[drraw-users] Old Subject from Wiki Wishlist

Gregory Davis greg at pando.com
Fri Aug 31 14:52:39 EDT 2007


The feature request below can be added with the following change. 
-- Gregory Davis 


line 3207 and 3208 
-values=>[@gt], -default=>param("${ds}_tType")), 
checkbox(-name=>"${ds}_tSTACK", -values=>'Y', -label=>'')), 

can be replaced with: 
-values=>[@gt], -default=>param("${ds}_tType")), br, 
checkbox(-name=>"${ds}_tSTACK", -values=>'Y', -label=>'<', -title=>'Whether to stack these elements on top of each other'), 
checkbox(-name=>"${ds}_STACK", -values=>'Y', -label=>'^', -title=>'Whether to stack these elements on top of the previous one')), 


from: 
http://web.taranis.org/pipermail/drraw-users/2007q1/000095.html 

On Jan 21, Greg Davis wrote: 
| Thanks for your help. About the regex stacking, i may not have been 
| clear on the question. What im trying to do is stack two template based 
| DSs on a single graph each template based DS would contain the data from 
| about 40 rrds for a total of 80 actual DSs. So lets say template DS A 
| would be 40 stacked areas in blue and green and Template DS B would be 
| another 40 stacked areas in red and yellow stacked on top of Template A. 
| When I add the second template DS it always begins the stack at 0. So 
| far I've had to Add template A and then use 40 regular DSs stacked on 
| top of template A's output. Is this the only way to do it? 

Which version of drraw are you using? 

With version 2.1.1, i'm able to do this by using "STACK" as 
the type in the drop down list, but this doesn't let one 
control what is stacked (it's based on the previous entry). 

With 2.2a2, you can't really do it. 

If you want a headache, read on. 
you could do this: 
Say your template DS A procudes a0, a1, ... a39 
Set Element to "UN,0,$,IF" and Formula to "+" 
This gives a new CDEF a which is a0 + a1 + ... + a39 (skipping 
NaNs). 

Now, if you template DS A produces b0, b1, ... b39 
Set the CDEF to "b0,$,EQ,a,0,IF,$,+" 

The problem is you "b0,$,EQ" compares the value of b0 with $ 
(which is b0 or b1 or ... b39), but you really want to compare 
the variable names, not their values.  So while it may work 
most of the time, there'll be cases when it doesn't. 
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