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有人可以告诉我什么时候你prepare()有一个语句（在我的例子中是一个更新语句）然后execute()它，然后我rowCount()在更新查询之后检查使用（见下面的代码），rowCount() > 0如果它是然后找到匹配并且更新发生了，但是我然后得到一个 else 声明。

为了确保我不会对语句中的语法错误和条件感到困惑，我想在下面的代码中询问（已详细评论特定区域） else 语句是否基本上意味着无法更新，因为找不到匹配项和/或可能的语法错误或其他错误？我认为这意味着我加粗的只是想确保我不会让自己感到困惑。

请忽略准备语句中的 SQL UPDATE 语法本身，因为它是错误的，稍后会处理。我认为代码解释得更好，并在我正在谈论的区域中提供了详细的评论。

// check if key is set and alphanumeric and equals 40 chars long
// we use sha1 so it will always be 40 chars long.
if(isset($_GET['key']) && ctype_alnum($_GET['key']) && strlen($_GET['key']) == 40){
$key = trim($_GET['key']);
}

// if key isset and valid
if(isset($key)){


try {
    // connect to database
    $dbh = sql_con();

    // checke if activation key matches and user_uid matches
    $stmt = $dbh->prepare("
            SELECT
              users_status.user_uid,
              users_status.user_activation_key
            FROM
              users_status
            JOIN
              users
            ON
              users_status.user_activation_key = ?
            AND
              users_status.user_uid = users.user_uid LIMIT 1");

    // execute query
    $stmt->execute(array($key));

    // if row count greater than 0 then match found
    if ( $stmt->rowCount() > 0 ) {

        // user verified; we now must update users status in users table to active = 1
        // and set the user_activation_key in the users_status to NULL
        $stmt = $dbh->prepare("
            UPDATE
              users.user_status,
              users_status.user_activation_key
            SET
              user_status = ".USER_STATUS_ACTIVE.",
              user_activation_key = NULL
            JOIN
              users
            ON
              users_status.user_activation_key = ?
            AND
              users_status.user_uid = users.user_uid LIMIT 1");

        // execute query
        $stmt->execute(array($key));

        if ( $stmt->rowCount() > 0 ) {

            echo 'account now activated';
            exit;

        } else {
            // update not sucessful
            // THIS IS THE BIT IM CONFUSED WITH;
            // IF RETURNED RESULT IS 0 (WHICH IT WILL BE IF I GET HERE WHEN RUNNING SCRIPT)
            // THEN I GUESS THAT MEANS THERE WAS NOT AN ERROR IN SQL SYNTAX BUT
            // CONDITION IN SQL STATEMENT COULD NOT BE MATCHED ? IS THAT CORRECT WHAT I AM THINKING ?
            // IF I AM CORRECT THEN OBVIOUSLY I WILL DISPLAY A MESSAGE TO USER AND EXIT HERE;
            // AS IF I AM THINKING RITE ANY SYNTAX ERROR WOULD BE CAUGHT BY CATCH BLOCK AND THIS ELSE STATEMENT
            // MEANS COULD NOT UPDATE BECAUSE NO MATCH IN UPDATE QUERY COULD BE FOUND ?
        }


    } // else no match found
    else {

        // no match found invalid key
        echo '<h1>Invalid Activation Link</h1>';

        $SiteErrorMessages =
        "Oops! Your account could not be activated. Please recheck the link in your email.
        The activation link could not be found or the account has already been activated.";

        SiteErrorMessages();

        include($footer_inc);
        exit;

    }

    // close database connection
    $dbh = null;

} // if any errors found log them and display friendly message
catch (PDOException $e) {
    ExceptionErrorHandler($e);
    require_once($footer_inc);
    exit;
}

} else {

// else key not valid or set
echo '<h1>Invalid Activation Link</h1>';

$SiteErrorMessages =
"Oops! Your account could not be activated. Please recheck the link in your email.
The activation link appears to be invalid.<br /><br />
If the problem persists please request a new one <a href='/member/resend-activation-email'>here</a>.";

SiteErrorMessages();

include($footer_inc);
exit;

}
4

2 回答 2

1

如果您要进行多次更新，那么您应该真正使用事务。

但请注意，默认 MyISAM 引擎不支持事务，因此ALTER TABLE tbl_name ENGINE=InnoDB如果您希望它工作，则需要：

$success = false;
$dbh->beginTransaction();
# perform your first query
if ($query->rowCount() == 1) {
   # something was updated/inserted/deleted
   # perform second query
   if ($query->rowCount() == 1) {
       $success = true;
   }
}

if ($success) $dbh->commit();
else $dbh->rollBack();

至于你的问题，你可能需要围绕你的 ? 使用单引号，因此将您的声明更改为：

users_status.user_activation_key = '?';

您可能没有得到结果的另一个原因是，如果您的 $key 是一个整数并且您使用该PreparedStatement::execute($array)方法绑定您的参数，您需要确保将值转换为正确的类型以使其工作，例如：

$query->execute(array((int)$key));

否则只需使用$query->bindParam($key)

于 2012-04-15T10:25:08.047 回答
1

你说的对：

if $stmt->rowCount() == 0

那么这意味着没有行被更新。

如果您在执行查询时遇到 sql 错误或错误，您会在执行时收到 FALSE 返回值或 PDO EXCEPTION

  execute(array($key));
于 2012-04-15T10:13:34.323 回答