Question

目前我有一个基本登录的网站，我只是想知道如何显示登录的唯一用户的名称，技能和描述。这是我到目前为止所做的。我只能找到有关如何将数据显示到表中的文章。这是更新后的代码：

   <?php
   include('session.php');
   require 'config.php';
   $sql = "SELECT * FROM profile";
   $result = $conn->query($sql);
   //echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
?>
<html>
<head>
  <link rel="stylesheet" type="text/css" href="profile.css">

</head>

<body>



<ul>
  <li><a href="welcome.php">Home</a></li>
  <li><a href="postjob2.php">PostJob</a></li>
  <li><a href="findjob.php">Find Job</a></li>
  <li><a href="hiw2.php">How It Works</a></li>
  <li><a href="notification.php">Notifications</a></li>
  <li><a href="message.php">Message</a></li>
  <li><a href="profile.php">profile</a></li>

</ul>

<h1>Welcome To Bid4MyJob</h1>
  <li><a href="editprofile.php">edit profile</a></li>

<div id="ProfilePage">
    <div id="LeftCol">
        <div id="Photo"></div>
        <div id="ProfileOptions">
        a
        </div>
    </div>

    <div id="Info">
        <p>
            <strong>Name:<?php echo  $row["name"]?></strong>
            <!--<span>James</span>-->
        </p>
        <p>
            <strong>Skill:<?php echo $row["skill"]?><</strong>
            <!--span>James</span>-->
        </p>
        <!-- <p>
            <strong>review:<?php /*echo $row["review"]*/?><</strong>
            <span>james</span>
        </p> -->
        <p>
            <strong>Description:<?php echo $row["description"]?><</strong>
            <span>James</span>
        </p>
        <!--<p>
            <strong>Name:</strong>
            <span>james</span>
        </p>-->
    </div>

    <!-- Needed because other elements inside ProfilePage have floats 
    <div style="clear:both"></div>-->
</div>
</body>
</html>

Answer 1

当用户成功登录时，创建一个cookie并存储他们的用户名或电子邮件，无论您在该cookie中使用了什么。您必须在login.php文件中执行此操作，以检查用户名和密码。

     if(login success)
   {
    setCookie("username",value of username that you got from 
    user,'time()+3600','/');
    echo "login successful";
   }

之后在Profile中你有这样的代码;

    <?php
$name=$_COOKIE['username'];
$sql=$conn->prepare("SELECT * from profile where username=?");
$stmt->bind_param('s', $name); // 's' specifies the variable type =>'string'

$stmt->execute();

$result = $stmt->get_result();
while ($row = $result->fetch_assoc()) {
    // do something with $row
}
?>

之后，您可以在div中显示信息，例如$ row [＆＃39;用户名＆＃39;]等。

Answer 2

如果$ row []来自sql请求，只需在您的请求中添加如下内容：

$user = (...) your current user logged
(...)
select (...) where user = $user

Answer 3

您可以使用<?php session_start(); //before anything else include('session.php'); require 'config.php'; $sql = "SELECT * FROM profile"; $result = $conn->query($sql); //assuming $row contains the information from $_SESSION["user"] = $row; ?> <html> <head> <link rel="stylesheet" type="text/css" href="profile.css"> </head> .... ....在登录后识别人员并存储有关此人的信息。会话变量跨多个页面保存特定用户的数据。会话ID存储在用户浏览器中，用于标识人员的<?php session_start(); if(isset($_SESSION["user"]) { //logged in $name = $_SESSION["user"]["name"]; } else { //not logged in } ?> <html> .... ....数据。在您的程序中，您可以轻松地执行以下操作：

USER_PARALLEL_EXECUTE_TASKS

现在在另一页上，您可以轻松地：

DBA_PARALLEL_EXECUTE_TASKS

PHP显示唯一数据

3 个答案: