For what I know you cannot do that from the TM settings, but it should be quite simple using regex. As my regex knowledge is not very much advanced, I would go for:

search expression: (\d{1,3})?(\d{3}) to (\d{1,3})?(\d{3}) currency

replace expression $1.$2 to $3.$4 currency

This covers numbers from 1000 to 999999, however creates a problem with numbers below 1000. So depending on what you have more than one approach will be needed.

But in case you have only this in your lines, the solution could be a combination of filtering and replacing. To make a better proposal I need more examples.

Hi Jerzy Czopik,

am still working on this one … so far I managed to get the regex refined so it would work with any numbers (lower than 1000 and higher than 999999) but it has the backdraw that I can't use back references to further replace the currency after the number (say add a space before the currency if the source string does not contain it):

String: 100 to 10000000010 USD

Regex: ([0-9](?=(?:[0-9]{3})+(?![0-9])))

Backreference: $&.               (whereas the dot after & can be replaced by the character you need as thousand separator e.g. dot, comma, space, non-breaking space, narrow non-breaking space) BTW $0. or $1. as backreference will also work

will give:

100 to 10.000.000.010 USD

Hi Jerzy Czopik,

am still working on this one … so far I managed to get the regex refined so it would work with any numbers (lower than 1000 and higher than 999999) but it has the backdraw that I can't use back references to further replace the currency after the number (say add a space before the currency if the source string does not contain it):

String: 100 to 10000000010 USD

Regex: ([0-9](?=(?:[0-9]{3})+(?![0-9])))

Backreference: $&.               (whereas the dot after & can be replaced by the character you need as thousand separator e.g. dot, comma, space, non-breaking space, narrow non-breaking space) BTW $0. or $1. as backreference will also work

will give:

100 to 10.000.000.010 USD

Anthony Rudd may know 

Sorry, this is beyond my knowledge.

Because regex has limited flexibility for replacement, it may be necessary to write multiple regexes to handle various number "sizes".

To avoid programming multiple regexes to handle the various formats, it may be better to program a more general regex with optional fields, and then “clean up” the results, as shown below.

General regex: \b(\d{1,3})(,(\d{3}))?(,(\d{3}))?(\.(\d+))?\b
Replace string: $1.$3.$5,$7
Test string: 1,234,567.89 12,345.6 12
Result string: 1.234.567,89  12.345.,6  12..,

 Whereby, the erroneous separators “.,” and “..,” can be cleaned up with the following Find regex:

(?<ho>(\d*\.\d+)*)(\.+(?<lo>,\d+)|\.+,(?<lo>))
and Replace string:
${ho}${lo}

Test string: 1.234.567,89  12.345.,6  12..,
Result string: 1.234.567,89  12.345,6  12

The “General regex” can be extended using the same scheme to handle additional leading “thousand blocks”.

Because my previous reply was considered to be spam, I am sending the content as screenshot.

Because regex replacement is very inflexible, multiple regexes may need to be coded to process different number "sizes".

Anthony Rudd

Anthony Rudd said:

Because my previous reply was considered to be spam, I am sending the content as screenshot.

Enter the info as code instead... just a tip:

PS: wish I knew as much about regex as you do!!!