如果你想检查字符串是否代表一个适合int类型的整数，我对jonas的答案做了一点修改，以便字符串表示大于integer的整数。MAX_VALUE或小于Integer。MIN_VALUE，现在将返回false。例如:"3147483647"将返回false，因为3147483647大于2147483647，同样，"-2147483649"也将返回false，因为-2147483649小于-2147483648。

public static boolean isInt(String s) {
  if(s == null) {
    return false;
  }
  s = s.trim(); //Don't get tricked by whitespaces.
  int len = s.length();
  if(len == 0) {
    return false;
  }
  //The bottom limit of an int is -2147483648 which is 11 chars long.
  //[note that the upper limit (2147483647) is only 10 chars long]
  //Thus any string with more than 11 chars, even if represents a valid integer, 
  //it won't fit in an int.
  if(len > 11) {
    return false;
  }
  char c = s.charAt(0);
  int i = 0;
  //I don't mind the plus sign, so "+13" will return true.
  if(c == '-' || c == '+') {
    //A single "+" or "-" is not a valid integer.
    if(len == 1) {
      return false;
    }
    i = 1;
  }
  //Check if all chars are digits
  for(; i < len; i++) {
    c = s.charAt(i);
    if(c < '0' || c > '9') {
      return false;
    }
  }
  //If we reached this point then we know for sure that the string has at
  //most 11 chars and that they're all digits (the first one might be a '+'
  // or '-' thought).
  //Now we just need to check, for 10 and 11 chars long strings, if the numbers
  //represented by the them don't surpass the limits.
  c = s.charAt(0);
  char l;
  String limit;
  if(len == 10 && c != '-' && c != '+') {
    limit = "2147483647";
    //Now we are going to compare each char of the string with the char in
    //the limit string that has the same index, so if the string is "ABC" and
    //the limit string is "DEF" then we are gonna compare A to D, B to E and so on.
    //c is the current string's char and l is the corresponding limit's char
    //Note that the loop only continues if c == l. Now imagine that our string
    //is "2150000000", 2 == 2 (next), 1 == 1 (next), 5 > 4 as you can see,
    //because 5 > 4 we can guarantee that the string will represent a bigger integer.
    //Similarly, if our string was "2139999999", when we find out that 3 < 4,
    //we can also guarantee that the integer represented will fit in an int.
    for(i = 0; i < len; i++) {
      c = s.charAt(i);
      l = limit.charAt(i);
      if(c > l) {
        return false;
      }
      if(c < l) {
        return true;
      }
    }
  }
  c = s.charAt(0);
  if(len == 11) {
    //If the first char is neither '+' nor '-' then 11 digits represent a 
    //bigger integer than 2147483647 (10 digits).
    if(c != '+' && c != '-') {
      return false;
    }
    limit = (c == '-') ? "-2147483648" : "+2147483647";
    //Here we're applying the same logic that we applied in the previous case
    //ignoring the first char.
    for(i = 1; i < len; i++) {
      c = s.charAt(i);
      l = limit.charAt(i);
      if(c > l) {
        return false;
      }
      if(c < l) {
        return true;
      }
    }
  }
  //The string passed all tests, so it must represent a number that fits
  //in an int...
  return true;
}

I recently (today) needed to figure out a quick way to do this and of course I was going to use the exception approach for ease when the monkey on the shoulder (conscience) woke up so it took me down this old familiar rabbit hole; no exceptions are not that much more expensive in fact sometimes exceptions are faster (old AIX multiprocessor systems) but regardless it’s to elegant so I did something that the younger me never did and to my amazement nobody here did either (apologize if someone did and I missed it I honestly did not find) : so what did I think we all missed; taking a look at how the JRE implemented it, yes they threw an exception but we can always skip that part.

