Question

由于我远离php的专家，这让我惊呆了，我似乎无法为它编写脚本。

让我们看看能否尽可能清楚地解释这一点。

假设我有table1和table2

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

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

当从table1复制到表2时，我想在它前面添加2个字段。（id和tournamentid）

我面临的问题是Table1每次都有不同数量的roundxpos / roundxscore。

基本上我想做的就是一旦锦标赛结束，我想删除这张桌子。但将其中的数据复制到另一个表中以存档结果。但每场锦标赛都有不同规模的比赛。

我希望有人能理解我想要实现的目标+ _ +。

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引用它们。

rounds: team_id, tournament_id, no, pos, score

Answer 2

你应该规范你的桌子，但同时...... 我假设主要问题是在代码中处理可变数量的roundxpos和roundxscore列。

希望这会有所帮助：

获取此查询的结果：从表1显示列字段LIKE'round％pos'
获取此查询的结果：从表1显示列字段LIKE'round％score'

通过循环字段名称

创建table2插入

//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 .= ")";

将一个表复制到另一个具有不同列数的表

2 个答案: