在java中将String转换为int数组

Question

I have one string

String arr= "[1,2]";

ie "[1,2]" is like a single string

How do convert this arr to int array in java

Answer 1

String arr = "[1,2]";
String[] items = arr.replaceAll("\\[", "").replaceAll("\\]", "").replaceAll("\\s", "").split(",");

int[] results = new int[items.length];

for (int i = 0; i < items.length; i++) {
    try {
        results[i] = Integer.parseInt(items[i]);
    } catch (NumberFormatException nfe) {
        //NOTE: write something here if you need to recover from formatting errors
    };
}

Answer 2

Using Java 8's stream library, we can make this a one-liner (albeit a long line):

String str = "[1, 2, 3, 4, 5, 6, 7, 8, 9, 0]";
int[] arr = Arrays.stream(str.substring(1, str.length()-1).split(","))
    .map(String::trim).mapToInt(Integer::parseInt).toArray();
System.out.println(Arrays.toString(arr));

substring removes the brackets, split separates the array elements, trim removes any whitespace around the number, parseInt parses each number, and we dump the result in an array. I've included trim to make this the inverse of Arrays.toString(int[]) , but this will also parse strings without whitespace, as in the question. If you only needed to parse strings from Arrays.toString , you could omit trim and use split(", ") (note the space).

Answer 3

    final String[] strings = {"1", "2"};
    final int[] ints = new int[strings.length];
    for (int i=0; i < strings.length; i++) {
        ints[i] = Integer.parseInt(strings[i]);
    }

Answer 4

It looks like JSON - it might be overkill, depending on the situation, but you could consider using a JSON library (eg http://json.org/java/ ) to parse it:

    String arr = "[1,2]";

    JSONArray jsonArray = (JSONArray) new JSONObject(new JSONTokener("{data:"+arr+"}")).get("data");

    int[] outArr = new int[jsonArray.length()]; 

    for(int i=0; i<jsonArray.length(); i++) {
        outArr[i] = jsonArray.getInt(i);
    }

Answer 5

Saul's answer can be better implemented splitting the string like this:

string = string.replaceAll("[\\p{Z}\\s]+", "");
String[] array = string.substring(1, string.length() - 1).split(",");

Answer 6

try this one, it might be helpful for you

String arr= "[1,2]";
int[] arr=Stream.of(str.replaceAll("[\\[\\]\\, ]", "").split("")).mapToInt(Integer::parseInt).toArray();

Answer 7

You can do it easily by using StringTokenizer class defined in java.util package.

void main()
    {
    int i=0;
    int n[]=new int[2];//for integer array of numbers
    String st="[1,2]";
    StringTokenizer stk=new StringTokenizer(st,"[,]"); //"[,]" is the delimeter
    String s[]=new String[2];//for String array of numbers
     while(stk.hasMoreTokens())
     {
        s[i]=stk.nextToken();
        n[i]=Integer.parseInt(s[i]);//Converting into Integer
       i++;
     }
  for(i=0;i<2;i++)
  System.out.println("number["+i+"]="+n[i]);
}

Output :-number[0]=1 number[1]=2

Answer 8

    String str = "1,2,3,4,5,6,7,8,9,0";
    String items[] = str.split(",");
    int ent[] = new int[items.length];
    for(i=0;i<items.length;i++){
        try{
            ent[i] = Integer.parseInt(items[i]);
            System.out.println("#"+i+": "+ent[i]);//Para probar
        }catch(NumberFormatException e){
            //Error
        }
    }

Answer 9

In tight loops or on mobile devices it's not a good idea to generate lots of garbage through short-lived String objects, especially when parsing long arrays.

The method in my answer parses data without generating garbage, but it does not deal with invalid data gracefully and cannot parse negative numbers. If your data comes from untrusted source, you should be doing some additional validation or use one of the alternatives provided in other answers.

public static void readToArray(String line, int[] resultArray) {
    int index = 0;
    int number = 0;

    for (int i = 0, n = line.length(); i < n; i++) {
        char c = line.charAt(i);
        if (c == ',') {
            resultArray[index] = number;
            index++;
            number = 0;
        }
        else if (Character.isDigit(c)) {
            int digit = Character.getNumericValue(c);
            number = number * 10 + digit;
        }
    }

    if (index < resultArray.length) {
        resultArray[index] = number;
    }
}

public static int[] toArray(String line) {
    int[] result = new int[countOccurrences(line, ',') + 1];
    readToArray(line, result);
    return result;
}

public static int countOccurrences(String haystack, char needle) {
    int count = 0;
    for (int i=0; i < haystack.length(); i++) {
        if (haystack.charAt(i) == needle) {
            count++;
        }
    }
    return count;
}

countOccurrences implementation was shamelessly stolen from John Skeet

Answer 10

String arr= "[1,2]";
List<Integer> arrList= JSON.parseArray(arr,Integer.class).stream().collect(Collectors.toList());
Integer[] intArr = ArrayUtils.toObject(arrList.stream().mapToInt(Integer::intValue).toArray());

Answer 11

If you prefer an Integer[] instead array of an int[] array:

Integer[]

String str = "[1,2]";
String plainStr = str.substring(1, str.length()-1); // clear braces []
String[] parts = plainStr.split(","); 
Integer[] result = Stream.of(parts).mapToInt(Integer::parseInt).boxed().toArray(Integer[]::new);

int[]

String str = "[1,2]";
String plainStr = str.substring(1, str.length()-1); // clear braces []
String[] parts = plainStr.split(","); 
int[] result = Stream.of(parts).mapToInt(Integer::parseInt).toArray()

This works for Java 8 and higher.