Java - checking if parseInt throws exception

Question

I'm wondering how to do something only if Integer.parseInt(whatever) doesn't fail.

More specifically I have a jTextArea of user specified values seperated by line breaks.

I want to check each line to see if can be converted to an int.

Figured something like this, but it doesn't work:

for(int i = 0; i < worlds.jTextArea1.getLineCount(); i++){
                    if(Integer.parseInt(worlds.jTextArea1.getText(worlds.jTextArea1.getLineStartOffset(i),worlds.jTextArea1.getLineEndOffset(i)) != (null))){}
 }

Any help appreciated.

Possible duplicate of What is a NumberFormatException and how can I fix it? — xenteros
– xenteros, Commented Oct 23, 2016 at 12:26

user2891462 · Accepted Answer · 2020-05-13 16:28:42Z

62

public static boolean isParsable(String input) {
    try {
        Integer.parseInt(input);
        return true;
    } catch (final NumberFormatException e) {
        return false;
    }
}

edited May 13, 2020 at 16:28

user2891462

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answered Jun 23, 2011 at 15:05

fmucar

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5 Comments

Amir Raminfar Over a year ago

Don't catch ALL exceptions. That's really bad practice.

user207421 Over a year ago

NumberFormatException. You may as well get it right. It's documented.

user4919188 Over a year ago

this will ONLY work with this specific try/catch statement or any other NumberFormatException.

Kerem Baydoğan Over a year ago

"Don't catch ALL exceptions. That's really bad practice." In this case, it is perfectly fine and also the right thing to do because what you are only doing is checking if anything went wrong. Catch all exceptions and log them.

DownloadPizza Over a year ago

Catching ALL exceptions is never the right thing to do, as the only Exceptions which are expected here are the ones thrown by parseInt. If anything else were to go wrong you would not want the program to keep running (lets say an illegalaccessexception because someone messed with reflection)

Kerem Baydoğan · Accepted Answer · 2011-06-23 15:15:04Z

28

Check if it is integer parseable

public boolean isInteger(String string) {
    try {
        Integer.valueOf(string);
        return true;
    } catch (NumberFormatException e) {
        return false;
    }
}

or use Scanner

Scanner scanner = new Scanner("Test string: 12.3 dog 12345 cat 1.2E-3");

while (scanner.hasNext()) {
    if (scanner.hasNextDouble()) {
        Double doubleValue = scanner.nextDouble();
    } else {
        String stringValue = scanner.next();
    }
}

or use Regular Expression like

private static Pattern doublePattern = Pattern.compile("-?\\d+(\\.\\d*)?");

public boolean isDouble(String string) {
    return doublePattern.matcher(string).matches();
}

edited Jun 23, 2011 at 15:15

answered Jun 23, 2011 at 15:09

Kerem Baydoğan

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1 Comment

Pascal Kesseli Over a year ago

+1 for verification without exceptions using Scanner.

Stefan Bossbaly · Accepted Answer · 2011-06-23 15:13:22Z

12

It would be something like this.

String text = textArea.getText();
Scanner reader = new Scanner(text).useDelimiter("\n");
while(reader.hasNext())
    String line = reader.next();

    try{
        Integer.parseInt(line);
        //it worked
    }
    catch(NumberFormatException e){
       //it failed
    }
}

edited Jun 23, 2011 at 15:13

answered Jun 23, 2011 at 15:07

Stefan Bossbaly

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Comments

Amir Raminfar · Accepted Answer · 2011-06-23 15:06:04Z

10

parseInt will throw NumberFormatException if it cannot parse the integer. So doing this will answer your question

try{
Integer.parseInt(....)
}catch(NumberFormatException e){
//couldn't parse
}

answered Jun 23, 2011 at 15:06

Amir Raminfar

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1 Comment

Luís Henriques Over a year ago

This should be the correct answer, as the accepted one is wrong.

dogbane · Accepted Answer · 2011-06-23 15:17:05Z

8

You can use a scanner instead of try-catch:

Scanner scanner = new Scanner(line).useDelimiter("\n");
if(scanner.hasNextInt()){
    System.out.println("yes, it's an int");
}

answered Jun 23, 2011 at 15:17

dogbane

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Comments

Siddhartha Ghosh · Accepted Answer · 2018-01-12 05:44:53Z

3

You could try

NumberUtils.isParsable(yourInput)

It is part of org/apache/commons/lang3/math/NumberUtils and it checks whether the string can be parsed by Integer.parseInt(String), Long.parseLong(String), Float.parseFloat(String) or Double.parseDouble(String).

See below:

https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/math/NumberUtils.html#isParsable-java.lang.String-

answered Jan 12, 2018 at 5:44

Siddhartha Ghosh

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Comments

Yet Another Geek · Accepted Answer · 2011-06-23 15:06:53Z

2

You can use the try..catch statement in java, to capture an exception that may arise from Integer.parseInt().

Example:

try {
  int i = Integer.parseint(stringToParse);
  //parseInt succeded
} catch(Exception e)
{
   //parseInt failed
}

answered Jun 23, 2011 at 15:06

Yet Another Geek

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3 Comments

josh.trow Over a year ago

Why the downvote? This is a valid answer, and though the Exception catching is too vague for good practice it will work.

Yet Another Geek Over a year ago

@josh.trow, throwing an exception when doing a parseint is also bad practice as it really is not something truly exceptional, one could either provide a parsable method (to check if it can be parsed) or convert it to a weird value (like Ruby and perl)

user207421 Over a year ago

So what is the exception for?

Anantha Sharma · Accepted Answer · 2011-06-23 15:11:21Z

2

instead of trying & catching expressions.. its better to run regex on the string to ensure that it is a valid number..

answered Jun 23, 2011 at 15:11

Anantha Sharma

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4 Comments

user207421 Over a year ago

Better how? Why reinvent the wheel? Why run the risk that the regex doesn't match what parseInt() does? Why waste the time and the money?

Anantha Sharma Over a year ago

when an exception there is an overhead (for the JVM) to prepare the stack trace & then continue with executing the program (this is a bit time consuming)... its always better to check the expression before parsing it..

user207421 Over a year ago

The overhead is insignificant, and the other issues I raised are not. Your blanket statement is not supportable.

Adam Over a year ago

7 years later and no-one has picked up on this one - accusations of 'blanket statements' is not helpful. If performance is critical, then it would be necessary to run comparisons of the two techniques. Neither preparing regex nor stacktraces are completely negligible. And in fact the regex could be cached.

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