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ISLEAF explanation

Experts,

I am a beginner with these functions and I'm hoping someone can explain the ISLEAF results in this query. The results are producing the correct hierarchy.

SELECT ETC_ID,    SYS_CONNECT_BY_PATH(etc_id, '$') ids,
                   SYS_CONNECT_BY_PATH(etc_name, '$') names,
                   CONNECT_BY_ISLEAF isleaf
              FROM ETC_TBL
              WHERE ETC_RETIRED_IND = 0
        CONNECT BY ETC_PARENT_ETC_ID = PRIOR etc_id
        START WITH ETC_PARENT_ETC_ID = 0

I attached a sample Excel output of a few records.
I am wondering why the first 2 records have ISLEAF = 0 (High level explanation of why I should expect that result)
isleaf-example.xlsx
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jvoconnell
Asked:
jvoconnell
1 Solution
 
slightwv (䄆 Netminder) Commented:
I assume you have been to the docs for this?

http://docs.oracle.com/cd/E11882_01/server.112/e26088/pseudocolumns001.htm#SQLRF50940

CONNECT_BY_ISLEAF Pseudocolumn

The CONNECT_BY_ISLEAF pseudocolumn returns 1 if the current row is a leaf of the tree defined by the CONNECT BY condition. Otherwise it returns 0. This information indicates whether a given row can be further expanded to show more of the hierarchy.
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jvoconnellAuthor Commented:
Thank you. I did check online before posting but was strictly limiting myself (out of frustration) to "isleaf".  On the specific link you posted, the "LEVEL" psuedocolumn helped clarify things. I appreciate your time, you helped me out once again.
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