我有两个 mysql 表（产品和类别）。我在两个表中都有一些模拟数据。现在我需要以某种方式将类别附加到产品上。例如 - ID 为 1 的产品应返回以下内容：
| product name | category | | Monitor | Technology |
我知道我以前已经这样做过，但今天我似乎找不到解决方案。
编辑 这是我到目前为止所拥有的。连接运行良好，我可以在表格中显示数据。
<?php
// Include database connection
include("connection.php");
// Create variables for later use
$db = $conn;
$tableName = "Produkte";
$columns= ['id_product', 'name_product'];
// Create variable to use in index.php
$fetchData = fetch_data($db, $tableName, $columns);
// The function below feteches data from the tables specified and checks if the colums are emtpy by any chance.
function fetch_data($db, $tableName, $columns) {
// Check db connection
if (empty($db)) {
$message= "Database connection error";
}
// Check if the columns variable is empty and not an array by any chance
elseif (empty($columns) || !is_array($columns)) {
$message="Product Name must be defined in an indexed array";
}
// Check if table name is empty
elseif (empty($tableName)) {
$message= "Table Name is empty";
}
// Else proceed as usual.
else {
$columnName = implode(", ", $columns);
// The query needs to be repalced. Today my SQL stuff is leaving me a bit.
$query = "SELECT p.".$columnName." AS product, c.name_category FROM $tableName p JOIN Kategorie c ON c.id_";
$result = $db->query($query);
if ($result== true) {
if ($result->num_rows > 0) {
$row= mysqli_fetch_all($result, MYSQLI_ASSOC);
$message= $row;
}
else {
$message= "No Data Found";
}
}
// Throw error if error occures
else{
$message= mysqli_error($db);
}
}
return $message;
}
表 products 只有 2 列。 id 列和 product_name 列。
基本技术有所不同：
// create DBquery using JOIN statement $query = " SELECT p.name AS product, c.name AS category FROM products p JOIN categories c ON c.id = p.category_id;"; // get DB data using PDO $stmt = $pdo->prepare($query); $stmt->execute(); // show table header printf('| product name | category |' . PHP_EOL); // loop for output result as table rows while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) { printf('| %-12s | %10s |' . PHP_EOL, $row['product'], $row['category']); }在线尝试