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[xsl] max value of an element


Subject: [xsl] max value of an element
From: "Ilyasov Jienbay" <jienbay@xxxxxxxxxxx>
Date: Tue, 12 Mar 2002 10:43:13 +0000

hello,

thank you guys, you help a lot.
and again, now i have another question and i wonder if there are some solutions for it.


i have
<memo>
<start>$_</start>
<end>$_</end>
</memo>
.. and in the time-table it looks like

1_____2_____3_____4_____5_____6_____7_____8_____9_____10_____11_____12....
     |start 2    |                                   |start10      |
     |end 4      |                                   |end 12       |
     |___________|________________________           |_____________|
           |start 3                      |
           |end 8                        |
           |_____________________________|
                       |start 5    |
                       |end 7      |
                       |___________|

and i want to have them like

1_____2_____3_____4_____5_____6_____7_____8_____9_____10_____11_____12....
|start 2 |start 3 | |start10 |
|end 4 |end 8 | |end 12 |
|___________|_______________________| |_____________|
the colspan of an element whose "start" is less than preceding-element's end is shortened by the "preceding::end 4" - "start 3".


if both "start 5" and "end 7" are less than preceding::end(8) then it is skipped.(well, linked in a preceding <memo>'s <td/>)

the <memo>s are sorted 1st by <start>, then by <end> , but this time order="descending"

it is a table so the space between "end 8" and "start 10" is to be calculated 10-8=2, and putt 2 <td>s between them. At this point i have only the preceding <memo>' "end"(which is = 7), and i have 10-7=3, i.e. 3 <td>s instead of desired 2.
The problem is i can not figure out which (double-booked)elements "end" is longest. Of cource there can be a lot of such elements.


awaiting for advices,

jian

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