INTERACT FORUM
More => Old Versions => Media Center 17 => Topic started by: locust on March 13, 2012, 09:16:57 pm
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I've just finished my images caption (I know it looks messy but it is this way so that certain elements don't appear on the same line)
I like the new ifelse feature. I've managed to make a caption that would otherwise be extremely difficult.
If the output of the day is 01 it outputs 01st
and if 02 it outputs 02nd,
and if 03 it ouputs 03rd,
and if 21 it ouputs 21st
and if 22 it ouputs 22nd
and if 23 it ouputs 23rt
and if 31 it ouputs 31st
and if day equals any other it outputs [day]th
The part I'd like help with is the date I've got it working there is just one little tweak I'd like to do.
The part that outputs the day to a double digit, if the number is from 01 - 09 remove the 0 digit from it. Would it be hard to do/worth while). I don't know I just don't like the look of left trailing zeros.
if(isempty([date]),,formatdate([date,0],dddd the formatdate([date,0],dd,),)ifelse(isequal(formatdate([date,0],dd,),01,2),st,isequal(formatdate([date,0],dd,),02,2),nd,isequal(formatdate([date,0],dd,),03,2),rd,
isequal(formatdate([date,0],dd,),21,2),st,isequal(formatdate([date,0],dd,),22,2),nd,isequal(formatdate([date,0],dd,),23,2),rd,isequal(formatdate([date,0],dd,),31,2),st,1,th)
formatdate([date,0],MMMM,) formatdate([date,0],yyyy,)
)if(isempty([caption]),,[caption])
if(isempty([places]),,[places])
if(isempty([people]),,replace([people],;,/ &))
Thanks
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Use one 'd' instead of two.
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You can do that??
This is all the wiki says
yy Returns the last two digits of the year
yyyy Returns the full year
MM Returns the month as two digits
MMM Returns the month as three letter abbreviations
MMMM Returns the month in full
dd Returns the day as two digits
ddd Returns the day as three letter abbreviations
dddd Returns the full day
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By the way, try this:
ifelse(isempty(formatdate([date,0],d)), No Day Defined,
regex(formatdate([date,0],d), /#^(1|21|31)$#/), [R1]st,
regex(formatdate([date,0],d), /#^(2|22)$#/), [R1]nd,
regex(formatdate([date,0],d), /#^(3|23)$#/), [R1]rd,
1, formatdate([date,0],d)th)
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You can do that??
This is all the wiki says
Break the rules, live on the edge.
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ok lol, it worked :)
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Btw. here's another way to get your English days listed from the date:
if(isempty(formatdate([date,0],d)), No Day Defined, listitem(NA;1st;2nd;3rd;4th;5th;6th;7th;8th;9th;10th;11th;12th;13th;14th;15th;16th;17th;18th;19th;20th;21st;22nd;23rd;24th;25th;26th;27th;28th;29th;30th;31st, formatdate([date,0],d), ;))
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Brian0001,
In another thread, Vagskal brings to my attention that you can simplify this expression by moving some of it to a calculated field. By assigning the value:
NA;1st;2nd;3rd;4th;5th;6th;7th;8th;9th;10th;11th;12th;13th;14th;15th;16th;17th;18th;19th;20th;21st;22nd;23rd;24th;25th;26th;27th;28th;29th;30th;31st
to a new calculated field, say [Days w-Suffix], you can then use the simpler expression:
if(isempty(formatdate([date,0],d)),
No Day Defined,
listitem([Days w-Suffix], formatdate([date,0],d), ;))
Essentially, [Days w-Suffix] becomes a global value you can use in your expressions to simplify them.
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Thanks again MrC, got my caption updated, very happy :)