Formatting hierachical queries with lpad function

问题

SELECT LPAD(last_name, LENGTH(last_name)+(LEVEL*2)-2,'_') 
AS org_chart
FROM employees 
START WITH last_name='King' 
CONNECT BY PRIOR employee_id=manager_id ;

LPAD(char1,n [,char2]) returns char1, left-padded to length n with the sequence 
of characters in char2.

This tells SQL to take the LAST_NAME and left-pad it with the '_' character till the length of the resultant string is equal to the value determined by

LENGTH(last_name)+(LEVEL*2)-2. 
For  LEVEL = 1. Hence, (2 * 1) - 2 = 2 - 2 = 0. 
For LEVEL = 2. Hence, (2 * 2) - 2 = 4 - 2 = 2 . So its gets padded with 2 '_'

characters and is displayed indented.

and also how to determine the formula that length(ename) to be added with level*2-2

the output is king doesn't get padded with '-'

This is the correct output

ORG_CHART

    KING
    __PAUL
    __JONES
    ____SCOTT
    ______ADAMS
    ____FORD
    ______SMITH
    __BLAKE
    ____ALLEN
    ____WARD
    ____MARTIN
    ____TURNER
    ____JAMES
    __CLARK
    ____MILLER

but formula lpad(king,4+(l*10)-10,'-')=>lpad(king,4,'-')

which means king should be padded with 4'-'

回答1:

You need to include the length of the field because you have to allow for the number of characters in that, plus the amount of indentation you want. Taking the third-level Greenberg for example, that is displayed as:

____Greenberg

... with four underscores. Level is 3 here, so (level * 2) - 2 is four. But if you only used that value you'd get:

lpad('Greenberg', 4, '_')

and the output of that is just:

Gree

You want the final output string, including the underscores, to be four characters longer than the name on its own. 'Greenberg' is 9 characters, and ____Greenberg is 13; so your padding length has to be 13, which is the length of the name plus the number of underscores you want to appear in front.

Another way to get the same effect is with:

SELECT LPAD('_', (LEVEL - 1) * 2, '_') || last_name AS org_chart
...

That makes the underscore padding separate from the name itself - it's based just on the level, and the name is just concatenated on the end.

For King, the level is 1. You said the formula is:

lpad(king,4+(l*10)-10,'-')=>lpad(king,4,'-')

Which is right, but 'King' is already four characters long, so padding it out to four characters has no effect. Your are padding it out to the final length of 4. lpad doesn't add four underscores regardless; it only adds underscores up to the requested length, which is 4 in this case.

I think you're just misinterpreting how the function workd. As the documentation says (emphasis added):

LPAD returns expr1, left-padded to length n characters with the sequence of characters in expr2.
...
The argument n is the total length of the return value as it is displayed on your terminal screen.

So:

select lpad('King',4,'_') from dual;

LPAD('KING',4,'_')
------------------
King

If you asked for a longer final length you'd get the number of underscores needed to pad 'King' out to that length:

select lpad('King',5,'_') from dual;

LPAD('KING',5,'_')
------------------
_King

If you want King to be indented as well, by two underscores; and subsequent levels to be indented more to match (so Kochhar gets 4 and Greenberg gets 6) then remove the -2 from the calculation.

来源：https://stackoverflow.com/questions/23270781/formatting-hierachical-queries-with-lpad-function

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sql

Oracle