PHP：将MySQL数据库表复制到具有附加列的新数据库表中

Question

我正在寻找一种方法将具有多个表的现有数据库复制到具有相同表和列+一些其他列的新数据库中。到现在为止还挺好。如果我只是将数据库复制到具有相同数量的表和列的新数据库，我这样做：

+---------+---------+---------+
| TABLE 1 |         |         |
+---------+---------+---------+
| Col1    | Col2    | Col3    |
| Value 1 | Value 2 | Value 3 |
| Value 1 | Value 2 | Value 3 |
| Value 1 | Value 2 | Value 3 |
+---------+---------+---------+

复制到：

+---------+---------+---------+
| TABLE 2 |         |         |
+---------+---------+---------+
| Col1    | Col2    | Col3    |
| Value 1 | Value 2 | Value 3 |
| Value 1 | Value 2 | Value 3 |
| Value 1 | Value 2 | Value 3 |
+---------+---------+---------+

码：

public function loadDB($db1,$db2){
    $this->db->prepare("use ".$db1."");
    $sqlshow = "SHOW TABLES ";
    $statement = $this->db->prepare($sqlshow);
    $statement->execute();
    $tables = $statement->fetchAll(PDO::FETCH_NUM);

    foreach($tables as $table){
        $sql[] = "INSERT INTO ".$db2.".".$table[0]." SELECT * FROM ".$db1.".".$table[0]."; ";
    }
    $sqlState = implode(' ', $sql);
    $insertStatement = $this->db->exec($sqlState);
    return $insertStatement?$insertStatement:false;
}

此代码有效，我的数据库已成功复制，我的表中包含所有表和值。我现在需要的是一个工作示例，说明如何将数据库复制到新数据库，其中所有表都有四个额外的列，如下所示：

+---------+---------+---------+
| TABLE 1 |         |         |
+---------+---------+---------+
| Col1    | Col2    | Col3    |
| Value 1 | Value 2 | Value 3 |
| Value 1 | Value 2 | Value 3 |
| Value 1 | Value 2 | Value 3 |
+---------+---------+---------+

复制到：

+---------+---------+---------+-----------+-----------+-----------+-----------+
| TABLE 2 |         |         |           |           |           |           |
+---------+---------+---------+-----------+-----------+-----------+-----------+
| Col1    | Col2    | Col3    | Counter   | LoadDay   | User     | UserNew      |
| Value 1 | Value 2 | Value 3 | NEW VALUE | NEW VALUE | NEW VALUE | NEW VALUE |
| Value 1 | Value 2 | Value 3 | NEW VALUE | NEW VALUE | NEW VALUE | NEW VALUE |
| Value 1 | Value 2 | Value 3 | NEW VALUE | NEW VALUE | NEW VALUE | NEW VALUE |
+---------+---------+---------+-----------+-----------+-----------+-----------+

代码（到目前为止我尝试过的）：

public function loadDB($db1,$db2,$condition){
    $this->db->prepare("use ".$db1."");
    $sqlshow = "SHOW TABLES ";
    $statement = $this->db->prepare($sqlshow);
    $statement->execute();
    $tables = $statement->fetchAll(PDO::FETCH_NUM);

    foreach($tables as $table){
        $sqlshow2 = "SHOW COLUMNS FROM ".$table[0]." ";
        $statement = $this->db->prepare($sqlshow2);
        $statement->execute();
        $columns = $statement->fetchAll(PDO::FETCH_NUM);

        foreach($columns as $column){
            $sql[] = "INSERT INTO ".$db2.".".$table[0]." SELECT ".$column[0]." FROM ".$db1.".".$table[0]."; ";
        }
        $sql[] .= "INSERT INTO ".$db2.".".$table[0]." (`Counter`, `LoadDay`, `User`, `UserNew`) VALUES ('1', '".date("Y-m-d H:i:s")."', '".$condition."', '".$condition."')";
    }
    $sqlState = implode(' ', $sql);
    var_dump($sqlState);
    $insertStatement = $this->db->exec($sqlState);
    return $insertStatement?$insertStatement:false;
}

数据库的创建正在工作（在我发布的代码中不可见）。我只是没有将值复制到新数据库中的新表中。我在这做错了什么？

Answer 1

您的最终SQL查询出错了。我建议将您的代码更改为以下内容：

foreach($tables as $table){
    $sqlshow2 = "SHOW COLUMNS FROM ".$table[0]." ";
    $statement = $this->db->prepare($sqlshow2);
    $statement->execute();
    $columns = $statement->fetchAll(PDO::FETCH_NUM);

    $sql[] = "INSERT INTO ".$db2.".".$table[0]." SELECT * , '1', '".date("Y-m-d H:i:s")."', '".$condition."', '".$condition."'" . " FROM ".$db1.".".$table[0]."; ";

}