php - 如何在PHP中动态创建具有多个字段的表

Question

这个问题已经有答案了:

How can I fix MySQL error #1064? (3 个回答)

已关闭 8 个月前。

$dbhost = 'localhost';
    $dbuser = 'root';
    $dbpass = '';
    $conn = mysql_connect($dbhost, $dbuser, $dbpass);
    if(! $conn )
    {
      die('Could not connect: ' . mysql_error());
    }

    echo 'Connected successfully<br />';



$name_of_field="id_student";


function field_dynamic($field_num,$field_name){

    for($i=1;$i<=$field_num;$i++){

        if($i!=$field_num){

        $field[$i]="".$field_name."_".$i." VARCHAR(5) NOT NULL,"."";

        }else{

        $field[$i]="".$field_name."_".$i." VARCHAR(5) NOT NULL";

        }

    }
    return $field;
}

$num=13;

$sql_mathematic = "CREATE TABLE mathematic(". 

            field_dynamic($num,$name_of_field).")";

        mysql_select_db("adrian");
        $retval_mathematic = mysql_query( $sql_mathematic, $conn );

        if(! $retval_mathematic )
            {
              die('Could not create table: ' . mysql_error());
            }

            echo "Table created successfully\n";

它不工作:

Connected successfully
Could not create table: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 1

请帮助我。

Answer 1

这可以实现您正在谈论的内容，但您不应该使用它。

for($i=1;$i<=$field_num;$i++){
    $field[$i]="".$field_name."_".$i." VARCHAR(5) NOT NULL";

}
return implode(',',$field);