@Hüseyin-Bıçkın ,
If you just want to truncate, and it’s always starting with 3-or-more digits after the decimal point, then it’s relatively easy:
To make sure you’re counting from the decimal point, we’re going to use \. to get the literal decimal point (because regex uses . to match any character.
The next important item is the “capture group” using parentheses (...), so that whatever is inside gets saved into memory. In the replacement, use ${1} to use the value from the first capture group, ${2} for the second, and so on…
The third important item is using * to indicate “0 or more of the previous token”.
These three concepts are sufficient for your task:
Find = (\.[0-9][0-9][0-9])[0-9]*(in)
this finds a literal . followed by 3 digits, then 0 or more digits then the literal in, storing the decimal and three digits into group#1 and the in into group 2 (not necessary, but good practice).
Replace = ${1}${2}
the first group contained the decimal and 3 digits; the second group contained the in, so this means the decimal, three digits, and in.
However, using the “quantity modifier” aka “multiplying operator”" {ℕ} allows you to have ℕ matches of whatever token or subset comes before, which simplifies your FIND expression to
Find = (\.[0-9]{3})[0-9]*(in)
To simplify again, there is a character escape sequence \d which is basically equivalent to [0-9] for your purposes.
Find = (\.\d{3})\d*(in)
If you want to do true rounding (if the 4th digit is 5 or higher, add 1, else just truncate), there is no universal regex that will do it; depending on your number of digits, you can craft a regex that will do it, but there is so much to keep track of that you don’t want to do it in regex. If you need rounding, it’s much better to do it in a scripting language like PythonScript: see “Batch Decimal Rounding” discussion and a oneliner PythonScript in the “Rounding numbers pythonscript …” discussion for some solutions.
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