在解析字符串之前，如何检查它是否是一个数字?
当前回答
下面是用于检查字符串是否为数字的类。它还修复数值字符串:
特点:
删除不必要的零["12.0000000" -> "12"] 删除不必要的零["12.0580000" -> "12.058"] 删除非数字字符["12.00sdfsdf00" -> "12"] 处理负字符串值["-12,020000" -> "-12.02"] 删除多个点["-12.0.20.000" -> "-12.02"] 没有额外的库，只有标准Java
给你…
public class NumUtils {
/**
* Transforms a string to an integer. If no numerical chars returns a String "0".
*
* @param str
* @return retStr
*/
static String makeToInteger(String str) {
String s = str;
double d;
d = Double.parseDouble(makeToDouble(s));
int i = (int) (d + 0.5D);
String retStr = String.valueOf(i);
System.out.printf(retStr + " ");
return retStr;
}
/**
* Transforms a string to an double. If no numerical chars returns a String "0".
*
* @param str
* @return retStr
*/
static String makeToDouble(String str) {
Boolean dotWasFound = false;
String orgStr = str;
String retStr;
int firstDotPos = 0;
Boolean negative = false;
//check if str is null
if(str.length()==0){
str="0";
}
//check if first sign is "-"
if (str.charAt(0) == '-') {
negative = true;
}
//check if str containg any number or else set the string to '0'
if (!str.matches(".*\\d+.*")) {
str = "0";
}
//Replace ',' with '.' (for some european users who use the ',' as decimal separator)
str = str.replaceAll(",", ".");
str = str.replaceAll("[^\\d.]", "");
//Removes the any second dots
for (int i_char = 0; i_char < str.length(); i_char++) {
if (str.charAt(i_char) == '.') {
dotWasFound = true;
firstDotPos = i_char;
break;
}
}
if (dotWasFound) {
String befDot = str.substring(0, firstDotPos + 1);
String aftDot = str.substring(firstDotPos + 1, str.length());
aftDot = aftDot.replaceAll("\\.", "");
str = befDot + aftDot;
}
//Removes zeros from the begining
double uglyMethod = Double.parseDouble(str);
str = String.valueOf(uglyMethod);
//Removes the .0
str = str.replaceAll("([0-9])\\.0+([^0-9]|$)", "$1$2");
retStr = str;
if (negative) {
retStr = "-"+retStr;
}
return retStr;
}
static boolean isNumeric(String str) {
try {
double d = Double.parseDouble(str);
} catch (NumberFormatException nfe) {
return false;
}
return true;
}
}
其他回答
// only int
public static boolean isNumber(int num)
{
return (num >= 48 && c <= 57); // 0 - 9
}
// is type of number including . - e E
public static boolean isNumber(String s)
{
boolean isNumber = true;
for(int i = 0; i < s.length() && isNumber; i++)
{
char c = s.charAt(i);
isNumber = isNumber & (
(c >= '0' && c <= '9') || (c == '.') || (c == 'e') || (c == 'E') || (c == '')
);
}
return isInteger;
}
// is type of number
public static boolean isInteger(String s)
{
boolean isInteger = true;
for(int i = 0; i < s.length() && isInteger; i++)
{
char c = s.charAt(i);
isInteger = isInteger & ((c >= '0' && c <= '9'));
}
return isInteger;
}
public static boolean isNumeric(String s)
{
try
{
Double.parseDouble(s);
return true;
}
catch (Exception e)
{
return false;
}
}
如果字符串可能包含小数，则可以使用BigDecimal:
try {
new java.math.BigInteger(testString);
} catch(NumberFormatException e) {
throw new RuntimeException("Not a valid number");
}
您可以使用Apache Commons Lang中的NumberUtils.isCreatable()。
因为NumberUtils。isNumber将在4.0中被弃用，所以使用NumberUtils.isCreatable()代替。
下面是用于检查字符串是否为数字的类。它还修复数值字符串:
特点:
删除不必要的零["12.0000000" -> "12"] 删除不必要的零["12.0580000" -> "12.058"] 删除非数字字符["12.00sdfsdf00" -> "12"] 处理负字符串值["-12,020000" -> "-12.02"] 删除多个点["-12.0.20.000" -> "-12.02"] 没有额外的库，只有标准Java
给你…
public class NumUtils {
/**
* Transforms a string to an integer. If no numerical chars returns a String "0".
*
* @param str
* @return retStr
*/
static String makeToInteger(String str) {
String s = str;
double d;
d = Double.parseDouble(makeToDouble(s));
int i = (int) (d + 0.5D);
String retStr = String.valueOf(i);
System.out.printf(retStr + " ");
return retStr;
}
/**
* Transforms a string to an double. If no numerical chars returns a String "0".
*
* @param str
* @return retStr
*/
static String makeToDouble(String str) {
Boolean dotWasFound = false;
String orgStr = str;
String retStr;
int firstDotPos = 0;
Boolean negative = false;
//check if str is null
if(str.length()==0){
str="0";
}
//check if first sign is "-"
if (str.charAt(0) == '-') {
negative = true;
}
//check if str containg any number or else set the string to '0'
if (!str.matches(".*\\d+.*")) {
str = "0";
}
//Replace ',' with '.' (for some european users who use the ',' as decimal separator)
str = str.replaceAll(",", ".");
str = str.replaceAll("[^\\d.]", "");
//Removes the any second dots
for (int i_char = 0; i_char < str.length(); i_char++) {
if (str.charAt(i_char) == '.') {
dotWasFound = true;
firstDotPos = i_char;
break;
}
}
if (dotWasFound) {
String befDot = str.substring(0, firstDotPos + 1);
String aftDot = str.substring(firstDotPos + 1, str.length());
aftDot = aftDot.replaceAll("\\.", "");
str = befDot + aftDot;
}
//Removes zeros from the begining
double uglyMethod = Double.parseDouble(str);
str = String.valueOf(uglyMethod);
//Removes the .0
str = str.replaceAll("([0-9])\\.0+([^0-9]|$)", "$1$2");
retStr = str;
if (negative) {
retStr = "-"+retStr;
}
return retStr;
}
static boolean isNumeric(String str) {
try {
double d = Double.parseDouble(str);
} catch (NumberFormatException nfe) {
return false;
}
return true;
}
}
正则表达式匹配
这里是另一个例子升级了“CraigTP”正则表达式匹配更多的验证。
public static boolean isNumeric(String str)
{
return str.matches("^(?:(?:\\-{1})?\\d+(?:\\.{1}\\d+)?)$");
}
只有一个负号——允许且必须在开头。 负号后面必须有数字。 只有一个小数符号。允许的。 小数点后必须有数字。
正则表达式测试
1 -- **VALID**
1. -- INVALID
1.. -- INVALID
1.1 -- **VALID**
1.1.1 -- INVALID
-1 -- **VALID**
--1 -- INVALID
-1. -- INVALID
-1.1 -- **VALID**
-1.1.1 -- INVALID
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