How do I create a Java string from the contents of a file?

JavaStringFileFile IoIo

Java Problem Overview


I've been using the idiom below for some time now. And it seems to be the most wide-spread, at least on the sites I've visited.

Is there a better/different way to read a file into a string in Java?

private String readFile(String file) throws IOException {
    BufferedReader reader = new BufferedReader(new FileReader (file));
    String         line = null;
    StringBuilder  stringBuilder = new StringBuilder();
    String         ls = System.getProperty("line.separator");

    try {
        while((line = reader.readLine()) != null) {
            stringBuilder.append(line);
            stringBuilder.append(ls);
        }

        return stringBuilder.toString();
    } finally {
        reader.close();
    }
}

Java Solutions


Solution 1 - Java

Read all text from a file

Java 11 added the readString() method to read small files as a String, preserving line terminators:

String content = Files.readString(path, StandardCharsets.US_ASCII);

For versions between Java 7 and 11, here's a compact, robust idiom, wrapped up in a utility method:

static String readFile(String path, Charset encoding)
  throws IOException
{
  byte[] encoded = Files.readAllBytes(Paths.get(path));
  return new String(encoded, encoding);
}

Read lines of text from a file

Java 7 added a convenience method to read a file as lines of text, represented as a List<String>. This approach is "lossy" because the line separators are stripped from the end of each line.

List<String> lines = Files.readAllLines(Paths.get(path), encoding);

Java 8 added the Files.lines() method to produce a Stream<String>. Again, this method is lossy because line separators are stripped. If an IOException is encountered while reading the file, it is wrapped in an UncheckedIOException, since Stream doesn't accept lambdas that throw checked exceptions.

try (Stream<String> lines = Files.lines(path, encoding)) {
  lines.forEach(System.out::println);
}

This Stream does need a close() call; this is poorly documented on the API, and I suspect many people don't even notice Stream has a close() method. Be sure to use an ARM-block as shown.

If you are working with a source other than a file, you can use the lines() method in BufferedReader instead.

Memory utilization

The first method, that preserves line breaks, can temporarily require memory several times the size of the file, because for a short time the raw file contents (a byte array), and the decoded characters (each of which is 16 bits even if encoded as 8 bits in the file) reside in memory at once. It is safest to apply to files that you know to be small relative to the available memory.

The second method, reading lines, is usually more memory efficient, because the input byte buffer for decoding doesn't need to contain the entire file. However, it's still not suitable for files that are very large relative to available memory.

For reading large files, you need a different design for your program, one that reads a chunk of text from a stream, processes it, and then moves on to the next, reusing the same fixed-sized memory block. Here, "large" depends on the computer specs. Nowadays, this threshold might be many gigabytes of RAM. The third method, using a Stream<String> is one way to do this, if your input "records" happen to be individual lines. (Using the readLine() method of BufferedReader is the procedural equivalent to this approach.)

Character encoding

One thing that is missing from the sample in the original post is the character encoding. There are some special cases where the platform default is what you want, but they are rare, and you should be able justify your choice.

The StandardCharsets class defines some constants for the encodings required of all Java runtimes:

String content = readFile("test.txt", StandardCharsets.UTF_8);

The platform default is available from the Charset class itself:

String content = readFile("test.txt", Charset.defaultCharset());

Note: This answer largely replaces my Java 6 version. The utility of Java 7 safely simplifies the code, and the old answer, which used a mapped byte buffer, prevented the file that was read from being deleted until the mapped buffer was garbage collected. You can view the old version via the "edited" link on this answer.

Solution 2 - Java

If you're willing to use an external library, check out Apache Commons IO (200KB JAR). It contains an org.apache.commons.io.FileUtils.readFileToString() method that allows you to read an entire File into a String with one line of code.

Example:

import java.io.*;
import java.nio.charset.*;
import org.apache.commons.io.*;

public String readFile() throws IOException {
    File file = new File("data.txt");
    return FileUtils.readFileToString(file, StandardCharsets.UTF_8);
}

Solution 3 - Java

A very lean solution based on Scanner:

Scanner scanner = new Scanner( new File("poem.txt") );
String text = scanner.useDelimiter("\\A").next();
scanner.close(); // Put this call in a finally block

Or, if you want to set the charset:

Scanner scanner = new Scanner( new File("poem.txt"), "UTF-8" );
String text = scanner.useDelimiter("\\A").next();
scanner.close(); // Put this call in a finally block

Or, with a try-with-resources block, which will call scanner.close() for you:

try (Scanner scanner = new Scanner( new File("poem.txt"), "UTF-8" )) {
    String text = scanner.useDelimiter("\\A").next();
}

Remember that the Scanner constructor can throw an IOException. And don't forget to import java.io and java.util.

Source: Pat Niemeyer's blog

Solution 4 - Java

import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;

Java 7

String content = new String(Files.readAllBytes(Paths.get("readMe.txt")), StandardCharsets.UTF_8);

Java 11

String content = Files.readString(Paths.get("readMe.txt"));

Solution 5 - Java

If you're looking for an alternative that doesn't involve a third-party library (e.g. Commons I/O), you can use the Scanner class:

private String readFile(String pathname) throws IOException {

    File file = new File(pathname);
    StringBuilder fileContents = new StringBuilder((int)file.length());        

    try (Scanner scanner = new Scanner(file)) {
        while(scanner.hasNextLine()) {
            fileContents.append(scanner.nextLine() + System.lineSeparator());
        }
        return fileContents.toString();
    }
}

Solution 6 - Java

Guava has a method similar to the one from Commons IOUtils that Willi aus Rohr mentioned:

import com.google.common.base.Charsets;
import com.google.common.io.Files;

// ...

String text = Files.toString(new File(path), Charsets.UTF_8);

EDIT by PiggyPiglet
Files#toString is deprecated, and due for removal Octobor 2019. Instead use Files.asCharSource(new File(path), StandardCharsets.UTF_8).read();

EDIT by Oscar Reyes

This is the (simplified) underlying code on the cited library:

InputStream in = new FileInputStream(file);
byte[] b  = new byte[file.length()];
int len = b.length;
int total = 0;

while (total < len) {
  int result = in.read(b, total, len - total);
  if (result == -1) {
    break;
  }
  total += result;
}

return new String( b , Charsets.UTF_8 );

Edit (by Jonik): The above doesn't match the source code of recent Guava versions. For the current source, see the classes Files, CharStreams, ByteSource and CharSource in com.google.common.io package.

Solution 7 - Java

import java.nio.file.Files;

.......

 String readFile(String filename) {
            File f = new File(filename);
            try {
                byte[] bytes = Files.readAllBytes(f.toPath());
                return new String(bytes,"UTF-8");
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return "";
    }

Solution 8 - Java

If you need a string processing (parallel processing) Java 8 has the great Stream API.

String result = Files.lines(Paths.get("file.txt"))
                    .parallel() // for parallel processing 
                    .map(String::trim) // to change line   
                    .filter(line -> line.length() > 2) // to filter some lines by a predicate                        
                    .collect(Collectors.joining()); // to join lines

More examples are available in JDK samples sample/lambda/BulkDataOperations that can be downloaded from Oracle Java SE 8 download page

Another one liner example

String out = String.join("\n", Files.readAllLines(Paths.get("file.txt")));

Solution 9 - Java

That code will normalize line breaks, which may or may not be what you really want to do.

Here's an alternative which doesn't do that, and which is (IMO) simpler to understand than the NIO code (although it still uses java.nio.charset.Charset):

public static String readFile(String file, String csName)
            throws IOException {
    Charset cs = Charset.forName(csName);
    return readFile(file, cs);
}

public static String readFile(String file, Charset cs)
            throws IOException {
    // No real need to close the BufferedReader/InputStreamReader
    // as they're only wrapping the stream
    FileInputStream stream = new FileInputStream(file);
    try {
        Reader reader = new BufferedReader(new InputStreamReader(stream, cs));
        StringBuilder builder = new StringBuilder();
        char[] buffer = new char[8192];
        int read;
        while ((read = reader.read(buffer, 0, buffer.length)) > 0) {
            builder.append(buffer, 0, read);
        }
        return builder.toString();
    } finally {
        // Potential issue here: if this throws an IOException,
        // it will mask any others. Normally I'd use a utility
        // method which would log exceptions and swallow them
        stream.close();
    }        
}

Solution 10 - Java

Gathered all the possible ways to read the File as String from Disk or Network.

  • Guava: Google using classes Resources, Files

     static Charset charset = com.google.common.base.Charsets.UTF_8;
     public static String guava_ServerFile( URL url ) throws IOException {
     	return Resources.toString( url, charset );
     }
     public static String guava_DiskFile( File file ) throws IOException {
     	return Files.toString( file, charset );
     }
    


  • APACHE - COMMONS IO using classes IOUtils, FileUtils

     static Charset encoding = org.apache.commons.io.Charsets.UTF_8;
     public static String commons_IOUtils( URL url ) throws IOException {
     	java.io.InputStream in = url.openStream();
     	try {
     		return IOUtils.toString( in, encoding );
     	} finally {
     		IOUtils.closeQuietly(in);
     	}
     }
     public static String commons_FileUtils( File file ) throws IOException {
     	return FileUtils.readFileToString( file, encoding );
     	/*List<String> lines = FileUtils.readLines( fileName, encoding );
     	return lines.stream().collect( Collectors.joining("\n") );*/
     }
    


  • Java 8 BufferReader using Stream API

     public static String streamURL_Buffer( URL url ) throws IOException {
     	java.io.InputStream source = url.openStream();
     	BufferedReader reader = new BufferedReader( new InputStreamReader( source ) );
     	//List<String> lines = reader.lines().collect( Collectors.toList() );
     	return reader.lines().collect( Collectors.joining( System.lineSeparator() ) );
     }
     public static String streamFile_Buffer( File file ) throws IOException {
     	BufferedReader reader = new BufferedReader( new FileReader( file ) );
     	return reader.lines().collect(Collectors.joining(System.lineSeparator()));
     }
    


  • Scanner Class with regex \A. which matches the beginning of input.

     static String charsetName = java.nio.charset.StandardCharsets.UTF_8.toString();
     public static String streamURL_Scanner( URL url ) throws IOException {
     	java.io.InputStream source = url.openStream();
     	Scanner scanner = new Scanner(source, charsetName).useDelimiter("\\A");
     	return scanner.hasNext() ? scanner.next() : "";
     }
     public static String streamFile_Scanner( File file ) throws IOException {
     	Scanner scanner = new Scanner(file, charsetName).useDelimiter("\\A");
     	return scanner.hasNext() ? scanner.next() : "";
     }
    


  • Java 7 (java.nio.file.Files.readAllBytes)

     public static String getDiskFile_Java7( File file ) throws IOException {
     	byte[] readAllBytes = java.nio.file.Files.readAllBytes(Paths.get( file.getAbsolutePath() ));
     	return new String( readAllBytes );
     }
    


  • BufferedReader using InputStreamReader.

     public static String getDiskFile_Lines( File file ) throws IOException {
     	StringBuffer text = new StringBuffer();
     	FileInputStream fileStream = new FileInputStream( file );
     	BufferedReader br = new BufferedReader( new InputStreamReader( fileStream ) );
     	for ( String line; (line = br.readLine()) != null; )
     		text.append( line + System.lineSeparator() );
     	return text.toString();
     }
    


Example with main method to access the above methods.

public static void main(String[] args) throws IOException {
	String fileName = "E:/parametarisation.csv";
	File file = new File( fileName );
	
	String fileStream = commons_FileUtils( file );
			// guava_DiskFile( file );
			// streamFile_Buffer( file );
			// getDiskFile_Java7( file );
			// getDiskFile_Lines( file );
	System.out.println( " File Over Disk : \n"+ fileStream );
	
	
	try {
		String src = "https://code.jquery.com/jquery-3.2.1.js";
		URL url = new URL( src );
		
		String urlStream = commons_IOUtils( url );
				// guava_ServerFile( url );
				// streamURL_Scanner( url );
				// streamURL_Buffer( url );
		System.out.println( " File Over Network : \n"+ urlStream );
	} catch (MalformedURLException e) {
		e.printStackTrace();
	}
}


@see

Solution 11 - Java

If it's a text file why not use apache commons-io?

It has the following method

public static String readFileToString(File file) throws IOException

If you want the lines as a list use

public static List<String> readLines(File file) throws IOException

Solution 12 - Java

Since JDK 11:

String file = ...
Path path = Paths.get(file);
String content = Files.readString(path);
// Or readString(path, someCharset), if you need a Charset different from UTF-8

Solution 13 - Java

To read a File as binary and convert at the end

public static String readFileAsString(String filePath) throws IOException {
    DataInputStream dis = new DataInputStream(new FileInputStream(filePath));
    try {
        long len = new File(filePath).length();
        if (len > Integer.MAX_VALUE) throw new IOException("File "+filePath+" too large, was "+len+" bytes.");
        byte[] bytes = new byte[(int) len];
        dis.readFully(bytes);
        return new String(bytes, "UTF-8");
    } finally {
        dis.close();
    }
}

Solution 14 - Java

With Java 7, this is my preferred option to read a UTF-8 file:

String content = new String(Files.readAllBytes(Paths.get(filename)), "UTF-8");

Since Java 7, the JDK has the new java.nio.file API, which provides many shortcuts, so 3rd party libraries are not always required for simple file operations.

Solution 15 - Java

Java attempts to be extremely general and flexible in all it does. As a result, something which is relatively simple in a scripting language (your code would be replaced with "open(file).read()" in python) is a lot more complicated. There doesn't seem to be any shorter way of doing it, except using an external library (like Willi aus Rohr mentioned). Your options:

  • Use an external library.
  • Copy this code into all your projects.
  • Create your own mini-library which contains functions you use often.

Your best bet is probably the 2nd one, as it has the least dependencies.

Solution 16 - Java

Using JDK 8 or above:

no external libraries used

You can create a new String object from the file content (Using classes from java.nio.file package):

public String readStringFromFile(String filePath) throws IOException {
	String fileContent = new String(Files.readAllBytes(Paths.get(filePath)));
	return fileContent;
}

Solution 17 - Java

There is a variation on the same theme that uses a for loop, instead of a while loop, to limit the scope of the line variable. Whether it's "better" is a matter of personal taste.

for(String line = reader.readLine(); line != null; line = reader.readLine()) {
    stringBuilder.append(line);
    stringBuilder.append(ls);
}

Solution 18 - Java

If you do not have access to the Files class, you can use a native solution.

static String readFile(File file, String charset)
        throws IOException
{
    FileInputStream fileInputStream = new FileInputStream(file);
    byte[] buffer = new byte[fileInputStream.available()];
    int length = fileInputStream.read(buffer);
    fileInputStream.close();
    return new String(buffer, 0, length, charset);
}

Solution 19 - Java

A flexible solution using IOUtils from Apache commons-io in combination with StringWriter:

Reader input = new FileReader();
StringWriter output = new StringWriter();
try {
  IOUtils.copy(input, output);
} finally {
  input.close();
}
String fileContents = output.toString();

It works with any reader or input stream (not just with files), for example when reading from a URL.

Solution 20 - Java

public static String slurp (final File file)
throws IOException {
    StringBuilder result = new StringBuilder();

    BufferedReader reader = new BufferedReader(new FileReader(file));

    try {
        char[] buf = new char[1024];

        int r = 0;

        while ((r = reader.read(buf)) != -1) {
            result.append(buf, 0, r);
        }
    }
    finally {
        reader.close();
    }

    return result.toString();
}

Solution 21 - Java

This one uses the method RandomAccessFile.readFully, it seems to be available from JDK 1.0 !

public static String readFileContent(String filename, Charset charset) throws IOException {
    RandomAccessFile raf = null;
    try {
        raf = new RandomAccessFile(filename, "r");
        byte[] buffer = new byte[(int)raf.length()];
        raf.readFully(buffer);
        return new String(buffer, charset);
    } finally {
        closeStream(raf);
    }
} 


private static void closeStream(Closeable c) {
    if (c != null) {
        try {
            c.close();
        } catch (IOException ex) {
            // do nothing
        }
    }
}

Solution 22 - Java

Be aware when using fileInputStream.available() the returned integer does not have to represent the actual file size, but rather the guessed amount of bytes the system should be able to read from the stream without blocking IO. A safe and simple way could look like this

public String readStringFromInputStream(FileInputStream fileInputStream) {
    StringBuffer stringBuffer = new StringBuffer();
    try {
        byte[] buffer;
        while (fileInputStream.available() > 0) {
            buffer = new byte[fileInputStream.available()];
            fileInputStream.read(buffer);
            stringBuffer.append(new String(buffer, "ISO-8859-1"));
        }
    } catch (FileNotFoundException e) {
    } catch (IOException e) { }
    return stringBuffer.toString();
}

It should be considered that this approach is not suitable for multi-byte character encodings like UTF-8.

Solution 23 - Java

You can try Scanner and File class, a few lines solution

 try
{
  String content = new Scanner(new File("file.txt")).useDelimiter("\\Z").next();
  System.out.println(content);
}
catch(FileNotFoundException e)
{
  System.out.println("not found!");
}

Solution 24 - Java

Based on @erickson`s answer, you can use:

public String readAll(String fileName) throws IOException {
    List<String> lines = Files.readAllLines(new File(fileName).toPath());
    return String.join("\n", lines.toArray(new String[lines.size()]));
}

Solution 25 - Java

User java.nio.Files to read all lines of file.

public String readFile() throws IOException {
		File fileToRead = new File("file path");
		List<String> fileLines = Files.readAllLines(fileToRead.toPath());
		return StringUtils.join(fileLines, StringUtils.EMPTY);
}

Solution 26 - Java

I cannot comment other entries yet, so I'll just leave it here.

One of best answers here (https://stackoverflow.com/a/326448/1521167):

private String readFile(String pathname) throws IOException {

File file = new File(pathname);
StringBuilder fileContents = new StringBuilder((int)file.length());
Scanner scanner = new Scanner(file);
String lineSeparator = System.getProperty("line.separator");

try {
    while(scanner.hasNextLine()) {        
        fileContents.append(scanner.nextLine() + lineSeparator);
    }
    return fileContents.toString();
} finally {
    scanner.close();
}
}

still has one flaw. It always puts new line char in the end of string, which may cause some weirds bugs. My suggestion is to change it to:

	private String readFile(String pathname) throws IOException {
	File file = new File(pathname);
	StringBuilder fileContents = new StringBuilder((int) file.length());
	Scanner scanner = new Scanner(new BufferedReader(new FileReader(file)));
	String lineSeparator = System.getProperty("line.separator");

	try {
		if (scanner.hasNextLine()) {
			fileContents.append(scanner.nextLine());
		}
		while (scanner.hasNextLine()) {
			fileContents.append(lineSeparator + scanner.nextLine());
		}
		return fileContents.toString();
	} finally {
		scanner.close();
	}
}

Solution 27 - Java

After Ctrl+F'ing after Scanner, I think that the Scanner solution should be listed too. In the easiest to read fashion it goes like this:

public String fileToString(File file, Charset charset) {
  Scanner fileReader = new Scanner(file, charset);
  fileReader.useDelimiter("\\Z"); // \Z means EOF.
  String out = fileReader.next();
  fileReader.close();
  return out;
}

If you use Java 7 or newer (and you really should) consider using try-with-resources to make the code easier to read. No more dot-close stuff littering everything. But that's mostly a stylistic choice methinks.

I'm posting this mostly for completionism, since if you need to do this a lot, there should be things in java.nio.file.Files that should do the job better.

My suggestion would be to use Files#readAllBytes(Path) to grab all the bytes, and feed it to new String(byte[] Charset) to get a String out of it that you can trust. Charsets will be mean to you during your lifetime, so beware of this stuff now.

Others have given code and stuff, and I don't want to steal their glory. ;)

Solution 28 - Java

Using this library, it is one line:

String data = IO.from(new File("data.txt")).toString();

Solution 29 - Java

Also if your file happens to be inside a jar, you can also use this:

public String fromFileInJar(String path) {
	try ( Scanner scanner 
            = new Scanner(getClass().getResourceAsStream(path))) {
		return scanner.useDelimiter("\\A").next();
	}
}

The path should start with / for instance if your jar is

my.jar/com/some/thing/a.txt

Then you want to invoke it like this:

String myTxt = fromFileInJar("/com/com/thing/a.txt");

Solution 30 - Java

In one line (Java 8), assuming you have a Reader:

String sMessage = String.join("\n", reader.lines().collect(Collectors.toList()));

Solution 31 - Java

Use code:

File file = new File("input.txt");
BufferedInputStream bin = new BufferedInputStream(new FileInputStream(
				file));
byte[] buffer = new byte[(int) file.length()];
bin.read(buffer);
String fileStr = new String(buffer);

fileStr contains output in String.

Solution 32 - Java

Scanner sc = new Scanner(new File("yourFile.txt"));
sc.useDelimiter("\\Z");

String s = sc.next();

Solution 33 - Java

in java 8 , there are a new Class

> java.util.stream.Stream

A stream represents a sequence of elements and supports different kind of operations to perform computations upon those elements

to Read more about it :

Oracle Documentation

Here an Example :

import java.nio.charset.StandardCharsets;
import java.nio.file.Files;
import java.nio.file.Paths;
import java.util.stream.Stream;

public Class ReadFile{
  public  static String readFile(String filePath) {
 StringBuilder  stringBuilder = new StringBuilder();
    String ls = System.getProperty("line.separator");
        try {

            try (Stream<String> lines = Files.lines(Paths.get(filePath), StandardCharsets.UTF_8)) {
                for (String line : (Iterable<String>) lines::iterator) {
                   

                      stringBuilder.append(line);
                      stringBuilder.append(ls);


                }
            }

        } catch (Exception e) {
            e.printStackTrace();
        }
        
      return stringBuilder.toString(); 
    

}

}

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
QuestionOscarRyzView Question on Stackoverflow
Solution 1 - JavaericksonView Answer on Stackoverflow
Solution 2 - JavaDaWilliView Answer on Stackoverflow
Solution 3 - JavaPablo GrisafiView Answer on Stackoverflow
Solution 4 - JavaJobinView Answer on Stackoverflow
Solution 5 - JavaDónalView Answer on Stackoverflow
Solution 6 - JavafinnwView Answer on Stackoverflow
Solution 7 - Javauser590444View Answer on Stackoverflow
Solution 8 - JavaAndrei NView Answer on Stackoverflow
Solution 9 - JavaJon SkeetView Answer on Stackoverflow
Solution 10 - JavaYashView Answer on Stackoverflow
Solution 11 - JavaHome in TimeView Answer on Stackoverflow
Solution 12 - JavaleventovView Answer on Stackoverflow
Solution 13 - JavaPeter LawreyView Answer on Stackoverflow
Solution 14 - JavaMoritz PetersenView Answer on Stackoverflow
Solution 15 - JavaClaudiuView Answer on Stackoverflow
Solution 16 - JavaSaikatView Answer on Stackoverflow
Solution 17 - JavaDan DyerView Answer on Stackoverflow
Solution 18 - JavaIlya GazmanView Answer on Stackoverflow
Solution 19 - JavawauView Answer on Stackoverflow
Solution 20 - JavaScott S. McCoyView Answer on Stackoverflow
Solution 21 - JavabarjakView Answer on Stackoverflow
Solution 22 - JavaHenryView Answer on Stackoverflow
Solution 23 - JavajamesjaraView Answer on Stackoverflow
Solution 24 - JavaMuskovetsView Answer on Stackoverflow
Solution 25 - JavaNitinView Answer on Stackoverflow
Solution 26 - JavaAjkView Answer on Stackoverflow
Solution 27 - JavaHaakon LøtveitView Answer on Stackoverflow
Solution 28 - JavasatnamView Answer on Stackoverflow
Solution 29 - JavaOscarRyzView Answer on Stackoverflow
Solution 30 - JavaMalcolm BoekhoffView Answer on Stackoverflow
Solution 31 - JavaDevram KandhareView Answer on Stackoverflow
Solution 32 - JavaMadhav Balakrishnan NairView Answer on Stackoverflow
Solution 33 - Javauser2058603View Answer on Stackoverflow