How to check if a String is a numeric type in Java

Mayank Tripathi

Post:397

Points:3117

Re: How to check if a String is a numeric type in Java

Posted on 2 years ago

This is generally done with a simple user-defined function (i.e. Roll-your-own "isNumeric" function).

Something like:

public static boolean isNumeric(String str)  
{  
  try  
  {  
    double d = Double.parseDouble(str);  
  }  
  catch(NumberFormatException nfe)  
  {  
    return false;  
  }  
  return true;  
}

However, if you're calling this function a lot, and you expect many of the checks to fail due to not being a number then performance of this mechanism will not be great, since you're relying upon exceptions being thrown for each failure, which is a fairly expensive operation.

An alternative approach may be to use a regular expression to check for validity of being a number:

public static boolean isNumeric(String str)
{
  return str.matches("-?\\d+(\\.\\d+)?");  //match a number with optional '-' and decimal.
}

Be careful with the above RegEx mechanism, though, as it'll fail if your using non-latin (i.e. 0 to 9) digits. For example, arabic digits. This is because the "\d" part of the RegEx will only match [0-9] and effectively isn't internationally numerically aware. (Thanks to OregonGhost for pointing this out!)

Or even another alternative is to use Java's built-in java.text.NumberFormat object to see if, after parsing the string the parser position is at the end of the string. If it is, we can assume the entire string is numeric:

public static boolean isNumeric(String str)
{
  NumberFormat formatter = NumberFormat.getInstance();
  ParsePosition pos = new ParsePosition(0);
  formatter.parse(str, pos);
  return str.length() == pos.getIndex();
}

Mayank Tripathi
Post:397
Points:3117
Re: How to check if a String is a numeric type in Java
Posted on 2 years ago

This is generally done with a simple user-defined function (i.e. Roll-your-own "isNumeric" function).

Something like:
```
public static boolean isNumeric(String str)  
{  
  try  
  {  
    double d = Double.parseDouble(str);  
  }  
  catch(NumberFormatException nfe)  
  {  
    return false;  
  }  
  return true;  
}
```
However, if you're calling this function a lot, and you expect many of the checks to fail due to not being a number then performance of this mechanism will not be great, since you're relying upon exceptions being thrown for each failure, which is a fairly expensive operation.

An alternative approach may be to use a regular expression to check for validity of being a number:
```
public static boolean isNumeric(String str)
{
  return str.matches("-?\\d+(\\.\\d+)?");  //match a number with optional '-' and decimal.
}
```
Be careful with the above RegEx mechanism, though, as it'll fail if your using non-latin (i.e. 0 to 9) digits. For example, arabic digits. This is because the "\d" part of the RegEx will only match [0-9] and effectively isn't internationally numerically aware. (Thanks to OregonGhost for pointing this out!)

Or even another alternative is to use Java's built-in java.text.NumberFormat object to see if, after parsing the string the parser position is at the end of the string. If it is, we can assume the entire string is numeric:
```
public static boolean isNumeric(String str)
{
  NumberFormat formatter = NumberFormat.getInstance();
  ParsePosition pos = new ParsePosition(0);
  formatter.parse(str, pos);
  return str.length() == pos.getIndex();
}
```
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HOW TO CHECK IF A STRING IS A NUMERIC TYPE IN JAVA

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HOW TO CHECK IF A STRING IS A NUMERIC TYPE IN JAVA

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