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  1. Commons JXPath
  2. JXPATH-5

asPath() returns a path to the last sibling

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Details

    • Bug
    • Status: Resolved
    • Major
    • Resolution: Fixed
    • 1.2 Final
    • 1.3
    • None
    • Operating System: All
      Platform: All

    • 33377

    Description

      The method call JXPathContext::iteratePointers() returns the correct number of
      pointers. However, the pointers do not always return correct path on asPath()
      method call on them. A pointer returns a path that corresponds to the last
      Node in the sibling Nodes of the node that the pointer points to.

      Here is an example:

      Class A{
      private List list;
      //getter and setter
      :
      }

      Here is a code snippet for creating an object graph that starts from
      instance of A.

      A a1 = new A();
      List list1 = new LinkedList();

      A a11 = new A();
      list1.add(a11);

      A a12 = new A();
      List list12 = new LinkedList();
      A a121 = new A();
      list12.add(a121);
      a12.setList(list12);
      list1.add(a12);

      A a13 = new A();
      list1.add(a11);

      a1.setList(list1);

      And the list attribute can have instances of A as elements in it.

      The JXPathContext correspong to a1 returns pointers when it's iteratePointers
      () method is called. And the pointers correspond to the following nodes.

      A[1]
      A[2]
      A[2]/A[1]
      A[3]

      This is absolutely as expected. However, asPath() method on each of these
      Pointers do not always return the correct path.

      Currently, the asPath() method calls on the corresponding Pointers return this
      output respectively.

      A[3]
      A[3]
      A[2]/A[1]
      A[3]

      This needs to be fixed.

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              Unassigned Unassigned
              krishna.patil@gmail.com Krishna Patil
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              Dates

                Created:
                Updated:
                Resolved: