CODEWITHSUNDEEP
kdb+
Terry
Terry

Reputation: 519

deleting all records for all tables in memory in Kdb+

I would like to delete all records for all tables in memory but still keep the schemas.

for example:

a:([]a:1 2;b:2 4);
b:([]c:2 3;d:3 5);

I wrote a function:

{[t] t::select from t where i = -1} each tables[]

this didnt work, so i tried

{[t] ![`t;enlist(=;`i;-1);0b;()]} each tables[]

didnt work either. Any idea or better ways?

Upvotes: 1

Views: 879

Answers (2)

Matt Moore
Matt Moore

Reputation: 2775

Mark's solution is best for doing what you want rather than functional form. Just adding to your question on t failing as putting kdb code in comments is awkward.

Your functional form fails not because of the t but because your last argument is not a symbol list `$(). Also you would want to delete where i is > -1, not =

q){[t] ![t;enlist(>;`i;-1);0b;`$()]} each `d`t`q
`d`t`q

q)d
date sym time bid1 bsize1 bid2 bsize2 bid3 bsize3 ask1 asize1 ask2 asize2 ask..
-----------------------------------------------------------------------------..
q)t
date sym time src price size
----------------------------
q)q
date sym time src bid ask bsize asize
-------------------------------------

Upvotes: 2

Mark Kelly
Mark Kelly

Reputation: 1780

If you pass a global table name as a symbol, it removes all the rows, leaving an empty table

q)delete from `a
`a

q)a
a b
---

q)meta a
c| t f a
-| -----
a| j    
b| j

To do it for all global tables in root name space

{delete from x} each tables[]

Your second attempt using function was close. You can achieve it via the following (functional form of the above):

![;();0b;`symbol$()] each tables[]
  • The first argument should be the symbol of the table for the same reason I mentioned before
  • The second argument should be an empty list as we want to delete all records (we do not want to delete where i=-1, as that would delete nothing)
  • The final argument (list of columns to delete) should be an empty symbol list instead of an empty general list.

Upvotes: 2

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