10年前年轻的我可能会觉得这有失他的身份，但话又说回来，他是一个大嘴巴的炫耀者，性情不好，有一种神的情结，所以就是这样。

我把这些放在这里，是为了方便将来来这里的人。以下是我的发现:

public static int parseInt(String s, int radix) throws NumberFormatException
{
    /*
     * WARNING: This method may be invoked early during VM initialization
     * before IntegerCache is initialized. Care must be taken to not use
     * the valueOf method.
     */

    if (s == null) {
        throw new NumberFormatException("null");
    }

    if (radix < Character.MIN_RADIX) {
        throw new NumberFormatException("radix " + radix +
                                        " less than Character.MIN_RADIX");
    }

    if (radix > Character.MAX_RADIX) {
        throw new NumberFormatException("radix " + radix +
                                        " greater than Character.MAX_RADIX");
    }

    int result = 0;
    boolean negative = false;
    int i = 0, len = s.length();
    int limit = -Integer.MAX_VALUE;
    int multmin;
    int digit;

    if (len > 0) {
        char firstChar = s.charAt(0);
        if (firstChar < '0') { // Possible leading "+" or "-"
            if (firstChar == '-') {
                negative = true;
                limit = Integer.MIN_VALUE;
            } else if (firstChar != '+')
                throw NumberFormatException.forInputString(s);

            if (len == 1) // Cannot have lone "+" or "-"
                throw NumberFormatException.forInputString(s);
            i++;
        }
        multmin = limit / radix;
        while (i < len) {
            // Accumulating negatively avoids surprises near MAX_VALUE
            digit = Character.digit(s.charAt(i++),radix);
            if (digit < 0) {
                throw NumberFormatException.forInputString(s);
            }
            if (result < multmin) {
                throw NumberFormatException.forInputString(s);
            }
            result *= radix;
            if (result < limit + digit) {
                throw NumberFormatException.forInputString(s);
            }
            result -= digit;
        }
    } else {
        throw NumberFormatException.forInputString(s);
    }
    return negative ? result : -result;
}

public class HelloWorld{

    static boolean validateIP(String s){
        String[] value = s.split("\\.");
        if(value.length!=4) return false;
        int[] v = new int[4];
        for(int i=0;i<4;i++){
            for(int j=0;j<value[i].length();j++){
                if(!Character.isDigit(value[i].charAt(j))) 
                 return false;
            }
            v[i]=Integer.parseInt(value[i]);
            if(!(v[i]>=0 && v[i]<=255)) return false;
        }
        return true;
    }

    public static void main(String[] argv){
        String test = "12.23.8.9j";
        if(validateIP(test)){
            System.out.println(""+test);
        }
    }
}

Integer.valueOf(string); 

大部分时间对我都管用!

这部分取决于你所说的“可以转换为整数”是什么意思。

如果你的意思是“可以在Java中转换为int型”，那么Jonas的回答是一个很好的开始，但并没有完全完成工作。例如，它会通过9999999999999999999999999999999999。我会在方法的末尾添加来自您自己的问题的正常try/catch调用。

逐字符检查将有效地拒绝“根本不是整数”的情况，留下“它是整数但Java不能处理它”的情况，由较慢的异常路由捕获。你也可以用手来做，但会复杂得多。

对@Jonas K answer进行了一点改进，这个函数将排除一个像“*”这样的操作符。

public boolean isInteger(String str) {
    // null pointer
    if (str == null) {
        return false;
    }
    int len = str.length();
    // empty string
    if (len == 0) {
        return false;
    }
    // one digit, cannot begin with 0
    if (len == 1) {
        char c = str.charAt(0);
        if ((c < '1') || (c > '9')) {
            return false;
        }
    }

    for (int i = 0; i < len; i++) {
        char c = str.charAt(i);
        // check positive, negative sign
        if (i == 0) {
            if (c == '-' || c == '+') {
                continue;
            }
        }
        // check each character matches [0-9]
        if ((c < '0') || (c > '9')) {
            return false;
        }
    }
    return true;
}