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What am I missing? NumberFormatException error

I want to read from a txt file which contains just numbers. Such file is in UTF-8, and the numbers are separated only by new lines (no spaces or any other things) just that. Whenever i call Integer.valueOf(myString), i get the exception.
This exception is really strange, because if i create a predefined string, such as "56\n", and use .trim(), it works perfectly. But in my code, not only that is not the case, but the exception texts says that what it couldn't convert was "54856". I have tried to introduce a new line there, and then the error text says it couldn't convert "54856
"
With that out of the question, what am I missing?
File ficheroEntrada = new File("C:\\in.txt");
FileReader entrada =new FileReader(ficheroEntrada);
BufferedReader input = new BufferedReader(entrada);
String s = input.readLine();
System.out.println(s);
Integer in;
in = Integer.valueOf(s.trim());
System.out.println(in);
The exception text reads as follows:
Exception in thread "main" java.lang.NumberFormatException: For input string: "54856"
at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:68)
at java.base/java.lang.Integer.parseInt(Integer.java:658)
at java.base/java.lang.Integer.valueOf(Integer.java:989)
at Quicksort.main(Quicksort.java:170)
The file in.txt consists of:
54856
896
54
53
2
5634

Well, aparently it had to do with Windows and those \r that it uses... I just tried executing it on a Linux VM and it worked. Thanks to everyone that answered!!

Try reading the file with Scanner class has use it's hasNextInt() method to identify what you are reading is Integer or not. This will help you find out what String/character is causing the issue
public static void main(String[] args) throws Exception {
File ficheroEntrada = new File(
"C:\\in.txt");
Scanner scan = new Scanner(ficheroEntrada);
while (scan.hasNext()) {
if (scan.hasNextInt()) {
System.out.println("found integer" + scan.nextInt());
} else {
System.out.println("not integer" + scan.next());
}
}
}

If you want to ensure parsability of a string, you could use a Pattern and Regex that.
Pattern intPattern = Pattern.compile("\\-?\\d+");
Matcher matcher = intPattern.matcher(input);
if (matcher.find()) {
int value = Integer.parseInt(matcher.group(0));
// ... do something with the result.
} else {
// ... handle unparsable line.
}
This pattern allows any numbers and optionally a minus before (without whitespace). It should definetly parse, unless it is too long. I don't know how it handles that, but your example seems to contain mostly short integers, so this should not matter.

Most probably you have a leading/trailing whitespaces in your input, something like:
String s = " 5436";
System.out.println(s);
Integer in;
in = Integer.valueOf(s.trim());
System.out.println(in);
Use trim() on string to get rid of it.
UPDATE 2:
If your file contains something like:
54856\n
896
54\n
53
2\n
5634
then use following code for it:
....your code
FileReader enter = new FileReader(file);
BufferedReader input = new BufferedReader(enter);
String currentLine;
while ((currentLine = input.readLine()) != null) {
Integer in;
//get rid of non-numbers
in = Integer.valueOf(currentLine.replaceAll("\\D+",""));
System.out.println(in);
...your code

Integer.parseInt failing 'For input string: "5000"'

I'll admit it, I'm stumped. It's not a double. It's not outside of the range of an integer. It's not NAN. It's not a non-integer in any way shape or form as far as I can tell.
Why would I get that error?
Here's the code that causes it:
String filename = "confA.txt";
//Make a new filereader to read in confA
FileReader fileReader = new FileReader(filename);
//Wrap into a bufferedReader for sanity's sake
BufferedReader bufferedReader = new BufferedReader(fileReader);
//Get the port number that B is listening to
int portNum = Integer.parseInt(bufferedReader.readLine());
It fails on that last line, stating:
java.lang.NumberFormatException: For input string: "5000"
Which is the number I want.
I've also attempted
Integer portNum = Integer.parseInt(bufferedReader.readLine());
But that didn't work either. Neither did valueOf().

Most probably there is some unprintable character somewhere in your file line. Please consider the following example (this was tested in Java 9 jshell)
jshell> String value = "5000\u0007";
value ==> "5000\007"
jshell> Integer.parseInt(value);
| java.lang.NumberFormatException thrown: For input string: "5000"
| at NumberFormatException.forInputString (NumberFormatException.java:65)
| at Integer.parseInt (Integer.java:652)
| at Integer.parseInt (Integer.java:770)
| at (#15:1)
Here the string contains the "bell" character at the end. It makes parse to fail while it is not printed in exception text. I think you have something similar. The simpliest way to verify this is to check
String line = bufferedReader.readLine();
System.out.println("line length: " + line.length());
The value other than 4 will support my idea.

I had the same problem. I was reading in a flat file with the Buffered reader and saving the contents as an ArrayList of type String, but on performing an integer.parse when retrieving a string value from the list, I realised that there was a whole lot of garbage in the string read from the file, as I got a java.lang.NumberFormatException.
This was the method I implemented with the code (called from my main method) to solve the problem:
`
// class level
private static final Pattern numericPattern = Pattern.compile("([0-9]+).([\\\\.]{0,1}[0-9]*)");
// in main method after reading in the file
String b = stripNonNumeric(stringvaluefromfile);
int a = Integer.parseInt(b);
public static String stripNonNumeric(String number) {
//System.out.println(number);
if (number == null || number.isEmpty()) {
return "0";
}
Matcher matcher = numericPattern.matcher(number);
// strip out all non-numerics
StringBuffer sb = new StringBuffer("");
while (matcher.find()) {
sb.append(matcher.group());
}
// make sure there's only one dot
int prevDot = -1;
for (int i = sb.length() - 1; i >= 0; i--) {
if (sb.charAt(i) == '.') {
if (prevDot > 0) {
sb.deleteCharAt(prevDot);
}
prevDot = i;
}
}
if (sb.length() == 0) {
sb.append("0");
}
return sb.toString();
}`

Reading From File With Comma Delimeter Error

Here's the .txt file i'm trying to read from
20,Dan,09/05/1990,3,Here
5,Danezo,04/09/1990,99,There
And here's how I'm doing it.. Whenever the .txt file has only one line, it seems to be reading from file fine. Whenever more than one line is being read, I get this error
Exception in thread "main" java.lang.NumberFormatException: For input string: "Danezo"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:580)
at java.lang.Integer.parseInt(Integer.java:615)
at AttackMonitor.readFromFile(AttackMonitor.java:137)
at AttackMonitor.monitor(AttackMonitor.java:57)
at MonsterAttackDriver.main(MonsterAttackDriver.java:14)
Java Result: 1
Here's the readfromfile code.
private void readFromFile() throws FileNotFoundException, IOException
{
monsterAttacks.clear();
Scanner read = new Scanner(new File("Attacks.txt"));
read.useDelimiter(",");
String fullDateIn = "";
int attackIdIn = 0;
int attackVictimsIn = 0;
String monsterNameIn= "";
String attackLocationIn= "";
while (read.hasNext())
{
attackIdIn = Integer.parseInt(read.next());
monsterNameIn = read.next();
fullDateIn = read.next();
attackVictimsIn = Integer.parseInt(read.next());
attackLocationIn = read.next();
monsterAttacks.add(new MonsterAttack(fullDateIn, attackIdIn, attackVictimsIn, monsterNameIn, attackLocationIn));
}
read.close();
}

What is happening is that at the end of each line there is a newline character, which is currently not a delimiter. So your code is attempting to read it as the first integer of the next line, which it is not. This is causing the parse exception.
To remedy this, you can try adding newline to the list of delimiters for which to scan:
Scanner read = new Scanner(new File("Attacks.txt"));
read.useDelimiter("[,\r\n]+"); // use just \n on Linux
An alternative to this would be to just read in each entire line from the file and split on comma:
String[] parts = read.nextLine().split(",");
attackIdIn = Integer.parseInt(parts[0]);
monsterNameIn = parts[1];
fullDateIn = parts[2];
attackVictimsIn = Integer.parseInt(parts[3]);
attackLocationIn = parts[4];

You can use the Biegeleisen suggestion. Or else you can do as follows.
In your while loop you are using hasNext as condition. Instead of that you can use while (read.hasNextLine()) and get the nextLine inside the loop and then split it by your delimiter and do the processing. That would be a more appropriate approach.
e.g
while (read.hasNextLine()) {
String[] values = scanner.nextLine().split(".");
// do your rest of the logic
}

Put the while loop content in a try catch, and catch for NumberFormatException. So whenever it falls to catch code, you can understand you tried to convert a string to int.
Could help more if your business is explained.

attackLocationIn = read.next(); This value takes as "Here\n 5" because there is no comma between Here and 5 and it has new line character.
so 2nd iteration attackIdIn = Integer.parseInt(read.next()); here read.next() value is "Danezo" it is String and you are trying parse to Integer. That's why you are getting this exception.
What I suggest is use BufferReader to read line by line and split each line with comma. It will be fast also.
Or another solution Add comma at end of each line and use read.next().trim() in your code. That's it it will work with minimal changes to your current code.

java.lang.NumberFormatException: For input string: "22"

public void loadFromFile(String filename) {
File file = new File(filename);
BufferedReader br;
try {
br = new BufferedReader(new FileReader(file));
numberOfAttributes = Integer.parseInt(br.readLine());
}
...
}
Above is my program: I am trying to read from a txt file where the first line is the number 22 and nothing more. I don't know why the program gives me an exception.

Try stripping any whitespace from the string:
numberOfAttributes = Integer.parseInt(br.readLine().trim());

I think you might have a UTF-8 BOM (byte-order mark) at the start of your file.
Here's a class that reproduces the error:
import java.io.*;
public class BomTest {
public static void main(String[] args) throws Exception {
File file = new File("example.txt");
// Write out UTF-8 BOM, followed by the number 22 and a newline.
byte[] bs = { (byte)0xef, (byte)0xbb, (byte)0xbf, (byte)'2', (byte)'2', 10 };
FileOutputStream fos = new FileOutputStream(file);
fos.write(bs);
fos.close();
BufferedReader r = new BufferedReader(new FileReader(file));
String s = r.readLine();
System.out.println(Integer.parseInt(s));
}
}
When I run this class, I get the following output:
luke#computer:~$ java BomTest
Exception in thread "main" java.lang.NumberFormatException: For input string: "22"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Integer.parseInt(Integer.java:481)
at java.lang.Integer.parseInt(Integer.java:514)
at BomTest.main(BomTest.java:15)
There isn't really an easy way to deal with UTF-8 BOMs in Java; it's best not to generate them in the first place. See also this answer.

