数据库问题：无法通过PHP和MYSQL更新_Php_Html_Mysql

数据库问题：无法通过PHP和MYSQL更新

php html mysql

数据库问题：无法通过PHP和MYSQL更新,php,html,mysql,Php,Html,Mysql,目前正在处理一个项目，无法解决此问题。URL中的$\u POST ID工作正常，但我无法使用该ID获取行。请帮助花了很多时间浏览代码并在不同的区域发表评论，以获得的效果数据库连接工作-为了隐私，我刚刚删除了$con <?php $con = mysqli_connect(blablabla; if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error();

目前正在处理一个项目，无法解决此问题。URL中的$\u POST ID工作正常，但我无法使用该ID获取行。请帮助

花了很多时间浏览代码并在不同的区域发表评论，以获得

的效果

数据库连接工作-为了隐私，我刚刚删除了$con

<?php
$con = mysqli_connect(blablabla;

if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " .
    mysqli_connect_error();

if (isset($_POST['update'])) {
    $id = $_POST['id'];
    $Catname = $_POST['Catname'];
   $Catdescription = $_POST['Catdescription'];
    $Catpicture = base64_encode(file_get_contents($_FILES['Catpicture'] 
['tmp_name']));
$result = mysqli_query($mysqli, "UPDATE CatadoptionDB SET Catname='$Catname',Catdescription='$Catdescription',Catpicture='$Catpicture' WHERE id=$id");

}
}
?>
<?php
$id = $_GET['id'];

$result = mysqli_query($mysqli, "SELECT * FROM CatadoptionDB WHERE id=$id");
while ($res = mysqli_fetch_array($result)) {
$Catname = $res['Catname'];
$Catdescription = $res['Catdescription'];
$Catpicture = $res['Catpicture'];
}
?>

<html>

<head>
<title>Edit Data</title>
</head>

<body>
<form name="form1" method="post" action="" enctype="multipart/form-data">
    <table border="0">
        <tr>
            <td>Catname</td>
            <td>
                <input type="text" name="Catname" value='<?php echo 
$Catname; ?>'>
            </td>
        </tr>
        <tr>
            <td>Catdescription</td>
            <td><textarea name="Catdescription" value='<?php echo 
$Catdescription; ?>'>
            </textarea>
            </td>
        </tr>
        <tr>
            <td>Catpicture</td>
            <td><input type="file" name="Catpicture" value='<?php echo 
$Catpicture; ?>'>
            </td>
        </tr>
        <tr>
            <td><input type="hidden" name="id" value='<?php echo 
$_GET['id']; ?>'>
            </td>
            <td><input type="submit" name="update" value="Update"></td>


        </tr>
    </table>
</form>
</body>

</html>

首先，表单方法设置为'post'，因此您应该通过$\u post['id']获取$id值
其次，根据您的代码，更新表只会在db connect error上运行，所以它永远不会更新
也许你可以试着做如下的小改变
<?php
$con = mysqli_connect();
if (mysqli_connect_errno()) die("Failed to connect to MySQL: " . mysqli_connect_error() );

$id = isset( $_GET['id']) ? $_GET['id'] : $_POST['id'];

if (isset($_POST['update'])) {
    $id = $_POST['id'];
    $Catname = $_POST['Catname'];
    $Catdescription = $_POST['Catdescription'];
    $Catpicture = base64_encode(file_get_contents($_FILES['Catpicture']
    ['tmp_name']));
    $result = mysqli_query($mysqli, "UPDATE KatadoptionDB SET Catname='$Catname',Catdescription='$Catdescription',Catpicture='$Catpicture' WHERE id=$id");

}
?>
<?php

$result = mysqli_query($mysqli, "SELECT * FROM CatadoptionDB WHERE id=$id");
while ($res = mysqli_fetch_array($result)) {
    $Catname = $res['Catname'];
    $Catdescription = $res['Catdescription'];
    $Catpicture = $res['Catpicture'];
}
?>

<html>

<head>
    <title>Edit Data</title>
</head>

<body>
<form name="form1" method="post" action="" enctype="multipart/form-data">
    <table border="0">
        <tr>
            <td>Catname</td>
            <td>
                <input type="text" name="Catname" value='<?php echo
                $Catname; ?>'>
            </td>
        </tr>
        <tr>
            <td>Catdescription</td>
            <td><textarea name="Catdescription" value='<?php echo
                $Catdescription; ?>'>
            </textarea>
            </td>
        </tr>
        <tr>
            <td>Catpicture</td>
            <td><input type="file" name="Catpicture" value='<?php echo
                $Catpicture; ?>'>
            </td>
        </tr>
        <tr>
            <td><input type="hidden" name="id" value='<?php echo $id; ?>'>
            </td>
            <td><input type="submit" name="update" value="Update"></td>
        </tr>
    </table>
</form>
</body>
</html>

如果要在textarea输入中显示$Catdescription值，请先更改，然后重试
 <td><textarea name="Catdescription" value='<?php echo
                $Catdescription; ?>'>
            </textarea>
            </td>


在这两个查询中，您都通过了$mysqli
而不是$con
。感谢您的帮助，但还不能完全正常工作。将$mysqli
更改为$con
现在显示输入值中的$catname
。但不显示$Catdescription，也不更新数据库。
  <td><textarea name="Catdescription" ><?php echo
                $Catdescription; ?>
            </textarea>
            </td>