirdNum);
if(thirdDigit >= 0 && thirdDigit <= 100){
System.out.println("third num: " +thirdNum);
}else{
System.out.println("third digit not in a range ");
}
}
}
我也不打算解释这个。如果您在使用此代码后遇到任何问题，则必须尝试。在评论中提出任何问题。
希望这会有所帮助！