【发布时间】：2021-03-05 05:24:27
【问题描述】：
我正在开发一个允许用户编辑汽车信息的网页。在主线中，有一个编辑按钮（输入 - 输入带有隐藏键值的文本），它将用户带到这个“编辑汽车信息”页面。最初，一旦第一次打开页面，这个隐藏值用于查询数据库、检索原始信息并将它们设置为字段的占位符。用户可以在输入字段中写入信息，然后按“提交编辑”按钮，然后更新数据库表中的行。但是，我收到一个错误，即隐藏值的名称未定义。我不明白当它对选择查询工作得很好时，它是如何为更新查询未定义的。任何人都可以对此有所了解吗？我该怎么办？这是错误的图片：
这是大陆代码：（这里设置了隐藏值）
<?php
$mysqli= new mysqli("localhost", "root","","Car_registration");
if(empty($_SESSION)) // if the session not yet started
session_start();
if(isset($_SESSION['username'])) { // if user already logged in
header("location: mainlanding_user.php"); //send to homepage
exit;
}
?>
<!DOCTYPE html>
<html>
<head>
<title> Car Registration: User's Mainlanding </title>
<link href="css/style3.css" rel="stylesheet">
</head>
<body>
<header>
<h1>Account Information</h1>
<img id="img1" src= "image/car.jpg" alt ="car image">
</header>
<nav id='nav'>
<form action="logout.php">
<input type="submit" value=" Logout " id="button">
</form>
</nav>
<h2>Profile </h2>
<div class='container1'>
<?php
$username="root";
$password="";
$database="Car_registration";
$mysqli= new mysqli("localhost",$username,$password,$database);
$query= "select * from driver where username='".$_SESSION['logged_username']."'";
$result = $mysqli->query($query);
while( $row = $result->fetch_assoc() ){
echo "<div id='container'>" ;
echo "<dl> <dt>First Name</dt> <dd>".$row['Fname'];
echo "</dd> <br> <dt>Last name</dt><dd>".$row['Lname'];
echo "</dd> <br> <dt>License Number</dt><dd>".$row['license_no'];
echo "</dd> <br> <dt>Age</dt><dd>".$row['Age'];
echo "</dd> <br> <dt>Birthday</dt><dd>".$row['bdate'];
echo "</dd> <br> <dt>City</dt><dd>".$row['City'];
echo "</dd></dl>";
echo "</div>";
$license_no = $row['license_no']; //used for finding cars
}
?>
<div class="align-me">
<div class="form-wrapper" action="search_plate_no.php">
<form class="center">
<input class="input-fields" name="search" type="text" placeholder="Search a plate number">
<input class="input-fields submit" name="find" type="submit" value="Search">
</form>
</div>
</div>
<h3> Registered Cars </h3>
<div class='container2'>
<?php
$username="root";
$password="";
$database="Car_registration";
$mysqli= new mysqli("localhost",$username,$password,$database);
$query= "select * from cars where license_no='".$license_no."'";
$result = $mysqli->query($query);
echo "<table border=1>
<tr>
<th>Plate No.</th>
<th>License No.</th>
<th>Car Type</th>
<th>Fines</th>
<th>City</th>
<th>Edit</th>
<th>Delete</th>
</tr>";
while ($temp = $result->fetch_assoc()){
?>
<tr>
<td><?php echo $temp['Plate_no']; ?></td>
<td><?php echo $temp['license_no']; ?></td>
<td><?php echo $temp['Car_type']; ?></td>
<td><?php echo $temp['Fines']; ?></td>
<td><?php echo $temp['city']; ?></td>
<td>
<form action = "edit_car.php" method="post">
<input type="hidden" name="id" value="<?php echo $temp['Plate_no']; ?>">
<input type="submit" name="edit" value="Edit">
</form>
</td>
<td>
<form action = "delete_car.php" method="post">
<input type="hidden" name="id" value="<?php echo $temp['Plate_no']; ?>">
<input type="submit" name="delete" value="Delete">
</form>
</td>
</tr>
<?php
}
?>
</table>
</div>
<form action="register_car.php">
<input type="submit" value=" Register Car " id="button2">
</form>
<footer>
<h4> All rights belong to Car Registration Inc. </h4>
<img id="img3" src= "image/license.png" alt ="license plates image">
</footer>
</body>
</html>
编辑汽车页面：（此处产生错误）
<!DOCTYPE html>
<html>
<head>
<title> Edit Car Information Page </title>
<link href="css/style2.css" rel="stylesheet">
</head>
<body>
<div class="container">
<header>
<h1>Edit Car Information </h1>
<img id="img1" src= "image/register.png" alt ="Registration image">
</header>
<?php
$username="root";
$password="";
$database="Car_registration";
$mysqli= new mysqli("localhost",$username,$password,$database);
$plate_no= $_POST["id"]; //This line causes an error
$_SESSION['plateNo'] = $plate_no;
$query= "select * from cars where Plate_no='".$plate_no."'";
$result = $mysqli->query($query);
while( $row = $result->fetch_assoc()){
$plate_no = $row['Plate_no'];
$car_type = $row['Car_type'];
}
?>
<main>
<h2> You can only edit the following information: </h2>
<form action="" method="post">
<label for="car_type_input">Car Type:</label>
<input type="text" placeholder="<?php echo $car_type?>" id="car_type_input" name="car_type_input"><br><br>
<div class="vertical-center">
<input type="submit" value=" Submit Edit " name="button1" id="button1">
</div>
</form>
<?php
$username="root";
$password="";
$database="Car_registration";
$mysqli= new mysqli("localhost",$username,$password,$database);
if( isset($_POST['button1']) ){ //If user changed field, take value. If not, keep old value.
if( !empty($_POST['car_type_input']) ){ //If there is user input
$car_type_2 = $_POST['car_type_input'];
$query= "update cars set Car_type='".$car_type_2."' WHERE Plate_no='".$_SESSION['plateNo']."'";
}
if ($mysqli->query($query))
echo "Fields updated successfuly!";
else
echo "Update Fields Failed!";
}
?>
</main>
<footer>
<h3> All rights belong to Car Registration Inc. </h3>
<img id="img3" src= "image/license.png" alt ="license plates image">
</footer>
</div>
</body>
</html>
【问题讨论】：
-
不要发文字图片
-
$plate_no= $_POST["id"]; //This line causes an error是什么错误？ -
错误在我链接的图片中。它是第 27 行的“未定义索引：id”
标签： php html mysql html-input