MySQL: using calculated counter column as key for sub-query - mysql

Question

Sorry, long pre-history, but it is needed to clarify the question.
In my org the computers have names like CNT30[0-9]{3}[1-9a-z], for example cnt300021 or cnt30253a.
Last symbol is a "qualifier", so single workplace may have equally named computers assigned to it, distinguished by this qualifier. For example cnt300021 may mean desktop computer on workplace #002, and cnt30002a may mean notebook assigned for same workplace. Workplaces are "virtual" and their existence made just for our (IT dept) convenience.
Each dept have its own unique range [0-9]{3}. For example, computers of accounting have names starting cnt302751 upto cnt30299z which gives them 25 unique workplaces max, with up to 35 computers per workplace. (IRL most users have one desktop PC, much lesser have desktop and notebook, and only 2 or 3 technicians have more than one notebook at their disposal).
Recently, doing some inventory of computers' passports (unsure about a term: a paper, which means for computer the same, what a passport means for human), I found that there some holes in sequential numbering. For example, we have cnt302531 and cnt302551, but have no cnt302541, which means that there's no workplace #254.
What I want to do? I want to find these gaps without manual searching. For this I need a cycle from 1 to MaxComp=664 (no more workplace numbers assigned yet)
That's what I could write using some pseudo-SQL-BASIC:
for a=0 to MaxComp
a$="CNT30"+right(a+1000,3)
'comparing only 8 leftmost characters, ignoring 9th one - the qualifier
b$=(select name from table where left(name,8) like a$)
print a$;b$
next a
That code should give me two colummns: possible names and existing ones.
But I can't figure out how to implement this in SQL-query. What I tried:
# because of qualifier there may be several computers with same
# 8 leftmost characters
select #cnum:=#cnum+1 as CompNum, group_concat(name separator ',')
# PCs are inventoried by OCS-NG Inventory software
from hardware
cross join (select #cnum:=0) cnt
where left(hardware.name,8)=concat('CNT30',right(#cnum+1000,3))
limit 100
But this construct returns exactly one row. And I can't understand, if it is possible without using the stored procedures, and what I did wrong if it is possible?

Answer 1

Found working path:
(at first I tried to use stored function)
CREATE FUNCTION `count_comps`(num smallint) RETURNS tinytext CHARSET utf8
BEGIN
return (select group_concat(name separator ',')
from hardware where left(hardware.name,8)=concat('CNT30',right(num+1000,3))
);
END
Then I tried hard to replicate function's results in subquery. And I did it! Note: the inner select returns exactly same results as function does
# Starting point. May be INcreased to narrow the results list
set #cnum:=0;
select
#cnum:=#cnum+1 as CompNum,
concat('CNT30',right(#cnum+1000,3)) as CalcNum,
# this
count_comps(#cnum) as hwns,
# and this gives equal results
(select group_concat(name separator ',')
from hardware where left(name,8)=calcnum
) hwn2
from hardware
# no more dummy tables here
# Ending point. May be DEcreased to narrow the results list
where #cnum<665;
So, the wrong part of "classical" approach was the usage of dummy table, which seems to be not necessary.
Partial results example (starting set #cnum:=479;, finishing where #cnum<530;):
CompNum, CalcNum, hwns, hwn2
'488', 'CNT30488', 'CNT304881', 'CNT304881'
'489', 'CNT30489', 'CNT304892', 'CNT304892'
'490', 'CNT30490', 'CNT304901,CNT304902,CNT304903', CNT304901,CNT304902,CNT304903'
'491', 'CNT30491', NULL, NULL
'492', 'CNT30492', NULL, NULL
'493', 'CNT30493', 'CNT304932', 'CNT304932'
'494', 'CNT30494', 'CNT304941', 'CNT304941'
I found that there no workplaces #491 and #492. On next adding the PCs for the 'October Region' dept (range 480-529), at least two of new PCs will get names CNT304911 and CNT304921, filling this gap.