在php登錄腳本錯誤sql查詢

Question

我想爲我的網站創建一個基本的登錄腳本，我卡住了。我的代碼如下：

<?php 
$myServer = "localhost"; 
$myUser = "root"; 
$myPass = "*******"; 
$myDB = "social_bookmarking"; 

//connection to the database 
libxml_use_internal_errors(true); 
$connect = mysqli_connect($myServer,$myUser, $myPass) 
or die("Couldn't connect to SQLServer on $myServer"); 

//select a database to work with 
$selected = mysqli_select_db($connect, $myDB) 
or die("Couldn't open database $myDB"); 

$email = $_POST['email']; 
var_dump($email); 
$password = $_POST['passwd']; 
var_dump($password); 
$query = mysqli_query($connect, "SELECT * FROM users WHERE email='$email'"); 
var_dump($query); 
if(!$query) 
{ 
    die("Query failed:"); 
} 
else 
{ 
    $row = mysqli_fetch_array($query, MYSQLI_NUM); 
    var_dump($row); 
    if($email == $row['email']) 
    { 
     if($email=='' || $password == '') 
     { 
      header("Location: index.php?id=Some fields are empty"); 
     } 
     else if ($email==$row['email'] && $password =$row['password']) { 
      # code... 
      header("Location: main.php?id=$email"); 
     } 
    } 
    else 
    { 
     //mysqli_query("alter table users auto_increment = 1"); 

    } 
} 
?> 

這裏是我的代碼返回：

string 'palade.radupaul@live.com' (length=24) 
string '*********' (length=10) 
object(mysqli_result)[2] 
public 'current_field' => null 
public 'field_count' => null 
public 'lengths' => null 
public 'num_rows' => null 
public 'type' => null 

array (size=5) 
    0 => string '1' (length=1) 
    1 => string 'Palade Radu' (length=11) 
    2 => string 'pa10der4du' (length=10) 
    3 => string 'Radu' (length=4) 
    4 => string 'palade.radupaul@live.com' (length=24) 

這是我的表：

http://i.stack.imgur.com/nDaBX.png

我不知道爲什麼都那些變量'null'。現在我得到var_dump ['row']下的未定義索引'email'。我在哪裏得到全部錯誤？

Answer 1

試試這個修改後的代碼

$myServer = "localhost"; 
$myUser = "root"; 
$myPass = "*******"; 
$myDB = "social_bookmarking"; 

//connection to the database 
libxml_use_internal_errors(true); 
$connect = mysqli_connect($myServer,$myUser, $myPass) 
or die("Couldn't connect to SQLServer on $myServer"); 

//select a database to work with 
$selected = mysqli_select_db($connect, $myDB) 
or die("Couldn't open database $myDB"); 

$email = trim($_POST['email']); 
$password = trim($_POST['passwd']); 

if($email=='' || $password == '') 
{ 
    header("Location: index.php?id=Some fields are empty"); 
    exit; 
} 
else 
{  
    $email = mysql_real_escape_string(stripslashes($email)); 
    $password = mysql_real_escape_string(stripslashes($password)); 

    $query = mysqli_query($connect, "SELECT * FROM users WHERE email='$email' AND password='$password' LIMIT 1 "); 
    if(!$query) 
    { 
     die("Query failed:"); 
    } 
    else 
    { 
     if($row = mysqli_fetch_array($query)) 
     { 
      var_dump($row); 
      if($email == $row['user']) 
      { 
        # code... 
        header("Location: main.php?id=$email"); 
       exit; 
      } 
      else 
      { 
       //mysqli_query("alter table users auto_increment = 1"); 

      } 
     } 
     else 
     { 
      header("Location: index.php?id=Invalid Email ID or Password "); 
      exit; 
     }  
    } 
} 

EDIT 2

$myServer = "localhost"; 
$myUser = "root"; 
$myPass = "*******"; 
$myDB = "social_bookmarking"; 

//connection to the database 
libxml_use_internal_errors(true); 
$connect = mysqli_connect($myServer,$myUser, $myPass) 
or die("Couldn't connect to SQLServer on $myServer"); 

//select a database to work with 
$selected = mysqli_select_db($connect, $myDB) 
or die("Couldn't open database $myDB"); 

$email = trim($_POST['email']); 
$password = trim($_POST['passwd']); 

if($email=='' || $password == '') 
{ 
    header("Location: index.php?id=Some fields are empty"); 
    exit; 
} 
else 
{  
    $email = mysql_real_escape_string(stripslashes($email)); 
    $password = mysql_real_escape_string(stripslashes($password)); 

    $query = mysqli_query($connect, "SELECT * FROM users WHERE email='$email' AND password='$password' LIMIT 1 "); 
    if(!$query) 
    { 
     die("Query failed:"); 
    } 
    else 
    { 
     if($row = mysqli_fetch_array($query)) 
     { 
      # code... 
      header("Location: main.php?id=$email"); 
      exit; 
     } 
     else 
     { 
      header("Location: index.php?id=Invalid Email ID or Password "); 
      exit; 
     }  
    } 
} 

現在就來試試上面的代碼，看看發生了什麼。