Question

大家好我正在尝试应用此查询但值不会保存在我的数据库中。我已经尝试了很多，但我无法解决它。请帮我。提前致谢

if($_POST) {
    $lecturer_id = $_SESSION["lecturer_id"];
    $game_name= $_POST['game_name'];
    mysql_query("insert into games(game_name, lecturer_id) values ('$game_name', '$lecturer_id')");
    $input_type = $_POST['type'];
    if($input_type=='multiple_choice') {
        $question= $_POST['quest'];
        $val1= $_POST['value1'];
        $val2= $_POST['value2'];
        $val3= $_POST['value3'];
        $val4= $_POST['value4'];
        //echo $question;
        // echo $val4;
        mysql_query("insert into subgames(game_id, input_id, statement, option1,    option2, option3, option4) values ((SELECT id 
        FROM games
        WHERE game_name = '$game_name'), '$input_type', '$question', '$val1', '$val2', '$val3', '$val4'");
    } elseif($input_type=='input_field') {
        $question= $_POST['quest'];
        $answer= $_POST['ans'];
    }
} else {
echo "Not Submitted";
}

Answer 1

您是否尝试检查mysql_error()?

添加此代码：

echo mysql_error();

然后发布错误。

Answer 2

    Step 1:-
     SHOW CREATE TABLE table_name:

    It will show table details

    | table_name | CREATE TABLE `table_name` (
     `id` int(20) unsigned NOT NULL auto_increment,
     `key_column` smallint(5) unsigned default '1',
      KEY `key_column` (`key_column`),
      CONSTRAINT `table_name_ibfk_1` FOREIGN KEY (`key_column`) REFERENCES 
     `second_table` (`id`) ON DELETE SET NULL
      ) ENGINE=InnoDB DEFAULT CHARSET=utf8 |

     Step2:-
     ALTER TABLE table_name DROP KEY `key_column`;

     Step3:-
     ALTER TABLE table_name DROP FOREIGN KEY `table_name_ibfk_1`;

SQL查询不会将数据发送到数据库

2 个答案: