Question

我正在尝试显示已登录的唯一用户，然后将其存储到我的数据库中的表中，例如，如果我以James身份登录，我想显示James＆＃39;来自数据库的数据并将他的id存储到另一个table.from我的理解$ _SESSION [＆＃39; login_user＆＃39;];应该返回唯一用户的用户名。

当我跑步时，我得到这个错误，这是输出：

注意：未定义的索引：第3行的C：\ xampp \ htdocs \ bid.php中的login_user

注意：未定义的变量：第8行的C：\ xampp \ htdocs \ bid.php中的SESSION 感谢您与我们联系

的login.php

<?php
   include("config.php");
   session_start();

   if($_SERVER["REQUEST_METHOD"] == "POST") {
      // username and password sent from form 

      $myusername = mysqli_real_escape_string($conn,$_POST['username']);
      $mypassword = mysqli_real_escape_string($conn,$_POST['password']); 

      $sql = "SELECT customer_id FROM customer WHERE email_adress = '$myusername' and password = '$mypassword'";
      $_SESSION['username'] = $myusername
      $result = mysqli_query($conn,$sql);
      $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
      $active = $row['customer_id'];

      $count = mysqli_num_rows($result);

      // If result matched $myusername and $mypassword, table row must be 1 row

 if($count >= 1) {
     $_SESSION['login_user'] = $myusername;

     header("location: index2.php");
  }else {
     $error = "Your Login Name or Password is invalid";
  }
   }
?>

bid.php：

<?php
    session_start();
    echo $_SESSION['username'];
require 'config.php';
//include "job.php" 
$jobid    = $_POST['job_id'];
$bid    = $_POST['bid'];
bidderid = $SESSION['username']
echo "$jobid";






$query   = "INSERT into bid (bid_amount,job_id) VALUES('" . $bid . "','" . $jobid . "')";
$success = $conn->query($query);

if (!$success) {
    die("Couldn't enter data: ".$conn->error);

}

echo "Thank You For Contacting Us <br>";

$conn->close();

 ?>

session.php文件

<?php
   include('config.php');
   session_start();

   $user_check = $_SESSION['login_user'];

   $ses_sql = mysqli_query($conn,"select email_adress from customer where email_adress = '$user_check' ");

   $row = mysqli_fetch_array($ses_sql,MYSQLI_ASSOC);

   $login_session = $row['email_adress'];

   if(!isset($_SESSION['login_user'])){
      header("location:login.php");
   }
?>

Answer 1

的login.php

export async function get(pathname) {
    const params = {
      headers: {
        'Content-Type': 'application/json',
      }
    };
    let response = await fetch(`${__ROOT_API__}${pathname}`, params);
    if (!validateStatusCode(response)) {
      var error = new Error(response.statusText)
      error.response = response
      throw error
    }
    return response.json();
}

现在，如果成功，$ _SESSION [＆＃39; login_user＆＃39;]将在bid.php和其他所有网页中提供。

在session.php中，请恢复为这些。

  11 |     let response = await fetch(`${__ROOT_API__}${pathname}`, params);
      12 |     if (!validateStatusCode(response)) {
    > 13 |       var error = new Error(response.statusText)
      14 |       error.response = response
      15 |       throw error
      16 |     }

      at _callee$ (src/services/project/project.js:13:19)
      at tryCatch (node_modules/regenerator-runtime/runtime.js:62:40)
      at Generator.invoke [as _invoke] (node_modules/regenerator-runtime/runtime.js:296:22)
      at Generator.prototype.(anonymous function) [as next] (node_modules/regenerator-runtime/runtime.js:114:21)
      at step (node_modules/babel-runtime/helpers/asyncToGenerator.js:17:30)
      at node_modules/babel-runtime/helpers/asyncToGenerator.js:28:13

请参阅此内容。

<?php
   session_start();
   // session should start here

   include("config.php");


   if($_SERVER["REQUEST_METHOD"] == "POST") {
      // username and password sent from form 

      $myusername = mysqli_real_escape_string($conn,$_POST['username']);
      $mypassword = mysqli_real_escape_string($conn,$_POST['password']); 

      $sql = "SELECT customer_id FROM customer WHERE email_adress = '$myusername' and password = '$mypassword'";

      $_SESSION['username'] = $myusername; // you missed a semi colon

      $result = mysqli_query($conn,$sql);
      $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
      $active = $row['customer_id'];

      $count = mysqli_num_rows($result);

      // If result matched $myusername and $mypassword, table row must be 1 row

 if($count >= 1) {

     $_SESSION['login_user'] = $myusername; // $myusername contains username of this user

     $_SESSION['login_data'] = $row; // $row contains user record from the db. including customer_id etc..


     header("location: index2.php");
  }else {
     $error = "Your Login Name or Password is invalid";
  }
   }
?>

希望这有帮助..快乐编码

PHP如何显示用户数据

1 个答案: