SQL - How do I make an auto computing column?





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I've created a table named House, which has the column 'area'.
I have another table named HouseFire, with the columns '%_Burnt' and 'area_burnt'.
I have values for both 'area' and '%_Burnt'. I want to make 'area_burnt' to automatically give me the value as soon as I input the % value in '%_burnt'.
I'm using heidisql






mysql sql mariadb auto-update heidisql







share|improve this question




















  • 2





    You probably need a VIEW: w3schools.com/sql/sql_view.asp

    – UjinT34
    Nov 23 '18 at 15:51








  • 1





    Tag your question with the database you are using.

    – Gordon Linoff
    Nov 23 '18 at 16:09











  • I doubt that it's even possible to create an auto computed column based on a field from another table.

    – LukStorms
    Nov 23 '18 at 16:12













  • This seems to be a UI problem, not a database problem. Adjust the tags accordingly.

    – Rick James
    Nov 24 '18 at 23:04




















0















I've created a table named House, which has the column 'area'.
I have another table named HouseFire, with the columns '%_Burnt' and 'area_burnt'.
I have values for both 'area' and '%_Burnt'. I want to make 'area_burnt' to automatically give me the value as soon as I input the % value in '%_burnt'.
I'm using heidisql






mysql sql mariadb auto-update heidisql







share|improve this question




















  • 2





    You probably need a VIEW: w3schools.com/sql/sql_view.asp

    – UjinT34
    Nov 23 '18 at 15:51








  • 1





    Tag your question with the database you are using.

    – Gordon Linoff
    Nov 23 '18 at 16:09











  • I doubt that it's even possible to create an auto computed column based on a field from another table.

    – LukStorms
    Nov 23 '18 at 16:12













  • This seems to be a UI problem, not a database problem. Adjust the tags accordingly.

    – Rick James
    Nov 24 '18 at 23:04
















0












0








0








I've created a table named House, which has the column 'area'.
I have another table named HouseFire, with the columns '%_Burnt' and 'area_burnt'.
I have values for both 'area' and '%_Burnt'. I want to make 'area_burnt' to automatically give me the value as soon as I input the % value in '%_burnt'.
I'm using heidisql






mysql sql mariadb auto-update heidisql







share|improve this question
















I've created a table named House, which has the column 'area'.
I have another table named HouseFire, with the columns '%_Burnt' and 'area_burnt'.
I have values for both 'area' and '%_Burnt'. I want to make 'area_burnt' to automatically give me the value as soon as I input the % value in '%_burnt'.
I'm using heidisql






mysql sql mariadb auto-update heidisql




mysql sql mariadb auto-update heidisql






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 24 '18 at 0:01







João Nunes

















asked Nov 23 '18 at 15:44









João NunesJoão Nunes

11




11








  • 2





    You probably need a VIEW: w3schools.com/sql/sql_view.asp

    – UjinT34
    Nov 23 '18 at 15:51








  • 1





    Tag your question with the database you are using.

    – Gordon Linoff
    Nov 23 '18 at 16:09











  • I doubt that it's even possible to create an auto computed column based on a field from another table.

    – LukStorms
    Nov 23 '18 at 16:12













  • This seems to be a UI problem, not a database problem. Adjust the tags accordingly.

    – Rick James
    Nov 24 '18 at 23:04
















  • 2





    You probably need a VIEW: w3schools.com/sql/sql_view.asp

    – UjinT34
    Nov 23 '18 at 15:51








  • 1





    Tag your question with the database you are using.

    – Gordon Linoff
    Nov 23 '18 at 16:09











  • I doubt that it's even possible to create an auto computed column based on a field from another table.

    – LukStorms
    Nov 23 '18 at 16:12













  • This seems to be a UI problem, not a database problem. Adjust the tags accordingly.

    – Rick James
    Nov 24 '18 at 23:04










2




2





You probably need a VIEW: w3schools.com/sql/sql_view.asp

– UjinT34
Nov 23 '18 at 15:51







You probably need a VIEW: w3schools.com/sql/sql_view.asp

– UjinT34
Nov 23 '18 at 15:51






1




1





Tag your question with the database you are using.

– Gordon Linoff
Nov 23 '18 at 16:09





Tag your question with the database you are using.

– Gordon Linoff
Nov 23 '18 at 16:09













I doubt that it's even possible to create an auto computed column based on a field from another table.

– LukStorms
Nov 23 '18 at 16:12







I doubt that it's even possible to create an auto computed column based on a field from another table.

– LukStorms
Nov 23 '18 at 16:12















This seems to be a UI problem, not a database problem. Adjust the tags accordingly.

– Rick James
Nov 24 '18 at 23:04







This seems to be a UI problem, not a database problem. Adjust the tags accordingly.

– Rick James
Nov 24 '18 at 23:04














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