Brooke, You wrote.
I'm handing this job back to you. How you worked out that coding for b1 is beyond me. How exactly does the scan define b1 as the next j's value for b?
It doesn't define it as the next j's, it only saves the last 'b' into 'b1' so I it can compare and only output if bigger than 'yesterday'.
I ran Bob's scan and looked at the output, It appeared to give a result for all issues on the current day. I thought that if I could simply put his code in the middle of a new loop, and have his loop date modified by the outer loop, no problem. (Ya right)!! After several tries I found it necessary to save the current 'b' to compare to the next pass, Bob's scan seems to do all that work,
But then I got into the time warp stuff, is (-2) today, tomorrow, the day after or before the day after the day before yesterday!! Finally I just ran the darn thing and kept changing the b,<,=,>b1 until I thought I had the the right output.
I assume from your post that what you are looking for is the 'highest' percentage b/b1 in for the duration of the outer loop?? Would you need a 'trigger point' (i.e.over/under some %) or just the max %?
Please believe me when I say that sometimes I don't even know how this stuff works!! |
