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因此，我从头开始演示网上商店，我似乎无法弄清楚为什么我的表单在提交时表单为空时不保留数据。当我编辑表单中的值时，它就在那里，可以成功更新到数据库，与删除一样，但是当我尝试添加一个新的时，它不会通过并将其添加到我设置的数据库中。我错过了什么吗？提前致谢！在php和MYSQL中的逻辑问题
require_once '../core/init.php';
include 'includes/head.php';
include 'includes/navigation.php';
// Get brands from database
$sql ="SELECT * FROM brand ORDER BY brand";
$results = $db->query($sql);
$errors = array();
//Edit brand
if (isset($_GET['edit']) && !empty($_GET['edit'])) {
$edit_id = (int)$_GET['edit'];
$edit_id = sanitize($edit_id);
$sql2 = "SELECT * FROM brand WHERE id = '$edit_id'";
$edit_result = $db->query($sql2);
$eBrand = mysqli_fetch_assoc($edit_result);
}
// Delete brand
if (isset($_GET['delete']) && !empty($_GET['delete'])) {
$delete_id = (int)$_GET['delete'];
$delete_id = sanitize($delete_id);
$sql = "DELETE FROM brand WHERE id = '$delete_id'";
$db->query($sql);
header('Location: brands.php');
}
// if add form is submmited
if (isset($_POST['add_submit'])) {
$brand = sanitize(mysqli_real_escape_string($db, $_POST['brand']));
// check if brand is blank
if ($_POST['brand'] == '') {
$errors[] .= 'You must enter a brand!';
}
// check if brand exists in database
$sql = "SELECT * FROM brand WHERE brand = '$brand'";
if (isset($_GET['edit'])) {
$sql = "SELECT * FROM brand WHERE brand = '$brand' AND id != '$edit_id'";
}
$result = $db->query($sql);
$count = mysqli_num_rows($result);
if ($count > 0) {
$errors[] .= $brand.' already exists. Please choose another brand
name...';
}
// display errors
if (!empty($errors)) {
echo display_errors($errors);
}else {
// add brand to database
$sql = "INSERT INTO brand (brand) VALUES '$brand'";
if (isset($_GET['edit'])) {
$sql = "UPDATE brand SET brand = '$brand' WHERE id = '$edit_id'";
}
$db->query($sql);
header('Location: brands.php');
}
}
?>
<h2 class="text-center">Brands</h2><hr>
<!-- Brand form -->
<div class="text-center">
<form class="form-inline" action="brands.php<?=((isset($_GET['edit']))?'?
edit='.$edit_id:'');?>" method="post">
<div class="form-group">
<?php
$brand_value = '';
if (isset($_GET['edit'])) {
$brand_value = $eBrand['brand'];
}else {
if (isset($_POST['brand'])) {
$brand_value = sanitize($_POST['brand']);
}
} ?>
<label for="brand"><?=((isset($_GET['edit']))?'Edit':'Add a');?> Brand:
</label>
<input type="text" name="brand" id="brand" class="form-control" value="<?
=$brand_value; ?>">
<?php if(isset($_GET['edit'])): ?>
<a href="brands.php" class="btn btn-default">Cancel</a>
<?php endif; ?>
<input type="submit" name="add_submit" value="<?=
((isset($_GET['edit']))?'Edit':'Add');?> brand" class="btn btn-success">
</div>
</form>
</div><hr>
<table class="table table-bordered table-striped table-condensed"
style="width:auto; margin:0 auto;">
<thead>
<th></th><th>Brand</th><th></th>
</thead>
<tbody>
<?php while($brand = mysqli_fetch_assoc($results)) : ?>
<tr>
<td><a href="brands.php?edit=<?=$brand['id'];?>" class="btn btn-xs btn-
default"><span class="glyphicon glyphicon-pencil"></span></a></td>
<td><?=$brand['brand']; ?></td>
<td><a href="brands.php?delete=<?=$brand['id'];?>" class="btn btn-xs
btn-default"><span class="glyphicon glyphicon-remove-sign"></span></a></td>
</tr>
<?php endwhile; ?>
调试，我的男人，调试。在这两种情况下，所有相关变量都充满了您期望的值吗？如果您在phpmyadmin或任何您使用的版本中进行了复制（来自调试，而不是手工填写），您的查询是否正确并正确工作？ – deg
你错过了INSERT SQL值部分的括号，你应该添加错误检查。 –
一些明智的代码缩进将是一个好主意。它可以帮助我们阅读代码，更重要的是，它可以帮助您**调试您的代码** [快速浏览编码标准]（http://www.php-fig.org/psr/psr-2/ ）为了您自己的利益。您可能会被要求在几周/几个月内修改此代码 ，最后您会感谢我。 – RiggsFolly