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因此，我從頭開始演示網上商店，我似乎無法弄清楚爲什麼我的表單在提交時表單爲空時不保留數據。當我編輯表單中的值時，它就在那裏，可以成功更新到數據庫，與刪除一樣，但是當我嘗試添加一個新的時，它不會通過並將其添加到我設置的數據庫中。我錯過了什麼嗎？提前致謝！在php和MYSQL中的邏輯問題
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; ?>
調試，我的男人，調試。在這兩種情況下，所有相關變量都充滿了您期望的值嗎？如果您在phpmyadmin或任何您使用的版本中進行了複製（來自調試，而不是手工填寫），您的查詢是否正確並正確工作？ – deg
你錯過了INSERT SQL值部分的括號，你應該添加錯誤檢查。 –
一些明智的代碼縮進將是一個好主意。它可以幫助我們閱讀代碼，更重要的是，它可以幫助您**調試您的代碼** [快速瀏覽編碼標準]（http://www.php-fig.org/psr/psr-2/ ）爲了您自己的利益。您可能會被要求在幾周/幾個月內修改此代碼 ，最後您會感謝我。 – RiggsFolly