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如何连接到表格并显示其内容? (MySQL 和 PHP)

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;
}

products 只有 2 列。 id 列和 product_name 列。

P粉738821035P粉738821035239 天前469

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  • P粉014293738

    P粉0142937382024-04-04 00:33:33

    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

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