Question

我有一个名为checklog.php的单独文件，其中包含我的会话详细信息

<?php
        session_start();
        if (!isset($_SESSION['logged'])){
            $_SESSION = array();
            header('location: login.php');
        }
?>

和我的主要上传页面，其中食谱被上传到数据库我试图将用户ID发布到数据库但是无济于事

<?php
require_once ("checklog.php");
require_once ("function.php");
include_once ("home_start_logged.php");
require_once ("db_connect.php"); 
//get form data//
$_SESSION['userid']== $_POST['userid'];
$upload = trim($_POST['Upload']);
$mealname = trim($_POST['mealname']);
$ingredients = trim($_POST['ingredients']);
$hours = trim($_POST['hours']);
$minutes = trim($_POST['minutes']);
$recipe = trim($_POST['recipe']);

echo $_SESSION['userid'];
if(trim($_POST['Submit']) =="Upload"){
if($db_server){
    //clean the input now we have a db connection//
    $mealname = clean_string($db_server, $mealname);
    $ingredients = clean_string($db_server, $ingredients);
    $hour = clean_string($db_server, $hour);
    $minutes = clean_string($db_server, $minutes);
    $recipe = clean_string($db_server, $recipe);
    $ingredients = clean_string($db_server, $ingredients);
    $image = clean_string($db_server,$image);
    mysqli_select_db($db_server, $db_database) ;


//check whether the recipe exists//
$query= "SELECT mealname FROM `recipename` WHERE mealname='$mealname'";
$result = mysqli_query($db_server, $query);
if ($row = mysqli_fetch_array($result)){
        $message = "Meal already exists. Please try again.";
    }else{
        //upload recipe to database//
        $query = "INSERT INTO `recipename` (
                      mealname, ingredients, hours, minutes, recipe, 
                      imagepath, userID) VALUES ('$mealname',
                      '$ingredients','$hours','$minutes','$recipe',
                      '$image','" . $_SESSION['userid'] . "')";
        echo query;
    mysqli_query($db_server, $query) or 
        die("Insert failed. ". mysqli_error($db_server));
    }

我的表单如下：

<form method="post" action="upload.php" enctype="multipart/form-data">
    <li>
Meal Name
    <input type="text" name="mealname" />
    </li>
    <li>
Ingredients
    <input type="text" name="ingredients" />
    </li>
    <li>
Cooking Time
    <input type="number" name="hours" placeholder="Hours" />
    <input type="number" name="minutes" placeholder="Minutes" />
     </li>
     <li>
 Recipe
 <input type="text" name="recipe"/>
     </li>
     <li>
Have you got a photo?
    <input type="file" name="image" id="image" size="10"/>
    </li>
  <input type="submit" id="submit" name="Submit" value="Upload" />  
  </form>

这是我定义$ _SESSION

的登录表单

<?php
require_once ("function.php");
    //get form data//
    $username = trim($_POST['username']);
    $password = trim($_POST['password']);
    //start session//
    if ($username&&$password) {
        session_start();
        require_once("db_connect.php"); 
        //clean the input now we have a db connection//
        $username = clean_string($db_server, $username);
        $password = clean_string($db_server, $password);
        $repeatpassword = clean_string($db_server, $repeatpassword) ;
        mysqli_select_db($db_server, $db_database) ;
        //check whether the username exists//
        $query="SELECT * FROM Users WHERE Username='$username'";
        $result=mysqli_query($db_server, $query) ;
        if ($row = mysqli_fetch_array($result)){
            $db_username = $row['Username'];
            $db_password = $row['Password'];
            $db_id = $row['userid'];

            if ($username==$db_username&&salt($password)==$db_password){
                $_SESSION['username']=$username;
                $_SESSION['userid']=$db_id;
                $_SESSION['logged']="logged";
                header ('Location: phpdatabase.php');
            }else{
                $message = "<h1>Incorrect Password!</h1>";
            }
        }else{
            $message = "<h1>That user does not exist!</h1>" .
                            "Please <a href='login.php'>try again</a>";
        }
        mysqli_free_result($result);
        require_once ("db_close.php");
    }else{
        $message = "<h1>Please enter a valid username/password</h1>";
    }
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<title>Fud.</title> 
<link rel="stylesheet" type="text/css" href="site.css" />

</head> 
    <h1>Login</h1>
    <form action='login.php' method='post'>
    Username:<input type='text' name='username'><br />
    Password: <input type='password' name='password'><br /> 
   <input type='submit' name='submit' value='Login' />
   <input name='reset' type='reset' value='reset' />

   <h4><a href='register.php'>Register</a></h4>

</form>
       <?php echo $message; ?>

Answer 1

在您的表单userid中没有显示任何内容，因此当您设置：

时

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

$_POST['userid']是空的，也许在其他地方你已经在$ _SESSION中设置了用户ID，如果是这种情况你不需要设置任何新内容，只需使用它。

Answer 2

第二个文件中没有session_start();。

要访问Session全局，您需要启动会话。

使用

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

不是

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

Answer 3

您正在将$_POST数组的值保存到$_SESSION，但请求表单中没有$_POST['userid']这样的元素。

解决方案：最好先登录他/她，对用户进行身份验证。用户登录后，您可以将会话中的用户ID保存在整个会话的每个页面中。

假设您的登录表单包含文本字段名称userid。在开始会话并验证用户后，将其放入登录检查文件中。

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

现在在您要将数据插入数据库删除行的页面中

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

试图将用户ID上传到数据库，Session将userid返回为null？

3 个答案: