Question

嘿我正在尝试使用我拥有的表格更新表格：

<?php
include ('includes/header.php'); 
require ('../db_con.php'); 

 // Check for a valid document ID, through GET or POST:
if ( (isset($_GET['id'])) && (is_numeric($_GET['id'])) ) { // From view_docs.php
    $id = $_GET['id'];

} elseif ( (isset($_POST['id'])) && (is_numeric($_POST['id'])) ) { // Form submission.
    $id = $_POST['id'];

} else { // No valid ID, kill the script.
    echo '<p class="error">This page has been accessed in error.</p>';
    exit();
}   

$q = "SELECT * FROM cats WHERE cat_id = $id";
$r = mysqli_query($dbc, $q);
?>

<form method="post" action="actions/update_cat.php">
    <input type="hidden" name="cat_id" value="<? echo $id ?>" />
    <input type="text" name="cat_name" />
    <br><br>
    <input type="text" name="cat_color" />
    <br><br>
    <input type="text" name="cat_icon" />
    <br><br>
    <input type="submit" value="submit" />
</form>

当我将其提交到我的动作脚本时，我知道它会传递输入表单中的值，但这是我的动作脚本 - 在任何人指出它之前，还没有防止SQL注入的措施：

<?php

require ('../../db_con.php'); 

$cat_id = $_POST['cat_id'];
$cat_name = $_POST['cat_name'];
$cat_color = $_POST['cat_color'];
$cat_icon = $_POST['cat_icon'];

$q = "UPDATE cats SET cat_name='cat_name', cat_color='cat_color', cat_icon='cat_icon'  WHERE cat_id='cat_id'";

if (mysqli_query($dbc, $q)) {
    echo "Record updated successfully";
} else {
    echo "Error updating record: " . mysqli_error($dbc);
}

mysqli_close($dbc);

?>

所以它向我报告已成功更新，但它没有更新表格？

Answer 1

您的查询在...中没有动态值

GetName

UPDATE报告成功但未更新表

1 个答案: