Split text string into 5 columns

@John,

Have you tried it?

Select Rows 11 to 16 and try running your code... your comment regarding the C-H is incorrect, it doesn't make any difference.

with C11:H16 to selected,
y = 11 and x = 6
so your loop will go from 6 to 11 with -1 steps... which will just exit the loop instantly and do nothing.

@Desree86, I do not use i in my code... so not sure why that is an error on that line. Have you tried this code:

sub macro_1()
dim i
for i = selection.row to selection.row + selection.rows.count
   Range("D" & i) = mid(Range("C" & i),1,3)
   'and so on
next
end sub

I've changed my variable to i to make mine and John's code more interchangeable.

Originally Posted by yudlugar

@John,

Have you tried it?

Select Rows 11 to 16 and try running your code... your comment regarding the C-H is incorrect, it doesn't make any difference.

with C11:H16 to selected,
y = 11 and x = 6
so your loop will go from 6 to 11 with -1 steps... which will just exit the loop instantly and do nothing.

@Desree86, I do not use i in my code... so not sure why that is an error on that line. Have you tried this code:

sub macro_1()
dim i
for i = selection.row to selection.row + selection.rows.count
   Range("D" & i) = mid(Range("C" & i),1,3)
   'and so on
next
end sub

I've changed my variable to i to make mine and John's code more interchangeable.

I tested it. You have it inverted y = selection.row (the first row in the selection) - 6 , x = selection.rows.count (the last row in the selection) - 11. It loops from the bottom of the selection to the top.