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Home/ Questions/Q 7563845
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Editorial Team
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Editorial Team
Asked: May 30, 20262026-05-30T13:46:06+00:00 2026-05-30T13:46:06+00:00

So imagine 2 big mysql tables imported from Excel. Table1 has the field SNameID,

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So imagine 2 big mysql tables imported from Excel.

Table1 has the field SNameID, and other fields describing SNameID (Age, Size).

SNameID Age Size

Table2 has the field TeacherID, and for each teacher there is/are student/s SNameID and the describing fields (Age, Size).

TeacherID SNameID Age Size

The idea is: the Table2 has the TeacherID data and the relationship information: this are his students. It’s quite a bad architecture … but I have to use both tables as they are further.

Please tell me, when some change occurs in the Age and/or Size in any of the table, can I make mysql automatically update the other table according to the SNameID?

I just can’t figure it out … and its the problem in many of my tables … All I need is a hint where or what to read – I can figure it our afterwards and will post a solution here as well.

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  1. Editorial Team
    Editorial Team
    2026-05-30T13:46:07+00:00Added an answer on May 30, 2026 at 1:46 pm

    You can setup a TRIGGER AFTER UPDATE to automatically sync changes in the data:

    DELIMITER |
    DROP TRIGGER IF EXISTS `teacher_update`|
    CREATE TRIGGER `teacher_update` AFTER UPDATE ON `teacher`
    FOR EACH ROW
    BEGIN
      UPDATE student s SET s.Age = NEW.Age, s.Size = NEW.Size WHERE s.SNameID = NEW.SNameID;
    END;
    |
    DROP TRIGGER IF EXISTS `student_update`|
    CREATE TRIGGER `student_update` AFTER UPDATE ON `student`
    FOR EACH ROW
    BEGIN
      UPDATE teacher t SET t.Age = NEW.Age, t.Size = NEW.Size WHERE t.SNameID = NEW.SNameID;
    END;
    |
    delimiter;
    

    However, what you REALLY should do is normalize this database design.

    Drop the Age and Size fields from the teacher table and when you SELECT, fetch the information from the student table using the foreign key.

    SELECT t.*, s.* FROM teacher t LEFT JOIN student s ON t.SNameID = s.SNameID
    
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