如何使用 mysql 在 php 中顯示每家餐廳的評論？

Question

我試圖顯示每家餐廳的評論數量。

我制作了 2 個包含我的數據的表格。

CREATE TABLE `restaurants` (
  `id` int(4) NOT NULL AUTO_INCREMENT,
  `name` varchar(200) NOT NULL,
  `city` varchar(200) NOT NULL,
  `country` varchar(200) NOT NULL,
  `score` int(1) NOT NULL DEFAULT '0',
  `reviews` int(1) NOT NULL DEFAULT '0',
  `slug` varchar(200) NOT NULL DEFAULT 'slug-test',
  `approved` int(1) NOT NULL DEFAULT '0',
  `description` text NOT NULL,
  `review` int(11) DEFAULT '0',
  `created_date` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
  `updated_date` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
  `img_url` varchar(255) NOT NULL,
  `category` varchar(255) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=65 DEFAULT CHARSET=latin1

CREATE TABLE `reviews` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `restaurant_id` int(11) NOT NULL,
  `review_text` text NOT NULL,
  `score` int(11) NOT NULL,
  PRIMARY KEY (`id`),
  KEY `restaurant_id` (`restaurant_id`),
  CONSTRAINT `reviews_ibfk_1` FOREIGN KEY (`restaurant_id`) REFERENCES `restaurants` (`id`) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1

我已經在列表中顯示了我的餐廳。

$query = "SELECT * FROM restaurants WHERE approved = 1 ORDER BY created_date DESC";
$result = $mysqli->query($query);

while($row = $result->fetch_array()) {
**HTML WITH MY VARIABLES**
}

這就是你目前顯示我的評論的方式（我的餐廳表中的硬編碼結果。

if ($reviews <= "0") {
echo "<a href='#' title='Write reviews'><p class='purple-def-color fz-12 mb-0'>write a review</p></a>";
} else if ($reviews > "1") {
echo "<p class='purple-def-color mb-0'>". $row['reviews'] ." reviews</p>";
} else {
echo "<p class='purple-def-color mb-0'>" . $row['reviews'] . " review</p>";
}

我嘗試使用 JOIN 查詢，但沒有成功以任何方式顯示評論數量。

Answer 1

嘗試左連接：

$query = 'SELECT * FROM restaurants LEFT JOIN reviews on reviews.restaurant_id = restaurants.id'

左連接返回左表（餐館）中的所有記錄，以及右表（評論）中的匹配記錄

Answer 2

根據您的表結構，您可以通過加入reviews表、選擇評論的 COUNT 條並按餐廳 ID 分組來獲取每家餐廳的評論計數。

按餐廳分組可讓您獲得每個餐廳的已加入評論記錄的聚合（例如count ）。

SELECT
    rs.*,
    COUNT(rv.`id`) as `reviewCount`
FROM `restaurants` rs
LEFT JOIN `reviews` rv
    ON (rv.`restaurant_id` = rs.`id`)
WHERE rs.`approved` = 1
GROUP BY rs.`id`
ORDER BY rs.`created_date` DESC;

然后，當您獲取 PHP 中的行時，您可以引用每一行的評論計數：

echo $row['reviewCount'];

---

為了進一步演示聚合函數，這里有一個示例，說明如何 select 每個餐廳的平均、 最低和最高評論分數：

SELECT
    rs.*,
    COUNT(rv.`id`) as `reviewCount`,
    AVG(rv.`score`) as `reviewAverageScore`,
    MIN(rv.`score`) as `reviewMinScore`,
    MAX(rv.`score`) as `reviewMaxScore`
FROM `restaurants` rs
LEFT JOIN `reviews` rv
    ON (rv.`restaurant_id` = rs.`id`)
WHERE rs.`approved` = 1
GROUP BY rs.`id`
ORDER BY rs.`created_date` DESC;