PHP PDO，需要确保我明白我的代码是什么？

Question

有人可以告诉我，当你prepare()声明（在我的情况下更新声明），然后​​它，然后我检查使用rowCount()后更新查询（见下面的代码）如果rowCount() > 0如果它是然后找到匹配和更新发生了，但是我得到了另一个声明。

为了确保我不会将自己与语句中的语法错误和条件混淆，我想问下我的代码（详细注释了具体的区域）else语句基本上意味着无法更新，因为匹配不是发现和/或可能的语法错误或一些其他错误？我认为这意味着我大胆表达的意思只是想确保我不会让自己感到困惑。

请在准备语句中忽略SQL UPDATE语法本身，因为它是错误的，将在稍后处理。我认为该代码解释更好，并在我所谈论的领域提供详细的评论。

// 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; 

} 

Answer 1

你是对的：

if $stmt->rowCount() == 0 

那么就意味着没有行已更新。

如果在执行查询您会收到一个假的返回值或者还当你执行

execute(array($key)); 

Answer 2

如果你正在做多个更新一个PDO EXCEPTION HEVE SQL错误或错误，你应该使用的交易。

不过请注意，交易不是在默认的MyISAM引擎支持，所以如果你想要这个工作，你需要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，你可能需要围绕你的？用单引号因此改变你的语句是：

users_status.user_activation_key = '?'; 

为什么你可能不会得到结果的另一个原因，是如果你的$键是整数并且使用PreparedStatement::execute($array)方法来绑定您的参数，您需要确保你投值到正确的类型，它的工作，例如：

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

否则只是用$query->bindParam($key)