Returning String value of Ljava.lang.Object

Question

I have the following code

TreePath tp = jTree.getSelectionPath();
String path = tp.getPath().toString();

This will set path as Ljava.lang.Object;@33530691, I understand this is because

The toString method for class Object returns a string consisting of the name of the class of which the object is an instance, the at-sign character `@', and the unsigned hexadecimal representation of the hash code of the object. In other words, this method returns a string equal to the value of:

getClass().getName() + '@' +Integer.toHexString(hashCode())

But how do you actually get this value as a String?

dan · Accepted Answer · 2012-08-31 09:57:34Z

1

getPath() returns an array of String values representing the components in the tree. If you need the String representation of the root, you will do getPath()[0];

answered Aug 31, 2012 at 9:57

dan

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1 Comment

Hip Hip Array Over a year ago

Thought .getPath returned Object and not Object[], should have payed more attention!!

Kumar Vivek Mitra · Accepted Answer · 2012-08-31 09:57:36Z

0

String's toString() method is by defalut overloaded, so the problem is not here with the String path, but try to see the TreePath tp.

I think you need to take a look at tp.

answered Aug 31, 2012 at 9:57

Kumar Vivek Mitra

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Comments

Thomas · Accepted Answer · 2012-08-31 09:58:40Z

0

I don't quite get your question. Isn't that already answered in the quote you posted?

getClass().getName() + '@' +Integer.toHexString(hashCode()) the default implementation of the toString() method, but you can override that, if you need a better representation (keep in mind that you need to extend TreePath in your case ).

Edit: I think, I now understand:

Since TreePath.getPath() returns an object array you get that java.lang.Object in the output.

answered Aug 31, 2012 at 9:58

Thomas

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Comments

RNJ · Accepted Answer · 2012-08-31 09:59:25Z

0

The object that is the component of the tree needs to have its toString() method overridden. You can override it with whatever you what it to return. I presume you are adding a custom object to the tree so you are in control of that object and its toString() method

answered Aug 31, 2012 at 9:59

RNJ

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mwikblom · Accepted Answer · 2012-08-31 10:01:58Z

0

If you want to get a comma separated string of the values you can do

String path = new ArrayList(tp.getPath()).toString();

answered Aug 31, 2012 at 10:01

mwikblom

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tibtof · Accepted Answer · 2012-08-31 10:02:16Z

0

Try something like this:

Object[] treeNodes = tp.getPath();
StringBuilder sb = new StringBuilder();
for(Object tn : treeNodes) {
    sb.append(File.separatorChar).append(tn.toString());
}
String path = sb.toString();

answered Aug 31, 2012 at 10:02

tibtof

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Peter Lawrey · Accepted Answer · 2012-08-31 10:03:53Z

0

You have an array of objects (which may all be Strings)

To toString these you need to use Arrays.toString()

String path = Arrays.toString(tp.getPath());

answered Aug 31, 2012 at 10:03

Peter Lawrey

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Mirvnillith · Accepted Answer · 2012-08-31 10:05:19Z

0

If it's for output/logging only, I guess Arrays.asList(tp.getPath()).toString() could suffice.

answered Aug 31, 2012 at 10:05

Mirvnillith

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Robin · Accepted Answer · 2012-08-31 10:06:14Z

The TreePath object represent the whole path to go from the root-node to the actual node. If you call getPath on it, it returns you an array containing the root-node and the actual selected node, and all nodes in between.

How to get the "correct String" depends on what you actually want to achieve:

If you want the string representation of the actual selected node, use the getLastPathComponent method
```
TreePath tp = jTree.getSelectionPath();
String path = tp.getLastPathComponent().toString();
```

If you want to have the complete path, you will have to loop over the array

TreePath tp = jTree.getSelectionPath();
StringBuilder pathBuilder = new StringBuilder();
for( Object node : tp.getPath() ){
  pathBuilder.append( node.toString() );
  //possibly append a separator as well
}
String path = pathBuilder.toString();

or simply use the Arrays#toString method on tp.getPath() instead of the toString method

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