Guest Sarah Posted June 10 Share Posted June 10 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 post Share on other sites
Newbies Al Quimby 1 Posted June 13 Newbies Share Posted June 13 (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 by Al Quimby Link to post Share on other sites
Newbies VolAnd 0 Posted June 23 Newbies Share Posted June 23 Quote Replace ( yourSerialNumberField; Length(yourSerialNumberField)-2; 0 ; " ") Link to post Share on other sites
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