正确连接表而无需重复数据

Question

我试图连接表以防止太多的数据库查询,但我不喜欢数据返回的方式.

查询:

SELECT person.name, dog.dog_name FROM person JOIN dog ON person.id = dog.person_id

返回看起来像这样:

Array
(
[0] => Array
    (
        [name] => Jim
        [dog_name] => Wolf
    )

[1] => Array
    (
        [name] => Jim
        [dog_name] => Frisky
    )

[2] => Array
    (
        [name] => Tina
        [dog_name] => Pokedot
    )

[3] => Array
    (
        [name] => Tina
        [dog_name] => Spiky
    )
)

是否可以让查询返回如下内容:

Array
(
[0] => Array
    (
        [name] => Jim
        [dog_name] => array(Wolf, Frisky)
    )

[1] => Array
    (
        [name] => Tina
        [dog_name] => array(Pokedot, Spiky)
    )
)

Answer 1

最接近的解决方案是:

SELECT person.name, GROUP_CONCAT(dog.dog_name) AS dog_names
FROM person JOIN dog ON person.id = dog.person_id
GROUP BY person.id

这将返回一个字符串,该字符串是逗号分隔的狗名称列表,而不是PHP数组.你必须explode()在应用程序代码中使用该字符串.

请注意,GROUP_CONCAT()的默认长度限制为1024个字符,并由配置选项控制group_concat_max_len.

我同意@KonstantinWeitz的评论,让RDBMS尽其所能,然后使用PHP代码将结果后处理成您想要的格式是值得的.

例如,这是我如何返回你描述的数组:

$peoplesdogs = array();
$stmt = $pdo->query("SELECT person.name, dog.dog_name FROM person JOIN dog ON person.id = dog.person_id");
while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
  $peoplesdogs[$row["name"]]["name"] = $row["name"];
  $peoplesdogs[$row["name"]]["dog_name"][] = $row["dog_name"];
}
return array_values($peoplesdogs);