How to read a specific line using the specific line number from a file in Java?

JavaFile Io

Java Problem Overview


In Java, is there any method to read a particular line from a file? For example, read line 32 or any other line number.

Java Solutions


Solution 1 - Java

For small files:

String line32 = Files.readAllLines(Paths.get("file.txt")).get(32)

For large files:

try (Stream<String> lines = Files.lines(Paths.get("file.txt"))) {
    line32 = lines.skip(31).findFirst().get();
}

Solution 2 - Java

Unless you have previous knowledge about the lines in the file, there's no way to directly access the 32nd line without reading the 31 previous lines.

That's true for all languages and all modern file systems.

So effectively you'll simply read lines until you've found the 32nd one.

Solution 3 - Java

Not that I know of, but what you could do is loop through the first 31 lines doing nothing using the readline() function of BufferedReader

FileInputStream fs= new FileInputStream("someFile.txt");
BufferedReader br = new BufferedReader(new InputStreamReader(fs));
for(int i = 0; i < 31; ++i)
  br.readLine();
String lineIWant = br.readLine();

Solution 4 - Java

Joachim is right on, of course, and an alternate implementation to Chris' (for small files only because it loads the entire file) might be to use commons-io from Apache (though arguably you might not want to introduce a new dependency just for this, if you find it useful for other stuff too though, it could make sense).

For example:

String line32 = (String) FileUtils.readLines(file).get(31);

http://commons.apache.org/io/api-release/org/apache/commons/io/FileUtils.html#readLines(java.io.File, java.lang.String)

Solution 5 - Java

You may try indexed-file-reader (Apache License 2.0). The class IndexedFileReader has a method called readLines(int from, int to) which returns a SortedMap whose key is the line number and the value is the line that was read.

Example:

File file = new File("src/test/resources/file.txt");
reader = new IndexedFileReader(file);

lines = reader.readLines(6, 10);
assertNotNull("Null result.", lines);
assertEquals("Incorrect length.", 5, lines.size());
assertTrue("Incorrect value.", lines.get(6).startsWith("[6]"));
assertTrue("Incorrect value.", lines.get(7).startsWith("[7]"));
assertTrue("Incorrect value.", lines.get(8).startsWith("[8]"));
assertTrue("Incorrect value.", lines.get(9).startsWith("[9]"));
assertTrue("Incorrect value.", lines.get(10).startsWith("[10]"));      

The above example reads a text file composed of 50 lines in the following format:

[1] The quick brown fox jumped over the lazy dog ODD
[2] The quick brown fox jumped over the lazy dog EVEN

Disclamer: I wrote this library

Solution 6 - Java

Although as said in other answers, it is not possible to get to the exact line without knowing the offset (pointer) before. So, I've achieved this by creating an temporary index file which would store the offset values of every line. If the file is small enough, you could just store the indexes (offset) in memory without needing a separate file for it.

The offsets can be calculated by using the RandomAccessFile

RandomAccessFile raf = new RandomAccessFile("myFile.txt","r"); 
//above 'r' means open in read only mode
ArrayList<Integer> arrayList = new ArrayList<Integer>();
String cur_line = "";
while((cur_line=raf.readLine())!=null)
{
    arrayList.add(raf.getFilePointer());
}
//Print the 32 line
//Seeks the file to the particular location from where our '32' line starts
raf.seek(raf.seek(arrayList.get(31));
System.out.println(raf.readLine());
raf.close();

Also visit the Java docs on RandomAccessFile for more information:

Complexity: This is O(n) as it reads the entire file once. Please be aware for the memory requirements. If it's too big to be in memory, then make a temporary file that stores the offsets instead of ArrayList as shown above.

Note: If all you want in '32' line, you just have to call the readLine() also available through other classes '32' times. The above approach is useful if you want to get the a specific line (based on line number of course) multiple times.

Solution 7 - Java

Another way.

try (BufferedReader reader = Files.newBufferedReader(
    	Paths.get("file.txt"), StandardCharsets.UTF_8)) {
    List<String> line = reader.lines()
  						      .skip(31)
  						      .limit(1)
  						      .collect(Collectors.toList());

    line.stream().forEach(System.out::println);
}

Solution 8 - Java

No, unless in that file format the line lengths are pre-determined (e.g. all lines with a fixed length), you'll have to iterate line by line to count them.

Solution 9 - Java

In Java 8,

For small files:

String line = Files.readAllLines(Paths.get("file.txt")).get(n);

For large files:

String line;
try (Stream<String> lines = Files.lines(Paths.get("file.txt"))) {
    line = lines.skip(n).findFirst().get();
}

In Java 7

String line;
try (BufferedReader br = new BufferedReader(new FileReader("file.txt"))) {
    for (int i = 0; i < n; i++)
        br.readLine();
    line = br.readLine();
}

Source: Reading nth line from file

Solution 10 - Java

If you are talking about a text file, then there is really no way to do this without reading all the lines that precede it - After all, lines are determined by the presence of a newline, so it has to be read.

Use a stream that supports readline, and just read the first X-1 lines and dump the results, then process the next one.

Solution 11 - Java

It works for me: I have combined the answer of https://stackoverflow.com/questions/5771366/reading-a-simple-text-file

But instead of return a String I am returning a LinkedList of Strings. Then I can select the line that I want.

public static LinkedList<String> readFromAssets(Context context, String filename) throws IOException {
    BufferedReader reader = new BufferedReader(new InputStreamReader(context.getAssets().open(filename)));
    LinkedList<String>linkedList = new LinkedList<>();
    // do reading, usually loop until end of file reading
    StringBuilder sb = new StringBuilder();
    String mLine = reader.readLine();
    while (mLine != null) {
        linkedList.add(mLine);
        sb.append(mLine); // process line
        mLine = reader.readLine();


    }
    reader.close();
    return linkedList;
}

Solution 12 - Java

Use this code:

import java.nio.file.Files;

import java.nio.file.Paths;

public class FileWork 
{

    public static void main(String[] args) throws IOException {

        String line = Files.readAllLines(Paths.get("D:/abc.txt")).get(1);
        
        System.out.println(line);  
    }

}

Solution 13 - Java

You can use LineNumberReader instead of BufferedReader. Go through the api. You can find setLineNumber and getLineNumber methods.

Solution 14 - Java

You can also take a look at LineNumberReader, subclass of BufferedReader. Along with the readline method, it also has setter/getter methods to access line number. Very useful to keep track of the number of lines read, while reading data from file.

Solution 15 - Java

public String readLine(int line){
		FileReader tempFileReader = null;
		BufferedReader tempBufferedReader = null;
		try { tempFileReader = new FileReader(textFile); 
		tempBufferedReader = new BufferedReader(tempFileReader);
		} catch (Exception e) { }
		String returnStr = "ERROR";
		for(int i = 0; i < line - 1; i++){
			try { tempBufferedReader.readLine(); } catch (Exception e) { }
		}
		try { returnStr = tempBufferedReader.readLine(); }  catch (Exception e) { }
		
		return returnStr;
	}

Solution 16 - Java

you can use the skip() function to skip the lines from begining.

public static void readFile(String filePath, long lineNum) {
    List<String> list = new ArrayList<>();
    long totalLines, startLine = 0;

    try (Stream<String> lines = Files.lines(Paths.get(filePath))) {
        totalLines = Files.lines(Paths.get(filePath)).count();
        startLine = totalLines - lineNum;
        // Stream<String> line32 = lines.skip(((startLine)+1));

        list = lines.skip(startLine).collect(Collectors.toList());
        // lines.forEach(list::add);
    } catch (IOException e1) {
        // TODO Auto-generated catch block
        e1.printStackTrace();
    }

    list.forEach(System.out::println);

}

Solution 17 - Java

EASY WAY - Reading a line using line number. Let's say Line number starts from 1 till null .

public class TextFileAssignmentOct {
	
	private void readData(int rowNum, BufferedReader br) throws IOException {
		int n=1;									//Line number starts from 1
		String row;
		while((row=br.readLine()) != null)	{		// Reads every line
			if (n == rowNum) {						// When Line number matches with which you want to read
				System.out.println(row);
			}
			n++;									//This increments Line number
		}
	}

	public static void main(String[] args) throws IOException {
		File f = new File("../JavaPractice/FileRead.txt");
		FileReader fr = new FileReader(f);
		BufferedReader br = new BufferedReader(fr);
		
		TextFileAssignmentOct txf = new TextFileAssignmentOct();
		txf.readData(4, br);	//Read a Specific Line using Line number and Passing buffered reader
	}
}

Solution 18 - Java

They are all wrong I just wrote this in about 10 seconds. With this I managed to just call the object.getQuestion("linenumber") in the main method to return whatever line I want.

public class Questions {

File file = new File("Question2Files/triviagame1.txt");

public Questions() {
        
}

public String getQuestion(int numLine) throws IOException {
    BufferedReader br = new BufferedReader(new FileReader(file));
    String line = "";
    for(int i = 0; i < numLine; i++) {
        line = br.readLine();
    }
    return line; }}

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestiontrinityView Question on Stackoverflow
Solution 1 - JavaaioobeView Answer on Stackoverflow
Solution 2 - JavaJoachim SauerView Answer on Stackoverflow
Solution 3 - JavaChris ThompsonView Answer on Stackoverflow
Solution 4 - JavaCharlie CollinsView Answer on Stackoverflow
Solution 5 - JavajramoyoView Answer on Stackoverflow
Solution 6 - JavaShankarDarugaView Answer on Stackoverflow
Solution 7 - JavaBrownRecluseView Answer on Stackoverflow
Solution 8 - Javab.rothView Answer on Stackoverflow
Solution 9 - JavaSidView Answer on Stackoverflow
Solution 10 - JavaUriView Answer on Stackoverflow
Solution 11 - JavaHuanView Answer on Stackoverflow
Solution 12 - JavaUsman YaqoobView Answer on Stackoverflow
Solution 13 - JavaRaghavendraView Answer on Stackoverflow
Solution 14 - JavaAnkur ShanbhagView Answer on Stackoverflow
Solution 15 - JavaCooper ScottView Answer on Stackoverflow
Solution 16 - JavaNeelay SolankiView Answer on Stackoverflow
Solution 17 - JavaAmit VermaView Answer on Stackoverflow
Solution 18 - Javauser3526115View Answer on Stackoverflow