Question

好，所以更像是有一种简单的方法输出到

[6, 8, 9, 10]
[1002]
empty array []
Bad input
Bad input
Bad input
Bad input

如果是这样的话？

这更像是挑剔那些最不礼貌的事情，在我开始做某些事情之前，要使输出变得“更细”会花费很多时间

所以我只想简单地知道它是否可以以简单的方式完成，如果是这样的话。

public class String2intArray {

  public static void main(String[] args) {
    System.out.println(Arrays.toString(str2ia("6,8 ,9 , 10")));
    System.out.println(Arrays.toString(str2ia("1002")));
    System.out.println(Arrays.toString(str2ia("")));
    System.out.println(Arrays.toString(str2ia(", 2")));
    System.out.println(Arrays.toString(str2ia("Ett,Två")));
    System.out.println(Arrays.toString((str2ia("1,,2"))));
    System.out.println(Arrays.toString(str2ia("1 , 2 ,")));
  }

  public static int[] str2ia(String s) {
    int arraylengt = 0;
    int arrayindex = 0;
    s = s.replaceAll("\\s", "");
    Scanner scan = new Scanner(s).useDelimiter(",");
    if(s.length() == 0) {
        int[] emptyArray = {};
        System.out.print("empty array ");
        return emptyArray;
    }
    try {
      if(!s.matches("(\\d+,)*\\d+")){
        throw new IllegalArgumentException("Bad input");            
      }
    } catch (IllegalArgumentException e) {
        System.out.print("Bad input");
        int[] emptyArray = {};
        return emptyArray;
    }
    while (scan.hasNext()) {
        if (scan.hasNextInt()) {
            int nextInt = scan.nextInt();

            arraylengt ++;
        } else {
            // discard next token
            scan.next();
        }
      }

      int[] intArray = new int[arraylengt];

      Scanner scan2 = new Scanner(s).useDelimiter(",");
      try {
        while (scan2.hasNext()) {
            if (scan2.hasNextInt()) {
                intArray[arrayindex] = scan2.nextInt();
                arrayindex++;

            } else {
                // discard next token
                throw new IllegalArgumentException("Bad input");
            }        
          }
        } catch (IllegalArgumentException e) {
            System.out.print("Bad input");
            int[] emptyArray = {};
            return emptyArray;

        }           
    return intArray;            
  }
}

Answer 1

这是通过对现有代码进行 2次更改来获得所需输出的一种方法：

首先：替换以下代码：

$request->query->get('date'); // GET
$request->request->get('date');  // POST

-with-

System.out.println(Arrays.toString(str2ia("6,8 ,9 , 10")));
System.out.println(Arrays.toString(str2ia("1002")));
System.out.println(Arrays.toString(str2ia("")));
System.out.println(Arrays.toString(str2ia(", 2")));
System.out.println(Arrays.toString(str2ia("Ett,Två")));
System.out.println(Arrays.toString((str2ia("1,,2"))));
System.out.println(Arrays.toString(str2ia("1 , 2 ,")));

-AND-

第二：替换以下代码：

// Define the input as an array of input strings and print result
// by iterating the input array elements
String [] inputStrings = {"6,8 ,9 , 10", "1002", "", ", 2", "Ett,Två", "1,,2", "1 , 2 ,"};
for (String input : inputStrings) {
    int [] result = str2ia(input);
    if (result == null) {
        System.out.println("");
    }
    else {
        System.out.println(Arrays.toString(result));
    }
}

-with-

...
} catch (IllegalArgumentException e) {
    System.out.print("Bad input");
    int[] emptyArray = {};
    return emptyArray;
}

Answer 2

创建您自己的代码以遍历数组：

int[] aArray = str2ia("6,8 ,9 , 10");
for (int number : aArray) {
  System.out.print(number);
}
System.out.println();

不使用[]打印错误信息

2 个答案: