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Read from relative path.

Hi

I have this trivial problem ... :) beleive it is major for me ... I just want to tell that to myself ....

I have a XML file in my Java project (command line based) . The XML file is in the folder <classes>/resources/db.xml.

Now I want to read this XML file.

I have tried this.getClass().getResourcesAsStream("db.xml"); /// bombs right on my face with malformed url exception
this.getClass().getResources("db.xml").getFile(); /// i feel even worse.

Can any one hash this out for me ...

RK

Madhavan Sundarraj

try the following

1. Give full path with the file as this.getClass().getResources("<classes>/resources/db.xml").getFile();

or

2. Include <classes>/resources in ur CLASSPATH

CEHJ

Try

URL url = getClass().getResource("/resources/db.xml");

and pass that to your parser

Mick Barry

assuming <classes> is in your classpath use:

InputStream in = this.getClass().getResourcesAsStream("/resources/db.xml");

Or use a relative name, being relative to the class that is making the call, eg. if class is in <classes> directory you would use:

InputStream in = this.getClass().getResourcesAsStream("resources/db.xml");

Webstorm

Hi rahulkothari,

Use
this.getClass().getResourceAsStream("resources/db.xml");
(no s at the end of Resource in getResourceAsStream)

also, the path you specify must be relative to your class directory.

Webstorm

You can also
replace this.getClass()
by
<the_class_name>.class
where <the_class_name> is the name of the class (usefull when "this" is not available from static methods)

CEHJ

Two other objects commonly passed to parsers are:

InputSource in = new InputSource(new InputStreamReader(getClass().getResourceAsStream("/resources/db.xml")));

and

Reader in = new InputStreamReader(getClass().getResourceAsStream("/resources/db.xml"));

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rahulkothari

ASKER

This is how it worked

this.getClass().getClassLoader().getResourceAsStream("resources/db.xml");

Since the resouces folder was in classpath no in the current package so getClass().getResourceAsStream("/resources/db.xml") did not work and also if we put the / then it will not work.

Thanx anyways ... for all the hints.

RK

CEHJ

So what happens with

URL url = getClass().getResource("db.xml");

?

rahulkothari

ASKER

well to be honest i tried it and it gave me back null so did not look into it again. But I am sure we have to get the resource from Classloader since it is this class which is aware of the entire classpath. The "this" is only aware of its current package I believe.

Let me know if you have a different point of view so that we can get more elegant solution.

RK

CEHJ

It calls the classloader anyway from what i can remember. It's a question of getting the path right

rahulkothari

ASKER

oh really ... let me try again.

RK

hint

AFAIK, getResourceAsStream only checks for resources relative to the current folder.i.e. the folder where the call is made. e.g. If you run it under mypackage.MyClass, it will only check for the resource relative to "mypackage".

The getSystemResourceAsStream on the otherhand, checks the entire classpath (i.e. System) for the resource file.