CODEWITHSUNDEEP
javavalidationtry-catch
John
John

Reputation: 75

How can I validate if a string is a string and not an integer in java?

I enter a list into a JTextArea, and when I push a button, it runs the method below. I need to check to see if str[i+1], str[i+2] is a String and not an int.

public void readData(JTextArea input) {
    String[] str = input.getText().split("\n");
    for(int i =1; i< str.length; i= i+3) {
        try {
            Integer.parseInt(str[i]);
            simulator.add(new Process(Integer.parseInt(str[i]), str[i+1],str[i+2])); 
        } catch(NumberFormatException e) {
            System.out.println("Please enter an integer only " +
                                str[i] + " is not an integer");
        }
    }
}

Upvotes: 1

Views: 491

Answers (5)

nem035
nem035

Reputation: 35481

You could have a function that tries to parse the string into an Integer or a Double and return true if it succeeded, or return false is an exception was thrown during parsing. Parsing into a Double should be enough since all integer values are decimal values without the .0

public static boolean isNumber(String s) {
    try {  
        Double.parseDouble(s);
        /* Use Integer.parseInt(s) instead, if
           you want to check if the String s
           is an Integer */
    } catch(NumberFormatException e) { // string is not a number
        return false; 
    }
    return true;
}

Then you can say if(!isNumber(str)) to check if the String str is not a number.

Alternatively, you could make the isNumber() be a isNotNumber() by swapping the return false and return true statements.

If you don't want to use exceptions, a different approach would be the following. Since we know a valid number can only contain digits and at most 1 dot for decimal point, we can iterate through the string and check for each character:

  • if it is not a digit and not a dot, return false
  • if it is a dot but a dot was already found, return false
  • otherwise it is valid number character and we do nothing

Here is a sample function:

public static boolean isNumber(String s) {
    int dotCount = 0;
    for(int i = 0; i < s.length(); i++) {
        if(s.charAt(i) != '.' && !Character.isDigit(s.charAt(i))) {
            return false;
        } else if(s.charAt(i) == '.') {
            if(dotCount == 1) {
                return false;
            }
            dotCount = 1;
        }
    }
    return true;
}

EDIT: based on @MadProgrammer's suggestions:

A more general approach that will accept values separated with commas such as 1,35 or any amount of spaces within the number string like with 123 456 . 333.

Approach:

Iterate through the string and check for each character:

  • if it is not a digit, dot, comma, or a space, return false
  • if it is a dot or a comma but one of them was already found, return false
  • otherwise it is valid number character and we do nothing

So the code would look something like:

public static boolean isNumber(String s) {
    int separatorCount = 0; // count dots and commas
    char currChar;
    s.trim();  // remove trailing and leading whitespace
    for (int i = 0; i < s.length(); i++) {
        currChar = s.charAt(i);
        if (currChar != '.' && currChar != ',' && currChar != ' '
                && !Character.isDigit(currChar)) {
            return false;
        } else if (currChar == '.' || currChar == ',') {
            if (separatorCount == 1) {
                return false;
            }
            separatorCount = 1;
        }
    }
    return true;
}

Another solution could use the NumberFormat's parse() method. However, this method only checks the beginning of the string (for example, for 12.3.3 it will return 12.3) so we have to return false if the returned string doesn't equal the input string as well as if the ParseException is thrown.

public static boolean isNumber(String s) {
    try {
        String newVal = NumberFormat.getInstance().parse(s).toString();
        if (!newVal.equals(s)) {
            return false;
        }
    } catch (ParseException e) {
        return false;
    }
    return true;
}

NOTE: All of the methods should probably have a check if(s == null) { return false; } for the input String s to prevent a NullPointerException

Upvotes: 3

Bohemian
Bohemian

Reputation: 424983

This is a simple test that asserts there is at least one non numeric char:

if (str.matches(".*[^\\d.].*"))

the regex translates as "somewhere in the input there's a character that's not a digit or a dot"

Upvotes: 0

Manjunath
Manjunath

Reputation: 1685

You can try this simple regular expression to check if a string represents a number or not:-

String str = "12345";
System.out.println(str.matches("\\d+"));

Upvotes: 2

Barun
Barun

Reputation: 1622

Regex seems the best option. Here's a regex that will parse float and integers and currency values with comma as well.

 String numberRex = "^([\\d]*\\.*\\d*|[\\d,]*\\.*\\d*)$";
 "1234,455.43".matches(numberRex); // true

Upvotes: 0

MadProgrammer
MadProgrammer

Reputation: 347184

Your rules are sparse, you don't specify if things like , or . are considered part of a number or not (1, 000.01 for example), also, you don't define if the value is allowed to contain numerical values or not, but...

You Could...

Try parsing each value (or the concatenation of the two) using Integer.parseInt, if they pass, then they are not String (or text) based values...

You Could...

Verify each character within the String to see if it contains more than just digits, using something like Character#isLetter

You Could...

Use a regular expression to determine if the value contain other content other than numbers.

Upvotes: 2

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