Question

我有一个问题，当我执行我的功能时，表中没有数据存储。我的数据库连接正常工作，因此与数据库本身通信时没有错误。这是我的代码：

PHP：

//function for creating a bet
function CreateBet() {

  //make variables global instead of local
  global $con;
  global $CharTick;
  global $hash;
  global $created_by;

  //infinite loop, no <= or >=
  for($cycle=0; ; $cycle++) {
    //define result - select coinflip table, gameid column and for what number to look
    $result = $con->query("SELECT * FROM coinflip WHERE gameid = $cycle");
    //check if table 'coinflip' in 'gameid' contains an already existing number, looking
    if($result->num_rows == 0) {
      //if that number is not used, create it
      $sql = "INSERT INTO coinflip (gameid, hash_sha1, hash_original, created_by) VALUES ($cycle, $hash, $CharTick, $created_by)";
      $con->query($sql);
      //stop the for loop
      break;
    }
  }
}


//execute the function for creating a bet
CreateBet();

用于创建与数据库的连接的PHP代码：

//connect to mariadb
$con = @mysqli_connect('localhost', 'root', '', 'testing');

//check if connection is NOT alright
if (!$con) {
    echo "Error: " . mysqli_connect_error();
    exit();
}
//echo if the connection was alright
echo 'Connected to MariaDB';

SQL：

CREATE TABLE `coinflip` (
  `gameid` varchar(128) NOT NULL,
  `winner` varchar(128) NOT NULL,
  `ticket` varchar(128) NOT NULL,
  `hash_sha1` varchar(128) NOT NULL,
  `hash_original` varchar(128) NOT NULL,
  `created_by` varchar(128) NOT NULL,
  `joined_by` varchar(128) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;


INSERT INTO `coinflip` (`gameid`, `winner`, `ticket`, `hash_sha1`, `hash_original`, `created_by`,         `joined_by`) VALUES
('0', '', '', '', '', '', ''),
COMMIT;

谢谢：）

Answer 1

您缺少“引用”您的字符串值：

$con->query($sql);

您应该测试if (!$con->query($sql)) { echo "Error: " . $con->error; }的结果以查看错误：

gameid

同样gameid似乎是一个整数，因此您可以对$con->query("SELECT * FROM coinflip WHERE gameid = '$cycle'");列使用整数类型，或者也可以在以下位置使用引号：

$result->num_rows > 0

最后，您可以使用parameterized queries来避免此类问题和可能的SQL注入。

无法将数据推送到MariaDB数据库

1 个答案: