[HELP] How to read Strings from the Resource file?

Hello everyone who is ready this,

I really need help with my plugin. I searched everywhere and I couldn't find it.
As the title says How can I read a String (or Boolean) from the Resource File? Is this possible? If true, how?

I hope you understand this. Please help me out!
Greetings,
Steffion!

 

What Resource file? Are you talking about a string in a config your plugin generated? If so, take a look at this:
http://wiki.bukkit.org/Configuration_API_Reference

From their examples:
Setting a string:

Code:

// setting a String
String stringValue = "Hello World!";
this.getConfig().set("path.to.string", stringValue);

And to get a string:

Code:

//Getting a string
this.getConfig.getString(String);

 

Code:

File pladata = new File(getDataFolder()+ File.separator + "filename.yml");
FileConfiguration plset = YamlConfiguration.loadConfiguration(pladata);
plset.getInt(for integers)
plset.getBoolean(for booleans)
plset.getString(for strings)
Just type plset. and wait

 

skipperguy12
Hello Skipper, Thanks for replying.

I don't mean the generated one but the one you see in the "src/" folder.

Vandrake
Is this the Generated file or the one from the "src/" folder?

EDIT by Moderator: merged posts, please use the edit button instead of double posting.

 

Just looked through my server files, can't seem to find a scr/ folder.

Where would it be?

 

skipperguy12
Its the folder where the plugin is written in. (So inside the jar.)
I think its impossible to read it but if it could please help!

 

OHH, you mean in eclipse! I'm pretty sure thats impossible, because its a jar. Either ways, if someone DOES figure out how to do it, you would still get an error if you try to alter anything: Java SE Binary is already in use or something like that. :L

 

skipperguy12 Sadly.... Still thanks for trying!

 

How about

Code:

new BufferedReader(new InputStreamReder(<yourmainclass>.class.getResourceAsStream("file.txt")));

then you can use the methods from the BufferedReader to get things from your file, it's that easy.

 

kumpelblase2
And how to get the string then ;#?

kumpelblase2 Because I want to get a specific config out of the resource and put it in a String. I hope you understand me ;#!

EDIT by Moderator: merged posts, please use the edit button instead of double posting.

 

http://docs.oracle.com/javase/1.5.0/docs/api/java/io/BufferedReader.html
I won't give you any more. Try out yourself, find what works for you.

 

Code:

/*Method to read lines*/
public ArrayList<String> getLines(String resource) {
  ArrayList<String> lines = new ArrayList<String>();
  Scanner s = new Scanner(getClass().getResourceAsStream(resource));
  while(s.hasNextLine()) {
    lines.add(s.nextLine());
  }
  s.close();
  return lines;
}
 
/*Getting lines from specified resource*/
ArrayList<String> lines = getLines(/*resourceName*/);
 
/*Getting string from specified line number*/
String line = lines.get(/*lineNumber*/);

I do prefer using Scanner over BufferedReader because I find it easier to use, however to make it even easier you get the line from the list Resources.readLines(URL url, Charset charset) returns.

Still, I'm not sure why you'd want to store a variable like this.

Edit - I forgot to close the Scanner in the method.