Guest Sarah Posted June 10, 2020 Share Posted June 10, 2020 I am carrying out a project to finalize a project for the college which consists of the following. An automatic serial number of two groups with numbers with a space in the middle of the third digit (ex: 159 023) ex: 159 023 159 024 159 025 Thanks for the help. Link to comment Share on other sites More sharing options...
Newbies Al Quimby Posted June 13, 2020 Newbies Share Posted June 13, 2020 (edited) If you are starting this database from scratch, set your Primary Key (type number) to a serial number (on creation) starting at 159 000 (or whatever starting number you need). From then on each new serial number will have that space between the first 3 digits and the last 3 digits. If you already have many records, create a calculation (number) field with this definition: Left ( yourSerialNumberField ; 3 ) & " " & Right (yourSerialNumberField ; 3 ) You can use this second field with a space between the first 3 and last 3 for your project. Edited June 14, 2020 by Al Quimby Link to comment Share on other sites More sharing options...
Newbies VolAnd Posted June 23, 2020 Newbies Share Posted June 23, 2020 Quote Replace ( yourSerialNumberField; Length(yourSerialNumberField)-2; 0 ; " ") Link to comment Share on other sites More sharing options...
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