mysqli致命错误：查询/准备语句中未使用索引

Question

我想要使用mysqli执行一个简单的准备好的语句，但它不会工作。

CREATE TABLE IF NOT EXISTS `account` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `email` varchar(100) COLLATE latin1_german2_ci NOT NULL,
  `password` varchar(100) COLLATE latin1_german2_ci NOT NULL,
  `created` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 COLLATE=latin1_german2_ci AUTO_INCREMENT=4 ;

$mysqli = new mysqli($server,$user,$pass,$db);

if(mysqli_connect_errno()) {
    echo "Connection Failed: " . mysqli_connect_errno();
    exit();
 }
 $user = "[e-mail]";
 $pass = "dada";
/* Create a prepared statement */
if($stmt = $mysqli -> prepare("SELECT * FROM account WHERE email=?
AND password=?")) {

  /* Bind parameters
     s - string, b - blob, i - int, etc */
  $stmt -> bind_param("ss", $user, $pass);

  /* Execute it */
  $stmt -> execute();

  /* Bind results */
  $stmt -> bind_result($result);

  /* Fetch the value */
  $stmt -> fetch();

  echo $user . "id of user is " . $result;

  /* Close statement */
  $stmt -> close();
}

/* Close connection */
$mysqli -> close();

致命错误：未捕获异常“mysqli_sql_exception”，消息为“查询/准备好的语句中没有使用索引SELECT*FROM ACCOUNCT WHERE EMAIL=？和password=？‘mysqli_stmt->execute()#1{main}

Answer 1

我认为你必须这么做：

CREATE TABLE IF NOT EXISTS `account` (
  `id` PRIMARY KEY int(11) NOT NULL AUTO_INCREMENT,
   // the rest

上面的代码将表的id字段设置为主键,这样它就不会重复自己,它仍然是表的索引。