从id = id的其他表中获取信息

Question

所以我有2张桌子。

train_information
user_train_information

当有人在表单中提交某些内容时。 它放在train_information表中，看起来像这样：

现在，当人们登录并从选择器中选择火车时。 他发生在数据库中：

在另一个页面上，我希望用户能够查看他们在一段时间内选择的所有内容。 所以我运行一个查询： SELECT * FROM user_train_information WHERE user_id=user_id;

这显示了表user_train_information但是可以显示train_information user_id = user_id吗？ 因为我希望用户显示他添加的火车。

编辑：

我现在拥有的：

function testingggg() {
        $sql = "SELECT *
FROM train_information
INNER JOIN user_train_information
ON train_information.train_id = user_train_information.train_id
WHERE user_train_information.user_id = user_id";
        $sth = $this->pdo->prepare($sql);
        $sth->bindParam("user_id", $_GET["user_id"], PDO::PARAM_INT);
        $sth->execute();
        return $sth->fetchAll();
    }

我在这里打电话给我：

<?php
        $summary = $database->testingggg();
    ?>
    <?php
        foreach ($summary as $result){
        echo $result['train_name']; 
    }
?>

我收到错误：

Answer 1

这就是我在这种情况下使用查询的方式，它的工作原理。 在你的情况下，某处有歧义。 还要确保$ _GET [“user_id”]中有适当的值。 请检查

   function testingggg() {
        $sql = "SELECT ti.train_name, ti.train_id, uti.user_id
                FROM train_information as ti
                JOIN user_train_information as uti
                ON ti.train_id = uti.train_id
                WHERE uti.user_id = :user_id";
        $sth = $this->pdo->prepare($sql);
        $sth->bindParam(":user_id", $_GET["user_id"], PDO::PARAM_INT);
        $sth->execute();
        return $sth->fetchAll();
    }

Answer 2

SELECT t.train_id
     , t.country_id train_country_id
     , t.train_name
     , t.number_of_bogies
     , t.number_of_axles
     , t.wheel_diameter_min
     , t.wheel_diameter_max
     , u.user_row_id
     , u.user_id
     , u.country_id user_country_id
  FROM user_train_information u
  JOIN train_information t
    ON t.train_id = u.train_id
 WHERE u.user_id = :user_id;

Answer 3

你试试这个。

    SELECT * FROM train_information INNER JOIN user_train_information 
     ON train_information.train_id = user_train_information.train_id
    WHERE user_train_information.user_id = $user_id;

Answer 4

SELECT * 
FROM   train_information 
       INNER JOIN user_train_information 
               ON train_information.train_id = user_train_information.train_id 
WHERE  user_train_information.user_id = user_id;