You will no doubt have to do some work on the empty values first - I cannot
remember but what happens when you do an #ArrayLen(ListToArray(yourlist))#?
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len(REreplace(myList, [^,]+, , all) + 1
should work. Just removing (or replacing with nothing) all the
non-delimiter characters, and then counting the delimiters, and adding
one for the fencepost problem.
cheers,
barneyb
On 8/22/06, Ian Skinner [EMAIL PROTECTED] wrote:
Counting all the
http://www.cflib.org/udf.cfm?id=961
-Original Message-
From: Ian Skinner [mailto:[EMAIL PROTECTED]
Sent: Tuesday, August 22, 2006 9:17 AM
To: CF-Talk
Subject: What is the easiest way to get the true length of a list.
Counting all the elements, including the empty ones?
Do I need to
I think there is even a handy UDF on cflib..
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Young Jedi,
Be careful with the Java split method... In your example, the length is
indeed 4... However, this:
cfset lst = jedi,homer,,, /
#ArrayLen(lst.split(,))#
Only has TWO array elements. It seems to only count list items until no more
valid items are found. This seems to be confirmed at
Do I need to modify the list to put some throwaway character to get
all the elements counted
I'm affraid so, however, and especially if your list is large, it could
be more efficient to
replace all commas with nothing and compare the length of the resulting
string.
This way, no list function
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