确定在Java中字符串是否是整数[复制]-DetermineifaStringisanIntegerinJava[duplicate]

188  

Using regular expression is better.

使用正则表达式比较好。

str.matches("-?\\d+");


-?     --> negative sign, could have none or one
\\d+   --> one or more digits

It is not good to use NumberFormatException here if you can use if-statement instead.

如果可以使用if语句，那么在这里使用NumberFormatException是不好的。

110  

Or you can enlist a little help from our good friends at Apache Commons : StringUtils.isNumeric(String str)

或者你也可以从我们在Apache Commons的好朋友那里得到一点帮助:StringUtils。isNumeric(String str)

61  

Or simply

或者简单地

mystring.matches("\\d+")

mystring.matches(“\ \ d +”)

though it would return true for numbers larger than an int

虽然它会返回大于整数的数。

43  

You want to use the Integer.parseInt(String) method.

您需要使用Integer.parseInt(String)方法。

try{
  int num = Integer.parseInt(str);
  // is an integer!
} catch (NumberFormatException e) {
  // not an integer!
}

12  

As an alternative to trying to parse the string and catching NumberFormatException, you could use a regex; e.g.

作为尝试解析字符串和捕获NumberFormatException的替代方法，您可以使用regex;如。

if (Pattern.compile("-?[0-9]+").matches(str)) {
    // its an integer
}

This is likely to be faster, especially if you precompile and reuse the regex. However, the catch is that Integer.parseInt(str) will still fail if str represents a number that is outside range of legal int values.

这可能会更快，特别是如果您预编译和重用regex。但是，如果str表示一个超出合法int值范围的数字，则parseint (str)仍然会失败。

8  

You can use Integer.parseInt(str) and catch the NumberFormatException if the string is not a valid integer, in the following fashion (as pointed out by all answers):

您可以使用Integer.parseInt(str)并捕获NumberFormatException，如果字符串不是一个有效整数，按照以下方式(所有答案都指出):

static boolean isInt(String s)
{
 try
  { int i = Integer.parseInt(s); return true; }

 catch(NumberFormatException er)
  { return false; }
}

However, note here that if the evaluated integer overflows, the same exception will be thrown. Your purpose was to find out whether or not, it was a valid integer. So its safer to make your own method to check for validity:

但是，请注意，如果评估的整数溢出，将抛出相同的异常。你的目的是找出它是否是一个有效的整数。因此，用你自己的方法来检查有效性更安全:

static boolean isInt(String s)  // assuming integer is in decimal number system
{
 for(int a=0;a