Question

我尝试从student_details表中获取一个id详细信息，我使用$ _get从url中获取id。示例：（http://localhost/School-project/admin/edit.php?id=2014-01）所以我的if id是2014-01，它是数据库表student_table中的主键。我没有收到任何错误，也没有输出数据。所以我复制下面的所有代码请帮帮我！因为我第二次问问题，也许我不擅长如何把我的代码放在下面。我还附上了桌子结构的按扣。 enter image description here

<?php

session_start();

// Viewer: Can this be abstracted away to a reusable function?
if ( !isset($_SESSION['username']) ) {
    header('Location: index.php');
    die();
}

?>

<?php include '../inc/header.php';

    $id=$_GET['id'];
    var_dump($id);

    $conn = mysqli_connect("localhost", "root", "", "smart_school");

    $queryString = "SELECT * FROM student_details WHERE id =$id";

    $result = mysqli_query($conn, $queryString);
    mysqli_close($conn);
    var_dump($result)

 ?>

<section id="admin">    
        <div class="container">
            <div class="row">
             <div class="col-md-12 admin-area">
                <h1>Admin Area</h1>
                <hr>

                    <div class="col-md-3" >
                        <div class="admin-menu">
                          <!-- Nav tabs -->
                          <ul class="list-group">
                            <li class="list-group-item " ><a href="admin.php">Admin Home</a></li>
                            <li class="list-group-item " ><a href="student-update.php">Student Update</a></li>      
                            <li class="list-group-item " ><a href="admin-profile.php">Profile</a></li>                     
                            <li class="list-group-item " ><a href="logout.php">Logout</a></li>
                          </ul>                   
                        </div>
                    </div>

                    <div class="col-md-8">
                        <div class="student-details">
                        <?php
                                    while ($row = mysqli_fetch_array($result)) {



                                        ?>
                                        <tr>
                                            <td class="success" ><?php echo $row["fname"] ?></td>
                                            <td class="success" ><?php echo $row["lname"] ?></td>
                                            <td class="success" ><?php echo $row["id"] ?></td>
                                            <td class="success" ><?php echo $row["age"] ?></td>
                                            <td class="success" ><?php echo $row["class"] ?></td>
                                            <td class="success" ><?php echo $row["group"] ?></td>
                                            <td class="success" ><?php echo $row["section"] ?></td>
                                            <td class="success" ><?php echo $row["year"] ?></td>

                                        </tr>
                                        <?php


                                    }
                                    ?>

                        </div>


                    </div>

            </div>
        </div>
    </div>
</section>
<?php include '../inc/footer.php' ?>

Answer 1

查询没有明显的错误，除了它对注入攻击是开放的。代码不检查产生的任何错误 - 它假定所有函数都返回true。

如果各种mysqli函数失败，则返回false，然后您可以处理错误并输出错误的详细信息，以帮助您识别问题。

以下代码是错误检查的一个示例。

$mysqli = new mysqli("localhost", "root", "", "smart_school");
// check connection 
if ($mysqli->connect_errno) {
    die("Connect failed: ".$mysqli->connect_error);
}
// define the query
$sql = "SELECT * FROM student_details WHERE id = ?";
// prepare the query
if (!$stmt = $mysqli->prepare($sql)) {
    // failed to prepare query;
    die("Prepare failed: ".$stmt->error);
}
// bind the parameter
if (!$res = $stmt->bind_param('s', $_GET['id'])) {
    // failed to bind
    die("Bind failed: ".$stmt->error);
}
// execute
if (!$res = $stmt->execute()) {
    // failed to execute
    die("Execute failed: ".$stmt->error);
}
if (!$result = $stmt->get_result()) {
    // failed to get results
    die("Get results failed: ".$stmt->error);
}

无法使用变量从表中调用数据

1 个答案: