用於註冊用戶PHP錯誤

Question

即時得到這個錯誤在一個基本的寄存器腳本：

Warning: mysql_result() expects parameter 1 to be resource, boolean given in /Applications/XAMPP/xamppfiles/htdocs/func/user.func.php on line 23 

多數民衆贊成讓我錯誤的register.php的部分是：

<?php 

include('init.php'); // user.func.php is included in this file 
include('template/header.php'); 
?> 

<h3>Register</h3> 

<?php 

    // Typical $_POST stuff here, down the line the next line is where the error happenes. Also, $register_email below is equal to $_POST['register_email']; 

      if(user_exists($register_email)) { ***THIS FUNCTION IS WHERE THE PROBLEM IS. THE ACTUAL FUNCTION IS DEFINED BELOW*** 
       $errors[] = 'That email has already been registered'; 
      } 

從功能user.func.php這給我的錯誤是：

function user_exists($email) { 
    $email = mysql_real_escape_string($email); 
    $query = mysql_query("SELECT COUNT(user_id) FROM users WHERE email = '$email'"); 
    return (mysql_result($query, 0) == 1) ? true : false; // ***THIS LINE RIGHT HERE*** 
} 

任何想法可能會導致此錯誤。這是一個令人討厭的錯誤。這不是我第一次得到那個。

UPDATE 感謝您的答案，我試過每一個，我得到了完全相同的錯誤。下面是完整的register.php至今：

<?php 

include('init.php'); 
include('template/header.php'); 
?> 

<h3>Register</h3> 

<?php 

if(isset($_POST['register_email'], $_POST['register_name'], $_POST['register_password'])) { 
    $register_email = $_POST['register_email']; 
    $register_name = $_POST['register_name']; 
    $register_password = $_POST['register_password']; 

    $errors = array(); 

    if(empty($register_email) || empty($register_name) || empty($register_password)) { 
     $errors[] = 'All fields required'; 
    } else { 
     echo 'OK'; 
     } 
     if(filter_var($register_email, FILTER_VALIDATE_EMAIL) == false) { 
      $errors[] = 'Email address is not valid'; 
     } 
     if(strlen($register_email) > 255 || strlen($register_name) > 35 || strlen($register_password) > 35) { 
      $errors[] = 'Ayo, quit tampering with the html'; 
     } 
     if(user_exists($register_email)) { 
      $errors[] = 'That email has already been registered'; 
     } 
} 

if(!empty($errors)) { 
    foreach($errors as $error) { 
     echo $error.'<br />'; 
    } 
} else { 

    } 

?> 

Answer 1

現在，我必須說，第一，我不是一個MySQL專家，我通常使用DB類（你也應該這樣。）但是，如果你是在說return (mysql_result($query, 0) == 1) ? true : false;行會給你一個錯誤。這意味着上面的行不起作用。這意味着它沒有返回資源。

你應該先調試功能..

function user_exists ($email) { 
    $email = mysql_real_escape_string($email); 
    if (!mysql_select_db("users")) { 
     echo 'Could not select "users" DB.<br />Error: ' . mysql_error(); 
    } 
    $query = mysql_query("SELECT COUNT(user_id) AS `count` FROM `users` WHERE `email` = '$email'"); 
    echo 'The count is currently: '$query['count']; 
    // return (mysql_result($query, 0) == 1) ? true : false; 
} 

如果它說，它無法選擇的用戶數據庫。那麼問題在於你的聯繫。正如我所說，我不是專業人士。但是，你應該將它連接這樣的：

$conn = mysql_connect('localhost', 'mysqluser', 'mypass'); 

現在你可以試試這個：

function user_exists ($email) { 
    global $conn; 
    $email = mysql_real_escape_string($email); 
    if (!mysql_ping($conn)) { 
     echo 'Could not ping the mysql. Connection is lost probably :('; 
    } 
    $query = mysql_query("SELECT COUNT(user_id) AS `count` FROM `users` WHERE `email` = '$email'", $conn); 
    echo 'The count is currently: ' . mysql_result($query, 0); 
    // return (mysql_result($query, 0) == 1) ? true : false; 
} 

如果代碼經過調試和連接是真棒！然後：

function user_exists ($email) { 
    global $conn; 
    if ($email) { 
     $query = mysql_query("SELECT COUNT(user_id) AS `count` FROM `users` WHERE `email` = '$email'", $conn); 
     if (mysql_result($query, 0)) { 
      return true; 
     } 
    } 
    return false; 
} 

或者：

function user_exists ($email) { 
    global $conn; 
    if ($email) { 
     $query = mysql_query("SELECT COUNT(user_id) AS `count` FROM `users` WHERE `email` = '$email'", $conn); 
     if ($result = mysql_fetch_array($query)) { 
      if ($result['count'] == 0) { 
       return true; 
      } 
     } 
    } 
    return false; 
} 

Answer 2

更改爲：

function user_exists($email) { 
    $email = mysql_real_escape_string($email); 
    $query = mysql_query("SELECT email FROM users WHERE email = '$email'"); 
    if (false === $query) return false; 
    return (mysql_num_rows($query) == 1); 
} 

Answer 3

如果你看看in the manual，mysql_query()可以返回的ressource（這是你所期望的）ORFALSE如果發生一個錯誤。

對於SELECT，SHOW，DESCRIBE，EXPLAIN和其他返回resultset的語句，mysql_query（）返回成功的資源，錯誤時返回FALSE。 對於其他類型的SQL語句，INSERT，UPDATE，DELETE，DROP等，mysql_query（）在成功時返回TRUE或在錯誤時返回FALSE。

Answer 4

使用

function user_exists($email) { 

    if(isset($email){ 
    $email = mysql_real_escape_string($email); 
    $query = mysql_query("SELECT COUNT(user_id) FROM users WHERE email = '$email'"); 
    $result = mysql_result($query,0); 
    if($result ===false) { 
    //error occur with the sql statement 
    //handel the error 
    } 
    else 
    return ($result == 1) ? true : false; // ***THIS LINE RIGHT HERE*** 
} 
} 

Answer 5

function user_exists($email) { 
    $email = mysql_real_escape_string($email); 
    $query = mysql_query("SELECT COUNT(user_id) FROM users WHERE email = '$email'"); 
    //return (mysql_result($query, 0) == 1) ? true : false; // ***THIS LINE RIGHT HERE*** 
    if($query) return (mysql_result($query, 0) != "") ? true : false; 
}