br.readLine() reads the entire line including the new line special character.Apart, form the solution suggested by James, you can use Scanner#nextInt().

try with numberOfAttributes = Integer.parseInt(br.readLine().trim());
public String trim()
Returns a copy of the string, with leading and trailing whitespace
omitted. If this String object represents an empty character sequence,
or the first and last characters of character sequence represented by
this String object both have codes greater than '\u0020' (the space
character), then a reference to this String object is returned.
Otherwise, if there is no character with a code greater than '\u0020'
in the string, then a new String object representing an empty string
is created and returned.

This happens because you have a space in the input line. Look at these:
int i1 = Integer.parseInt("22 ");
int i2 = Integer.parseInt("22a");
int i3 = Integer.parseInt("2 2");
int i4 = Integer.parseInt("22\n");
All of them generate exception. I suggest you to trim, tokenize or substitute. But in general, it doesn't sound to me a good solution to read a number from a file in that way.
If you really need to store data, why don't you create an object ad hoc and serialize/deserialize it?

You might have a null character in your string. Remove it using a regEx "\d+".
NumberFormatException is raised because the input string is not in expected number format. Generally, you can see 'the wrong string input' in the error message and can easily identify the bug. But in your case, the catch is that the error message does not display the string input completely (because it does not displays the null character).
Check the below output and the code.
public class TestParseInt{
private static final Pattern pattern = Pattern.compile("\\d+");
public static void main(String []args){
String a = "22\0";
try {
System.out.println("Successfull parse a: " + Integer.parseInt(a));
} catch(NumberFormatException e) {
System.out.println("Error:" +e.getMessage());
}
try {
Matcher matcher = pattern.matcher(a);
if(matcher.find()) {
System.out.println("Succesfull parse a: " +
Integer.parseInt(matcher.group(0)));
}
} catch(NumberFormatException e) {
System.out.println("Error" + e.getMessage());
}
}
}
Output:
Error:For input string: "22"
Succesfull parse a: 22

Trying to parse a string to int from a file. Get NumberFormatException: For input string: "", eaven tho string appears to be an good int string. Java

I have been trying to figure this out for couple of hours now and I hope one of you can help me. I have an file (actually two but thats not important) that have some rows and columns with numbers and blank spaces between. And I'm trying to read those with BufferedReader. And that works great. I can print out the strings & chars however I want. But when I try to parse those strings and chars I get the following error:
Exception in thread "main" java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at java.lang.Integer.parseInt(Unknown Source)
at FileProcess.processed(FileProcess.java:30)
at DecisionTree.main(DecisionTree.java:16)
From what I have found with google I think the error is located in how I read my file.
public class ReadFiles {
private BufferedReader read;
public ReadFiles(BufferedReader startRead) {
read = startRead;
}
public String readFiles() throws IOException {
try {
String readLine = read.readLine().trim();
String readStuff = "";
while(readLine != null) {
readStuff += (readLine + "\n");
readLine = read.readLine();
}
return readStuff;
}
catch(NumberFormatException e) {
return null;
}
}
And for the parsing bit
public class FileProcess {
public String processed() throws IOException {
fileSelect fs = new fileSelect();
ReadFiles tr = new ReadFiles(fs.traning());
String training = tr.readFiles();
ReadFiles ts = new ReadFiles(fs.test());
String test = ts.readFiles();
List liste = new List(14,test.length());
String[] test2 = test.split("\n");
for(int i = 0; i<test2[0].length(); i++) {
char tmp = test.charAt(i);
String S = Character.toString(tmp).trim();
//int i1 = Integer.parseInt(S);
System.out.print(S);
}
This isn't the actual code for what I planning to do with the output, but the error appears at the code that is commented out. So my string output is as following:
12112211
Which seems good to parse to integer. But it does not work. I tried to manually see what's in the char position 0 and 1, for 0 I get 1, but for 1 I get nothing aka "". So how can I remove the ""? I hope you guys can help me out, and let me know if you need more info. But I think I have covered what's needed.
Thanks in advance :)
Yeah, and another thing: If I replace "" with "0" it works, but then I get all those zeros which I can't find a clever way to remove. But is it possible to maybe skip them while parsing or something? My files only hold 1 and 2, so it wouldn't interfere with anything if it is possible.

The string "" will be returned if you have 2 of the splitting characters next to each other (i.e. \n\n) or if there is a whitespace character being passed into the trim() call so ignore empty strings and carry on.

You could use the Scanner class to parse for ints, skipping Whitespace:
sc = new java.util.Scanner (line);
sc.nextInt ();
Another idea is to trim the line, split, and parse the parts:
lin = line.trim ();
String [] words = lin.split (" +");
for (String si : words)
Integer.parseInt (si);