登錄過程解析錯誤：語法錯誤

好傢伙，請幫我在登錄過程中的代碼有一致的錯誤消息登錄過程解析錯誤：語法錯誤

Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING)

$_SESSION['userid'] = $row['userid'];

這裏是我的loginprocess的完整代碼：

session_start(); 
$message = ""; 

if(count($_POST) > 0){ 
    $conn = mysql_connect("localhost", "root", ""); 
    mysql_select_db("etransmittal", $conn); 
    $result = mysql_query("SELECT * FROM tbl_userlist WHERE username = '" . $_POST["username"] . "' AND user_password = '" . $_POST["password"] . "'); 
    $row = mysql_fetch_array($result); 

    if(is_array($row)){ 
     $_SESSION['userid'] = $row['userid']; 
     $_SESSION['username'] = $row['username']; 
    } 
    else{ 
     $message = "Invalid username or password"; 
    } 
} 

if(isset($_SESSION['userid'])){ 
    header("location: userhomepage.php"); 
}

來源

2015-12-02 pvegetah

在您的sql語句的最後加上雙引號 –

您是否擁有語法突出顯示的編輯器？即使堆棧溢出代碼格式化程序清楚地顯示錯誤。 –

順便說一句，如果您輸入「'或1 = 1 - '作爲用戶，則不需要密碼。它是一個錯誤還是一個功能？ ;-P –

您有一個雙引號缺失，我已修復它：

$result = mysql_query("SELECT * FROM tbl_userlist WHERE username = '" . $_POST["username"] . "' AND user_password = '" . $_POST["password"] . "'");

來源

2015-12-02 07:11:31 thepiyush13

<?php 
    if($_POST['submit_button']=='Login' ) { 

         $userdata = "select * from usertbl WHERE username = '".$_POST['username']."' "; 
         $res = mysql_query($userdata); 
         $userdata = mysql_fetch_array($res); 
         $userDetail = $userdata[0]; 

         if( ($userDetail['username'] == $_POST['username']) && ($userDetail['password'] == md5($_POST['password']))){ 

           $_SESSION['userid'] = $userDetail['id']; 
           $_SESSION['username'] = $userDetail['username']; 
           $_SESSION['useremail'] = $userDetail['email']; 


           header("Location:Dashboard.php"); 

         } else { 
           header("Location:index.php"); 
         } 
       } 
?> 

<form action="" method="POST"> 
    <input type="text" name="username" > 
    <input type="text" name="password" > 
    <input type="submit" name="submit_button" value="login"> 
</form>

來源

2015-12-02 09:22:56

登錄過程解析錯誤：語法錯誤

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