TREE Construction Example

Suppose that the SYSIN file contains the following statements:

LEVEL=20

USER -

POS=0

LENGTH=1

DEFAULT=L

POS=10

LENGTH=2

DEFAULT=03

PARENT -

PLEVEL=10

POS=0

LENGTH=6

DEFAULT=CDTREE

LEVEL=30

PARENT -

PLEVEL=20

POS=0

DEFAULT=L

LENGTH=1

POS=10

LENGTH=2

TRANSLATE=TABLE1

USER -

POS=0

LENGTH=1

DEFAULT=L

POS=1

LENGTH=4

DEFAULT=CKJ

SYNONYM -

POS=0

LENGTH=4

DEFAULT=USER

POS=1

LENGTH=4

SYNONYM -

POS=0

LENGTH=5

DEFAULT=SYN1-

POS=1

LENGTH=4

The Recipient Tree is constructed as follows:

For each line in REPORT file:

1. For a user at level 20 with the name Lxy, where xy are the contents of columns 10 and 11 in the report line:
   • If column 10 is blank, the name is ‘L0y’
   • If column 11 is blank, the name is ‘Lx3’
   • If both are blank, the name is ‘L03’
2. The parent of this user is at level 10 with the name CDTREE.
3. For a user at level 30 with the name Labcd, where abcd are the contents of columns 4 through 7 in the report line:

   Default CKJ is used in a manner similar to default 03 in item 1 above.

4. An attempt is made to determine a level 20 parent from the value ‘L’ plus the contents of columns 10 and 11. If this does not succeed, then the file referenced by DDNAME TABLE1 is scanned line by line until a match is found for the user name and the parent name are taken from the file.

   Two synonyms are created: ‘USERabcd’ and ‘SYN1-abcd’, where abcd represents the contents of columns 1 through 4 in the report line.