[英]How to Render options of an form with the values fetched from the database with php?
[英]PHP how to submit <select> options values which are fetched from mysql
到目前为止,我正在获取特定数据并将其显示在选择标记中作为我的代码的选项:
<?php
$db = mysqli_connect('localhost', 'root', '', 'e-shop');
$sql = 'SELECT * FROM categories';
$result2 = mysqli_query($db,$sql);
?>
<select name='category' >
<?php while($row1 = mysqli_fetch_array($result2)):;?>
<option value="<?php echo $row1[0];?>"><?php echo
$row1[1];?></option>
<?php endwhile;?>
</select>
我几乎没有上下文,我正在为我的书籍表单获取类别,但我不知道如何将该值分配给类别
这是我提交的有效代码:
<?php
$db = mysqli_connect('localhost', 'root', '', 'e-shop');
if (isset($_POST['add-product'])) {
$title = $_POST['title'];
$description = $_POST['description'];
$price = $_POST['price'];
$categoryy = $_POST['category'];
$target_dir = "../img/";
$fileName = basename($_FILES["image"]["name"]);
$target_file = $target_dir . $fileName;
move_uploaded_file($_FILES["image"]["tmp_name"], $target_file);
$db_dir = "../img/";
$db_file = $db_dir . $fileName;
if($_FILES["image"]["name"]){
$sql = sprintf("INSERT INTO products (title, category, picture, description, price) VALUES ('%s', '%s', '%s', '%s', '%s')", $title, $categoryy, $db_file, $description, $price);
} else {
$sql = sprintf("INSERT INTO products (title, description, category, price) VALUES ('%s', '%s', '%s', '%s')", $title, $description, $categoryy, $price);
}
mysqli_query($db, $sql);
mysqli_close($db);
}
?>
我想我需要一个名字来选择标签,我试过但没有用,
显然它确实有效,但我给了错误的行索引,该值是 id 而不是类别名称! 并命名标签作品
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