Question

当你prepare()一个声明（在我的情况下是一个更新声明）然后execute()时，有人能告诉我，然后在更新查询后使用rowCount()检查（参见下面的代码））如果rowCount() > 0如果是，则找到匹配并进行更新，但我得到了一个else语句。

为了确保我不会在语句中混淆语法错误和条件，我想在下面的代码中询问（详细说明具体区域）else语句是否基本上意味着无法更新，因为匹配是未找到和/或可能的语法错误或其他一些错误？我认为这意味着我大胆地想要确保我不会让自己感到困惑。

请忽略prepare语句中的SQL UPDATE语法本身，因为它错了，稍后会处理。我认为代码更好地解释了我正在谈论的领域中的详细评论。

// 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;

}

Answer 1

你是对的：

if $stmt->rowCount() == 0

然后就意味着没有更新任何行。

如果您在执行查询时遇到sql错误或错误，则在执行时会收到FALSE返回值或PDO EXCEPTION

  execute(array($key));

Answer 2

如果您正在进行多次更新，则应该真正使用交易。

请注意，默认的MyISAM引擎不支持事务，因此如果您希望这样做，则需要ALTER TABLE tbl_name ENGINE=InnoDB：

$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，你可能需要包围你的？使用单引号，所以将您的陈述更改为：

users_status.user_activation_key = '?';

您可能无法获得结果的另一个原因是，如果您的$ key是一个整数并且您使用PreparedStatement::execute($array)方法绑定参数，则需要确保将值转换为适当的类型它起作用，例如：

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

否则只需使用$query->bindParam($key)

PHP PDO，需要确保我理解我编码的内容？

2 个答案: