Question

我有一个登录脚本（从互联网上下载），我已经适应了我的数据库。 i但是从登录到我系统中的页面的链接与登录用户的详细信息有关。

登录系统所基于的用户表具有以下字段

`id` int(11) NOT NULL AUTO_INCREMENT,  
`username` varchar(18) NOT NULL,  
`first_name` varchar(32) NOT NULL,  
`last_name` varchar(32) NOT NULL,  
`gender` varchar(15) NOT NULL DEFAULT 'undisclosed',  
`bio` text NOT NULL,  
`image_location` varchar(125) NOT NULL DEFAULT 'avatars/default_avatar.png',  
`password` varchar(512) NOT NULL,  
`email` varchar(1024) NOT NULL,  
`email_code` varchar(100) NOT NULL,  
`time` int(11) NOT NULL,  
`confirmed` int(11) NOT NULL DEFAULT '0',  
`generated_string` varchar(35) NOT NULL DEFAULT '0',  
`ip` varchar(32) NOT NULL,  
`EmployeeID` int(11) DEFAULT '0',  
PRIMARY KEY (`id`)

保存用户详细信息的表包含以下字段

CREATE TABLE IF NOT EXISTS `chidren` (
 `ChildID` int(11) NOT NULL AUTO_INCREMENT,
 `EmployeeID` int(11) DEFAULT '0',
 `ChildName` varchar(50) DEFAULT NULL,
 `DateOfBirth` datetime DEFAULT NULL,
 `Mother` varchar(50) DEFAULT NULL,
 `Comment` longtext,
 `Clerk` varchar(50) DEFAULT NULL,
 `Picture` longblob,
 `Pic` longblob,
 PRIMARY KEY (`ChildID`),
 KEY `ChildID` (`ChildID`),
 KEY `EmployeeID` (`EmployeeID`)

您可以注意到，使用EmployeeID字段连接了两个表。

用户将从中导入到他/她的页面的登录脚本是

<?php
require 'core/init.php';
$general->logged_in_protect();

if (empty($_POST) === false) {

    $username = trim($_POST['username']);
    $password = trim($_POST['password']);

    if (empty($username) === true || empty($password) === true) {
        $errors[] = 'Sorry, but we need your username and password.';
    } else if ($users->user_exists($username) === false) {
        $errors[] = 'Sorry that username doesn\'t exists.';
    } else if ($users->email_confirmed($username) === false) {
        $errors[] = 'Sorry, but you need to activate your account. 
                     Please check your email.';
    } else {
        if (strlen($password) > 18) {
            $errors[] = 'The password should be less than 18 characters, without spacing.';
        }
        $login = $users->login($username, $password);
        if ($login === false) {
            $errors[] = 'Sorry, that username/password is invalid';
        }else {
            session_regenerate_id(true);// destroying the old session id and creating a new one
            $_SESSION['id'] =  $login;
            header('Location: home.php');
            exit();
        }
    }
} 
?>
<!doctype html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <link rel="stylesheet" type="text/css" href="css/style.css" >
    <title>Login</title>
</head>
<body>  
    <div id="container">
    <?php include 'includes/menu.php'; ?>

        <h1>Login</h1>

        <?php 
        if(empty($errors) === false){
            echo '<p>' . implode('</p><p>', $errors) . '</p>';  
        }
        ?>

        <form method="post" action="">
            <h4>Username:</h4>
            <input type="text" name="username" value="<?php if(isset($_POST['username'])) echo htmlentities($_POST['username']); ?>" />
            <h4>Password:</h4>
            <input type="password" name="password" />
            <br>
            <input type="submit" name="submit" />
        </form>
        <br>
        <a href="confirm-recover.php">Forgot your username/password?</a>

    </div>
</body>
</html>

登录脚本附带的主页是这样的（我已将其包括在内以防万一它有助于解决我的问题）

  <?php 
    require 'core/init.php';
    $general->logged_out_protect();

    $username   = htmlentities($user['username']); // storing the user's username after clearning for any html tags.

    ?>
    <!doctype html>
    <html lang="en">
    <head>
    <meta charset="UTF-8">
    <link rel="stylesheet" type="text/css" href="css/style.css" >
    <title>Home</title>
    </head>
    <body>  
    <div id="container">
        <?php include 'includes/menu.php'; ?>
        <h1>Hello <?php echo $username, '!'; ?></h1>
    </div>
    </body>
    </html>

这是我的页面，遗憾的是失败了。

<html>
<head>
<link rel="stylesheet" type="text/css" href="mystyle.css" />
        <title></title>
</head>
<body>

<?php
require 'core/init.php';
    $general->logged_out_protect();

    // query db and get date only for the user that logged in. Used GROUP BY because one employee will have more than one child 
    $EmployeeID = $_GET['EmployeeID'];
    $result = mysql_query("SELECT * FROM children WHERE EmployeeID=$EmployeeID
    GROUP BY holder.EmployeeID")

    or die(mysql_error());

    // display data in table

        echo "<table border='1' cellpadding='10'>";
        echo "<tr> <th>Child Name</th> <th>Mother</th> <th>Date of Birth</th>  ";

    while($row = mysql_fetch_array( $result )) {

    // echo out the contents of each row into a table
    echo "<tr>";
    //echo '<td>' . $row['EmployeeID'] . '</td>';
    echo '<td>' . $row['ChildName'] . '</td>';              
    echo '<td>' . $row['Mother'] . '</td>';
    echo '<td>' . $row['DateOfBirth'] . '</td>';
    //the following two fields link to files exactly the same as this one. Again the linking is by EmployeeID
    echo '<td><a href="arm/spouse.php?EmployeeID=' . $row['EmployeeID'] . '">SPOUSE DETAILS</a></td>';
    echo '<td><a href="arm/employeedatails.php?EmployeeID=' . $row['EmployeeID'] . '">WORK DETAILS</a></td>';
    echo "</tr>"; 
        } 

        // close table>
        echo "</table>";
    ?>

    <p>Click on any of the above to see your other details</p>
    </body>
    </html>

我的问题是我无法正确编写页面代码，因此它只能绘制有关已登录用户的数据。事实上，在我的所有尝试中，页面只是给我错误，下面是代码我尝试了该页面，我使用了GET功能，以便页面只显示有关特定用户的数据。

在你说之前，是的，我在我的页面中没有使用过mysqli（这会是错误的原因吗？）但是后来我对mysqli完全是绿色的，只是一个带有mysql的begginer。但我很感激，如果帮助将在mysqli，因为我发现它更安全。我是从MSACCESS迁移的人（我只需要迁移到我的程序在网上）。

Answer 1

最后我明白了 $ query =“SELECT * FROM children WHERE EmployeeID =：employeeID;”; //构造查询，使其接受准备好的变量。 $ statement = $ db-＆gt; prepare（$ query）; //准备查询。 $ statement-＆gt; execute（array（'：employeeID'=＆gt; $ EmployeeID））; //这就是我想要的它将用户连接到他的数据 $陈述式＆GT;调用setFetchMode（PDO :: FETCH_ASSOC）; //设置获取模式。

    while ($row = $statement->fetch())
    {
        $ChildName = $row['ChildName'];
        $Mother = $row['Mother'];
        $DateOfBirth = $row['DateOfBirth'];
        echo "Child Name: $ChildName";
        echo "<br />Mother: $Mother";
        echo "<br />Date of Birth: $DateOfBirth";
    }

使用GET功能链接页面

1 个答案: