I have a chunk of html code that should be outputted as a .html file, in java. The pre-written html code is the header and table for a page, and i need to generate some data, and output it to the same .html file. Is there an easier way to print the html code than to do prinln() line by line? Thanks
You can look at some Java libraries for parsing HTML code. A quick Google search tuns up a few. Read in the HTML and then use their queries to manipulate the DOM as needed and then spit it back out.
e.g. http://jsoup.org/
Try using a templating engine, MVEL2 or FreeMarker, for example. Both can be used by standalone applications outside of a web framework. You lose time upfront but it will save you time in the long run.
JSP (Java Server Pages) allows you to write HTML files which have some Java code easily embedded within them. For example
<html><head><title>Hi!</title></head><body>
<% some java code here that outputs some stuff %>
</body></html>
Though that requires that you have an enterprise Java server installed. But if this is on a web server, that might not be unreasonable to have.
If you want to do it in normal Java, that depends. I don't fully understand which part you meant you will be outputting line by line. Did you mean you are going to do something like
System.out.println("<html>");
System.out.println("<head><title>Hi!</title></head>");
System.out.println("<body>");
// etc
Like that? If that's what you meant, then don't do that. You can just read in the data from the template file and output all the data at once. You could read it into a multiline text string of you could read the data in from the template and output it directly to the new file. Something like
while( (strInput = templateFileReader.readLine()) != null)
newFileOutput.println(strInput);
Again, I'm not sure exactly what you mean by that part.
HTML is simply a way of marking up text, so to write a HTML file, you are simply writing the HTML as text to a file with the .html extension.
There's plenty of tutorials out there for reading and writing from files, as well as getting a list of files from a directory. (Google 'java read file', 'java write file', 'java list directory' - that is basically everything you need.) The important thing is the use of BufferedReader/BufferedWriter for pulling and pushing the text in to the files and realising that there is no particular code science involved in writing HTML to a file.
I'll reiterate; HTML is nothing more than <b>text with tags</b>.
Here's a really crude example that will output two files to a single file, wrapping them in an <html></html> tag.
BufferedReader getReaderForFile(filename) {
FileInputStream in = new FileInputStream(filename);
return new BufferedReader(new InputStreamReader(in));
}
public void main(String[] args) {
// Open a file
BufferedReader myheader = getReaderForFile("myheader.txt");
BufferedReader contents = getReaderForFile("contentfile.txt");
FileWriter fstream = new FileWriter("mypage.html");
BufferedWriter out = new BufferedWriter(fstream);
out.write("<html>");
out.newLine();
for (String line = myheader.readLine(); line!=null; line = myheader.readLine()) {
out.write(line);
out.newLine(); // readLine() strips 'carriage return' characters
}
for (String line = contents.readLine(); line!=null; line = contents.readLine()) {
out.write(line);
out.newLine(); // readLine() strips 'carriage return' characters
}
out.write("</html>");
}
To build a simple HTML text file, you don't have to read your input file line by line.
File theFile = new File("file.html");
byte[] content = new byte[(int) theFile.length()];
You can use "RandomAccessFile.readFully" to read files entirely as a byte array:
// Read file function:
RandomAccessFile file = null;
try {
file = new RandomAccessFile(theFile, "r");
file.readFully(content);
} finally {
if(file != null) {
file.close();
}
}
Make your modifications on the text content:
String text = new String(content);
text = text.replace("<!-- placeholder -->", "generated data");
content = text.getBytes();
Writing is also easy:
// Write file content:
RandomAccessFile file = null;
try {
file = new RandomAccessFile(theFile, "rw");
file.write(content);
} finally {
if(file != null) {
file.close();
}
}
Related
I have a.txt list trying to move the first line to the last line in Java
I've found scripts to do the following
Find "text" from input file and output to a temp file. (I could set
"text" to a string buffRead.readLine ??) and then...
delete the orig file and rename the new file to the orig?
Please for give me I am new to Java but I have done a lot of research and can't find a solution for what I thought would be a simple script.
Because this is Java and concerns file IO, this is a non-trivial setup. The algorithm itself is simple, but the symbols required to do so are not immediately evident.
BufferedReader reader = new BufferedReader(new FileReader("fileName"));
This gives you an easy way to read the contents of the file fileName.
PrintWriter writer = new PrintWriter(new FileWriter("fileName"));
This gives you a simple way to write to the file. The API to do so is the exact same as System.out when you use a PrintWriter, thus my choice to use one here.
At this point its a simple matter of reading the file and echoing it back in the correct order.
String text = reader.readLine();
This saves the first line of the file to text.
while (reader.ready()) {
writer.println(reader.readLine());
}
While reader has text remaining in it, print the lines into the writer.
writer.println(text);
Print the line that you saved at the start.
Note that if your program does anything else (and it's just a good habit anyway), you want to close your IO streams to avoid leaking resources.
reader.close();
writer.close();
Alternatively, you could also wrap the entire thing in a try-with-resources to perform the same cleanup automatically.
Scanner fileScanner = new Scanner(myFile);
fileScanner.nextLine();
This will return the first line of text from the file and discard it because you don't store it anywhere.
To overwrite your existing file:
FileWriter fileStream = new FileWriter("my/path/for/file.txt");
BufferedWriter out = new BufferedWriter(fileStream);
while(fileScanner.hasNextLine()) {
String next = fileScanner.nextLine();
if(next.equals("\n") out.newLine();
else out.write(next);
out.newLine();
}
out.close();
Note that you will have to be catching and handling some IOExceptions this way. Also, the if()... else()... statement is necessary in the while() loop to keep any line breaks present in your text file.
Add the same line to the last line of this file have a look into this link https://stackoverflow.com/a/37674446/6160431
I want to open a .java text file and insert a new package name at the top of the file. After some research to this I found out that I have to create a new file insert the new Package name and copy everything from my old file. So far so good.
1) 1st Problem either I am not doing it correct, but how am I able to copy line endings in the correct way? I kinda have the feeling that by copying I lose them, not sure but that since I was more focused on my other Problem
2) The file I am trying to rename just doesn't get renamed, therefore resulting in the problem, that all files in which I want to insert the new package are getting named the same and therefore I am left with the last file with the wrong name. Resulting in a loss of all my other files
private static String insertPackage(File file) throws IOException {
//creating package name
File parent = new File(file.getParent());
String packageName = "package " + parent.getName() + ";\n";
//copy old file
File buffer = new File(parent.getPath()+"\\buffer.java");
Charset charset = Charset.forName("UTF-8");
BufferedReader br = Files.newBufferedReader(file.toPath(),charset);
BufferedWriter bw = Files.newBufferedWriter(buffer.toPath(),charset);
bw.write(packageName,0,packageName.length());
String line;
while((line = br.readLine()) != null){
if(!line.startsWith("package ")) {
bw.write(line, 0, line.length());
}
}
//rename new file
// String filepath = file.getPath();
// File rename = new File(filepath);
boolean renamed = true;
if(file.delete()) {
renamed = buffer.renameTo(file);
}
bw.flush();
bw.close();
br.close();
if(!renamed){
return file.getPath();
}
return "";
}
The return value is just there to have a name in case something is failing. So right now everything.
This is just the function to insert the package name, delete the old package name and, copy everything else into the new file. After that it should delete the old file and rename the new one.
You are loosing the line breaks, because readLine strips them, but you never write any out to the file. You will need to add a \n (or \r\n depending on your needs) to every line you write to the new file. You can use BufferedWriter#newLine for writing the platform-dependent line separator.
The new file does not get renamed, because your file streams are still open for both files when you try to rename it. Moreover File#renameTo is highly platform dependent (as noted in the documentation) and might not be able to overwrite existing files, I recommend you switch over to Files#move, where you can specify StandardCopyOption.REPLACE_EXISTING.
I can also recommend to fully switch over to the NIO API and not use java.io.File at all, instead use Path directly
Note also that it is a good choice to use try-with-resources instead of manually closing streams, especially when dealing with multiple streams at once.
Usually, when I read text files, I do it like this:
File file = new File("some_text_file.txt");
Scanner scanner = new Scanner(new FileInputStream(file));
StringBuilder builder = new StringBuilder();
while(scanner.hasNextLine()) {
builder.append(scanner.nextLine());
builder.append('\n');
}
scanner.close();
String text = builder.toString();
There may be better ways, but this method has always worked for me perfectly.
For what I am working on right now, I need to read a large text file (over 700 kilobytes in size). Here is a sample of the text when opened in Notepad (the one that comes standard with any Windows operating system):
"lang"
{
"Language" "English"
"Tokens"
{
"DOTA_WearableType_Daggers" "Daggers"
"DOTA_WearableType_Glaive" "Glaive"
"DOTA_WearableType_Weapon" "Weapon"
"DOTA_WearableType_Armor" "Armor"
However, when I read the text from the file using the method that I provided above, the output is:
I could not paste the output for some reason. I have also tried to read the file like so:
File file = new File("some_text_file.txt");
Path path = file.toPath();
String text = new String(Files.readAllBytes(path));
... with no change in result.
How come the output is not as expected? I also tried reading a text file that I wrote and it worked perfectly fine.
It looks like encoding problem. Use a tool that can detect encoding to open the file (like Notepad++) and find how it is encoded. Then use the other constructor for Scanner:
Scanner scanner = new Scanner(new FileInputStream(file), encoding);
Or you can simply experiment with it, trying different encodings. It looks like UTF-16 to me.
final Scanner scanner = new Scanner(new FileInputStream(file), "UTF-16");
I am trying to write the result of the following method to a webpage. I think it is possible to do, I am just having trouble figuring out exactly how to do it. Any help would be very appreciated.
Thanks.
...
System.out.println("Final Register");
for (int i=0; i < ch.length; i++)
{
System.out.println(cd.getDenomLiteralAt(i)+" "+cd.getDenomNumAt(i));
}
}
...
In java there are many ways to write data on a file. To write text data the easiest way is by using a BufferedWriter.
See demo below
FileWriter fWriter = null;
BufferedWriter writer = null;
try {
fWriter = new FileWriter("fileName.html");
writer = new BufferedWriter(fWriter);
writer.write("<span>This iss your html content here</span>");
writer.newLine(); //this is not actually needed for html files - can make your code more readable though
writer.close(); //make sure you close the writer object
} catch (Exception e) {
//catch any exceptions here
}
All you need to do is know the path to the public HTML file of your webserver.
Run the Java code and write to a file below that path. It's no different than writing to any other file on a machine, except this one the world can see. Only thing is, if your Java will be creating the file, you should be sure to set protective enough permissions for the file. 0755 is moderately safe.
I have to implement an application that permits printing the content of all files within a tar.gz file.
For Example:
if I have three files like this in a folder called testx:
A.txt contains the words "God Save The queen"
B.txt contains the words "Ubi maior, minor cessat"
C.txt.gz is a file compressed with gzip that contain the file c.txt with the words "Hello America!!"
So I compress testx, obtain the compressed tar file: testx.tar.gz.
So with my Java application I would like to print in the console:
"God Save The queen"
"Ubi maior, minor cessat"
"Hello America!!"
I have implemented the ZIP version and it works well, but keeping tar library from apache ant http://commons.apache.org/compress/, I noticed that it is not easy like ZIP java utils.
Could someone help me?
I have started looking on the net to understand how to accomplish my aim, so I have the following code:
GZIPInputStream gzipInputStream=null;
gzipInputStream = new GZIPInputStream( new FileInputStream(fileName));
TarInputStream is = new TarInputStream(gzipInputStream);
TarEntry entryx = null;
while((entryx = is.getNextEntry()) != null) {
if (entryx.isDirectory()) continue;
else {
System.out.println(entryx.getName());
if ( entryx.getName().endsWith("txt.gz")){
is.copyEntryContents(out);
// out is a OutputStream!!
}
}
}
So in the line is.copyEntryContents(out), it is possible to save on a file the stream passing an OutputStream, but I don't want it! In the zip version after keeping the first entry, ZipEntry, we can extract the stream from the compressed root folder, testx.tar.gz, and then create a new ZipInputStream and play with it to obtain the content.
Is it possible to do this with the tar.gz file?
Thanks.
surfing the net, i have encountered an interesting idea at : http://hype-free.blogspot.com/2009/10/using-tarinputstream-from-java.html.
After converting ours TarEntry to Stream, we can adopt the same idea used with Zip Files like:
InputStream tmpIn = new StreamingTarEntry(is, entryx.getSize());
// use BufferedReader to get one line at a time
BufferedReader gzipReader = new BufferedReader(
new InputStreamReader(
new GZIPInputStream(
inputZip )));
while (gzipReader.ready()) { System.out.println(gzipReader.readLine()); }
gzipReader.close();
SO with this code you could print the content of the file testx.tar.gz ^_^
To not have to write to a File you should use a ByteArrayOutputStream and use the public String toString(String charsetName)
with the correct encoding.