S
scb_55_55
Not sure if this is the right group.
In Word XP (SP3 - latest securty, W-XP SP2, 1.4GHz, Latitude D600),
I've mailmerged somce values from Access XP. The merge numbers are ok
(formatted Single with 10 decimal places in Access).
When performing a simple calc (division), Word is displaying the right
number of decimal places. The equation looks like:
=1.02154*.45 + 1.104*.2 + 1.01613*.1 + 1.04167*.25
The 5 decimal place numbers are all calculated (from mailmerge values)
and formated as:
{bookmark \# "#,###.#####"}
When performing the next calculation, Word seem to be losing one
decimal place for each new figure in the equation.
Should be:
=.45965 + .2208 + .10161 + .26042 (.2208 is only 4 places anyhow)
Word is returning (this is with formatting as {bookmark \#
"#,###.#####"}):
=.45965 + .2208 + .101 + .26
If I changed the format (for the last component) to {bookmark \#
"#,##0.00000"}, Word returns:
=.45965 + .2208 + .101 + .26000
Any ideas?
TIA
In Word XP (SP3 - latest securty, W-XP SP2, 1.4GHz, Latitude D600),
I've mailmerged somce values from Access XP. The merge numbers are ok
(formatted Single with 10 decimal places in Access).
When performing a simple calc (division), Word is displaying the right
number of decimal places. The equation looks like:
=1.02154*.45 + 1.104*.2 + 1.01613*.1 + 1.04167*.25
The 5 decimal place numbers are all calculated (from mailmerge values)
and formated as:
{bookmark \# "#,###.#####"}
When performing the next calculation, Word seem to be losing one
decimal place for each new figure in the equation.
Should be:
=.45965 + .2208 + .10161 + .26042 (.2208 is only 4 places anyhow)
Word is returning (this is with formatting as {bookmark \#
"#,###.#####"}):
=.45965 + .2208 + .101 + .26
If I changed the format (for the last component) to {bookmark \#
"#,##0.00000"}, Word returns:
=.45965 + .2208 + .101 + .26000
Any ideas?
TIA