无法用php更新mysql数据库吗？

Posted 2021-04-30

我正在尝试更新数据库中的日期，但是由于某种原因它无法工作，这是代码。

这是我的php页面，我在其中放置代码以显示数据，并且有一个EDIT按钮可以将我发送到page_edit，我可以在其中键入另一个名称并更新它。我会选择一个随机行并在其中编辑名称。

主页（main.php）

<?php
if ($mysqli->connect_error) {die("Connection failed: " . $mysqli->connect_error);} 
$sql = "SELECT * FROM users ORDER BY name";
$result = $mysqli->query($sql);
while ($rows = mysqli_fetch_array($result)) {   
echo '
<div class="admin-content-con">
<header class="clearfix">
</header>
<table class="table table-striped">
<thead>
<tr>
<th>No</th>
<th>Name</th>
<th>Link</th>
<th>Actions</th>
</tr>
</thead>
<tbody>';            
        echo "<tr>";
        echo "<td>" .$rows ['id']."</td>";
        echo "<td>" .$rows ['name']."</td>";
        echo "<td>" .$rows ['link']."</td>";
        echo "<td> 
        <a href='page_edit.php?id=".$rows["id"]."'class='btn btn-xs btn-warning '  role='button'>edit</a> 
        <a href='delete.php?id=".$rows["id"]."'class='btn btn-xs btn-danger'  role='button' onclick='return checkDelete()'>del</a></td>";     
        echo "</tr>";       
        echo '</tbody>
              </table>
              </div>';} ?>

页面编辑：（page_edit.php）。

       <div class="settings-row">
       <h3>Name</h3>
       <p>This is permanently show on the left corner of navigation</p>

       <form action="name.php" method="post" >   
       <div class="form-group">
       <input type="text" class="form-control" name="name">
       <input type="submit" value="Submit" name="submit">

       </div>
       </form>
       </div>

更新数据库（name.php）的PHP后端代码。

<?php
include '*'; 
$dbname = "*";
$conn = mysqli_connect("localhost", "root", "root", $dbname);
if (!$conn) {die("Connection failed: " . mysqli_connect_error());}
$sql ="UPDATE users SET name='".$_POST['name']."' WHERE id='$id'";
if (mysqli_query($conn, $sql)) {
    mysqli_close($conn);
    header('Location: main.php'); 
    exit;
} else {
  echo "Error update record";
}
?>

答案

您需要将id传递到编辑页面。然后，您需要在更新页面上输入id。

<div class="settings-row">
       <h3>Name</h3>
       <p>This is permanently show on the left corner of navigation</p>

       <form action="name.php" method="post" >   
       <div class="form-group">
       <input type="text" class="form-control" name="name">
       <input type="submit" value="Submit" name="submit">
       <input type="hidden" value="<?php echo htmlspecialchars($_GET['id']);?>" name="id">
       </div>
       </form>
       </div>

然后

<?php
include '*'; 
$dbname = "*";
$conn = mysqli_connect("localhost", "root", "root", $dbname);
if (!$conn) {die("Connection failed: " . mysqli_connect_error());}
$id = $_POST['id'];
$sql ="UPDATE users SET name='".$_POST['name']."' WHERE id='$id'";
if (mysqli_query($conn, $sql)) {
    mysqli_close($conn);
    header('Location: main.php'); 
    exit;
} else {
  echo "Error update record";
}
?>

请参见How can I prevent SQL injection in PHP?，了解如何编写SQL。

大约，

$sql ="UPDATE users SET name=? WHERE id=?";

然后准备，绑定和执行。

另外为PHP和mysqli驱动程序使用错误报告。您的$id应该已报告为未定义变量。

1. How do I get PHP errors to display?
2. How to display errors for my MySQLi query?

另一答案

page_edit.php应该看起来像下面-

//Add hidden input type in form for get value of id

<div class="settings-row">
           <h3>Name</h3>
           <p>This is permanently show on the left corner of navigation</p>

           <form action="name.php" method="post" >
           <input type="hidden" class="form-control" name="id" value="<?php echo $_GET['id']; ?>">   
           <div class="form-group">
           <input type="text" class="form-control" name="name">
           <input type="submit" value="Submit" name="submit">

       </div>
       </form>
       </div>

而且page.php应该是这样的

<?php
include '*'; 
$dbname = "*";
$conn = mysqli_connect("localhost", "root", "root", $dbname);
if (!$conn) {die("Connection failed: " . mysqli_connect_error());}
$id = $_POST['id'];
$sql ="UPDATE users SET name='".$_POST['name']."' WHERE id='$id'";
if (mysqli_query($conn, $sql)) {
    mysqli_close($conn);
    header('Location: main.php'); 
    exit;
} else {
  echo "Error update record";
}
?>