Copy one table to another with different colum count

Question

As i am far from an expert in php this got me stunned and i can't seem to make a script for it.

Lets see if i can explain this as clearly as possible.

Lets say i have table1 and table2

Table1 = (teamid, name, round1pos, round1score, round2pos, round2score)

Table2 = (id, tournamentid, teamid, name, round1pos, round1score, round2pos, round2score...till round10pos/round10score)

When copied from table1 to table 2 i want to add 2 fields in front of it. (id and tournamentid)

The problem i am facing is that Table1 has a different amount of roundxpos/roundxscore each time.

Basically what i want to do is once a tournament is over i want to delete the table. but copy the data inside it to another table to archive the results. but each tournament can have a different size of rounds.

I hope someone can understand what i am trying to achieve +_+.

Answer 1

您应该创建第三个表，从现有表中删除roundX列，并使用其ID对其进行引用。

rounds: team_id, tournament_id, no, pos, score

Answer 2

You should normalize your tables but in the meantime.... I'm assuming the main problem is handling the variable number of roundxpos and roundxscore columns in the code.

Hope this helps:

• Get the results of this query: SHOW COLUMNS FROM Table1 WHERE Field LIKE 'round%pos'
• Get the results of this query: SHOW COLUMNS FROM Table1 WHERE Field LIKE 'round%score'

• Create the table2 inserts by looping through field names

   //example, assuming results are fetched into an associative array $query = "INSERT INTO table2 (tournamentid, teamid, name)"; foreach($roundpos_fields as $f){ $query .= "," . $f['Field']; } foreach($roundscore_fields as $f){ $query .= "," . $f['Field']; } $query .= "( SELECT $tournamentid, '' teamid, name"; foreach($roundpos_fields as $f){ $query .= "," . $f['Field']; } foreach($roundscore_fields as $f){ $query .= "," . $f['Field']; } $query .= ")";