I have a String
and I want to know if the String
is a number, it could be also a negative number
String test1 = "abcd"; // Here it must show that it's not a number
String test2 = "abcd-123"; // Here it must show that it's not a number
String test3 = "123"; // Here it must show that it's a number
String test4 = "-.12"; // Here it must show that it's a number
String test5 = "-123"; // Here it must show that it's a number
String test6 = "123.0; // Here it must show that it's a number
String test7 = "-123.00"; // Here it must show that it's a number
String test8 = "-123.15"; // Here it must show that it's a number
String test9 = "09"; // Here it must show that it's a number
String test10 = "0.0"; // Here it must show that it's a number
I used StringUtils#isNumber
and NumberUtils#isNumber
, but they don't help, negative numbers , "09" show as not a number
You can try this:
try {
double value = Double.parseDouble(test1);
if(value<0)
System.out.println(value + " is negative");
else
System.out.println(value + " is possitive");
} catch (NumberFormatException e) {
System.out.println("String "+ test1 + "is not a number");
}
According to the Javadoc of Double.valueOf
:
To avoid calling this method on an invalid string and having a NumberFormatException be thrown, the regular expression below can be used to screen the input string:
final String Digits = "(\\p{Digit}+)";
final String HexDigits = "(\\p{XDigit}+)";
// an exponent is 'e' or 'E' followed by an optionally
// signed decimal integer.
final String Exp = "[eE][+-]?"+Digits;
final String fpRegex =
("[\\x00-\\x20]*"+ // Optional leading "whitespace"
"[+-]?(" + // Optional sign character
"NaN|" + // "NaN" string
"Infinity|" + // "Infinity" string
// A decimal floating-point string representing a finite positive
// number without a leading sign has at most five basic pieces:
// Digits . Digits ExponentPart FloatTypeSuffix
//
// Since this method allows integer-only strings as input
// in addition to strings of floating-point literals, the
// two sub-patterns below are simplifications of the grammar
// productions from section 3.10.2 of
// The Java™ Language Specification.
// Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
"((("+Digits+"(\\.)?("+Digits+"?)("+Exp+")?)|"+
// . Digits ExponentPart_opt FloatTypeSuffix_opt
"(\\.("+Digits+")("+Exp+")?)|"+
// Hexadecimal strings
"((" +
// 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
"(0[xX]" + HexDigits + "(\\.)?)|" +
// 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
"(0[xX]" + HexDigits + "?(\\.)" + HexDigits + ")" +
")[pP][+-]?" + Digits + "))" +
"[fFdD]?))" +
"[\\x00-\\x20]*");// Optional trailing "whitespace"
if (Pattern.matches(fpRegex, myString))
Double.valueOf(myString); // Will not throw NumberFormatException
else {
// Perform suitable alternative action
}
The regular expression is substantial, but comprehensive and well-documented. You can trim it as you like.
尝试使用 catch 语句将输入解析为围绕它的整数,如果它不转换它应该返回 input 是一个字符如果它转换它返回 input 是一个数字
public static String checknumeric(String str){
String numericString = null;
String temp;
if(str.startsWith("-")){ //checks for negative values
temp=str.substring(1);
if(temp.matches("[+]?\\d*(\\.\\d+)?")){
numericString=str;
}
}
if(str.matches("[+]?\\d*(\\.\\d+)?")) {
numericString=str;
}
return numericString;
}
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