We Keep Coding

Remove a character followed by whitespace each newline of a string

I am writing a program to edit a rtf file. The rtf file will always come in the same format with
Q XXXXXXXXXXXX
A YYYYYYYYYYYY
Q XXXXXXXXXXXX
A YYYYYYYYYYYY
I want to remove the Q / A + whitespace and leave just the X's and Y's on each line. My first idea is to split the string into a new string for each line and edit it from there using str.split like so:
private void countLines(String str){
String[] lines = str.split("\r\n|\r|\n");
linesInDoc = lines;
}
From here my idea is to take each even array value and get rid of Q + whitespace and take each odd array value and get rid of A + whitespace. Is there a better way to do this? Note: The first line somteimes contains a ~6 digit alphanumeric. I tihnk an if statement for a 2 non whitespace chars would solve this.
Here is the rest of the code:
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.StringWriter;
import java.io.Writer;
import javax.swing.JEditorPane;
import javax.swing.text.BadLocationException;
import javax.swing.text.EditorKit;
public class StringEditing {
String[] linesInDoc;
private String readRTF(File file){
String documentText = "";
try{
JEditorPane p = new JEditorPane();
p.setContentType("text/rtf");
EditorKit rtfKit = p.getEditorKitForContentType("text/rtf");
rtfKit.read(new FileReader(file), p.getDocument(), 0);
rtfKit = null;
EditorKit txtKit = p.getEditorKitForContentType("text/plain");
Writer writer = new StringWriter();
txtKit.write(writer, p.getDocument(), 0, p.getDocument().getLength());
documentText = writer.toString();
}
catch( FileNotFoundException e )
{
System.out.println( "File not found" );
}
catch( IOException e )
{
System.out.println( "I/O error" );
}
catch( BadLocationException e )
{
}
return documentText;
}
public void editDocument(File file){
String plaintext = readRTF(file);
System.out.println(plaintext);
fixString(plaintext);
System.out.println(plaintext);
}

Unless I'm missing something, you could use String.substring(int) like
String lines = "Q XXXXXXXXXXXX\n" //
+ "A YYYYYYYYYYYY\n" //
+ "Q XXXXXXXXXXXX\n" //
+ "A YYYYYYYYYYYY\n";
for (String line : lines.split("\n")) {
System.out.println(line.substring(6));
}
Output is
XXXXXXXXXXXX
YYYYYYYYYYYY
XXXXXXXXXXXX
YYYYYYYYYYYY
If your format should be more general, you might prefer
System.out.println(line.substring(1).trim());

A BufferedReader will handle the newline \n for you.
You can use a matcher to validate that the line is in the desired format.
If the line is fixed length, simply use the substring
final String bodyPattern = "\\w{1,1}[ \\w]{5,5}\\d{12,12}";
try (BufferedReader br = new BufferedReader(new FileReader(fileName))) {
String line;
while ((line = br.readLine()) != null) {
if (line.matches(bodyPattern)) {
//
myString = line.substring(6);
}
}
}
//catch Block
You can adjust the regex pattern to your specific requirements

easily doable by a regex (assuming 'fileText' is your whole file's content)
removedPrefix = fileText.replaceAll("(A|Q) *(.+)\\r", "$2\\r");
The regex means a Q or A for start, then some (any amount of) spaces, then anything (marked as group 2), and a closing line. This doesn't do anything to the first line with the digits. The result is the file content without the Q/A and the spaces. There are easier ways if you know the exact number of spaces before your needed text, but this works for all, and greatly flexible.
If you process line by line it's
removedPrefix = currentLine.replaceAll("(A|Q) *(.+)", "$2");
As simple as that

Integer.parseInt() issues

I'm having a bit of a problem trying to get my code to work. I am working on a project for my computer science class, and I have to get my program to read the file and perform some math. When I tried doing this, the code was not working. I then checked with a friend who wrote the exact same code, and it did not work.
The input .txt file that the program reads looks like this:
2/3,4/5
-1/6,2/4
1/1,1/1
The code I have written looks like this:
import javax.swing.JFileChooser;
import java.util.*;
public class ProjectTest
{
public static void main(String[] args) throws Exception
{
JFileChooser chooserRational = new JFileChooser();
int returnValRational = chooserRational.showOpenDialog(null);
if(returnValRational == JFileChooser.APPROVE_OPTION)
{
System.out.println("You chose to open this file: " + chooserRational.getSelectedFile().getName());
Scanner input = new Scanner(chooserRational.getSelectedFile());
while(input.hasNext() == true)
{
String line = input.nextLine();
String[] output = line.split(",");
String[] output1 = output[0].split("/");
String[] output2 = output[1].split("/");
String a = output1[0];
String b = output1[1];
String c = output2[0];
String d = output2[1];
int int1 = Integer.parseInt(a);
int int2 = Integer.parseInt(b);
int int3 = Integer.parseInt(c);
int int4 = Integer.parseInt(d);
System.out.println(int1 + " " + int2 + " " + int3 + " " + int4);
}
input.close();
}
}
}
When I output just the Strings a, b, c, and d, the code works perfectly fine and outputs the values perfectly. When the code sees Integer.parseInt(a), however, it gives me an error that looks like this:
Exception in thread "main" java.lang.NumberFormatException: For input string: "?2"
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 ProjectTest1.main(ProjectTest1.java:33)
Any help would be greatly appreciated.

Because your data file contains an UTF-8 BOM.
You have two alternatives: edit your source data file to remove the BOM, or you can add some code to deal with the BOM. For the first option use Notepad++ and remove the BOM. For the second alternative:
Scanner input = new Scanner(chooserRational.getSelectedFile());
if (input.nextByte() == 0xFE) {
input.nextByte();
input.nextByte();
} else {
input = new Scanner(chooserRational.getSelectedFile());
}

You should replace
String line = input.nextLine();
with
String line = input.next();
since you have multiples groups of data on the same line.
Edit :
I ran your code and did not get the same exception as you. I had a NumberFormatException due to the nextLine call, I now fixed it and it runs with no error. I think like the others that you have an encoding problem. Search on the internet how to display invisible characters on your preferred text editor.

How to split string by new lines in JAVA?

I want to split string by new lines in Java.I am using following regex -
str.split("\\r|\\n|\\r\\n");
But still it is not splitting string by new lines.
Input -
0
0
0
0
Output = String [] array = {"0000"} instead I want = String [] array = {"0","0","0","0"}.
I have read various solutions on stack overflow but nothing works for me.
Code is -
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.text.DecimalFormat;
public class Input {
public static void main(String[] args) {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line;
String text = "";
try {
while((line=br.readLine())!=null){
text = text + line;
}
} catch (IOException e) {
e.printStackTrace();
}
String [] textarray = text.trim().split("[\\r\\n]+");
for(int j=0;j<textarray.length;j++)
System.out.println(textarray[j]);
// System.out.print("");
// for(int i=((textarray.length)-1);i>=0;i--){
// long k = Long.valueOf(textarray[i]).longValue();
// System.out.println(k);
//// double sqrt = Math.sqrt(k);
//// double value = Double.parseDouble(new DecimalFormat("##.####").format(sqrt));
//// System.out.println(value);
////
//// }
}

When you call br.readLine(), the newline characters are stripped from the end of the string. So if you type 0 + ENTER four times, you are trying to split the string "0000".
You would be better to read items in from stdin and store them in an expandable data structure, such as a List<String>. No need to split things if you've already read them separately.

Updated Answer:
If you are reading the inputstreamfrom the keyboard, the \n may not be put into the data correctly. In that case, you may want to choose a new sentinel value.
Original Answer:
I believe you need to create a sentinel value. So if \n is your sentinel value, you could do something like this:
Load the inputstream into a string variable
Go character by character through the string variable checking to see if \n is in the input (you could use a for loop and the substing(i, i+2)
If it is found, then you could add it to an array

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);