Joining multiple times in kdb - kdb

Question

I have two tables
table 1 (orders) columns: (date,symbol,qty)
table 2 (marketData) columns: (date,symbol,close price)
I want to add the close for T+0 to T+5 to table 1.
{[nday]
value "temp0::update date",string[nday],":mdDates[DateInd+",string[nday],"] from orders";
value "temp::temp0 lj 2! select date",string[nday],":date,sym,close",string[nday],":close from marketData";
table1::temp
} each (1+til 5)
I'm sure there is a better way to do this, but I get a 'loop error when I try to run this function. Any suggestions?

Answer 1

See here for common errors. Your loop error is because you're setting views with value, not globals. Inside a function value evaluates as if it's outside the function so you don't need the ::.
That said there's lots of room for improvement, here's a few pointers.
You don't need the value at all in your case. E.g. this line:
First line can be reduced to (I'm assuming mdDates is some kind of function you're just dropping in to work out the date from an integer, and DateInd some kind of global):
{[nday]
temp0:update date:mdDates[nday;DateInd] from orders;
....
} each (1+til 5)
In this bit it just looks like you're trying to append something to the column name:
select date",string[nday],":date
Remember that tables are flipped dictionaries... you can mess with their column names via the keys, as illustrated (very noddily) below:
q)t:flip `a`b!(1 2; 3 4)
q)t
a b
---
1 3
2 4
q)flip ((`$"a","1"),`b)!(t`a;t`b)
a1 b
----
1 3
2 4
You can also use functional select, which is much neater IMO:
q)?[t;();0b;((`$"a","1"),`b)!(`a`b)]
a1 b
----
1 3
2 4

Answer 2

Seems like you wanted to have p0 to p5 columns with prices corresponding to date+0 to date+5 dates.
Using adverb over to iterate over 0 to 5 days :
q)orders:([] date:(2018.01.01+til 5); sym:5?`A`G; qty:5?10)
q)data:([] date:20#(2018.01.01+til 10); sym:raze 10#'`A`G; price:20?10+10.)
q)delete d from {c:`$"p",string[y]; (update d:date+y from x) lj 2!(`d`sym,c )xcol 0!data}/[ orders;0 1 2 3 4]
date sym qty p0 p1 p2 p3 p4
---------------------------------------------------------------
2018.01.01 A 0 10.08094 6.027448 6.045174 18.11676 1.919615
2018.01.02 G 3 13.1917 8.515314 19.018 19.18736 6.64622
2018.01.03 A 2 6.045174 18.11676 1.919615 14.27323 2.255483
2018.01.04 A 7 18.11676 1.919615 14.27323 2.255483 2.352626
2018.01.05 G 0 19.18736 6.64622 11.16619 2.437314 4.698096