Php Mysql查询需要优化和更新

这是我的Codeigniter查询

 $this->db->select('o.*,seo.price as seo_price,SUM(domain.price) as domain_price,SUM(bes.price) as bes_price, SUM(op.price) as payment_price');
    $this->db->from('order as o');
    $this->db->join("order_package_seo as seo", "seo.order_id = o.id", "left");
    $this->db->join("order_package_domain as domain", "domain.order_id = o.id", "left");
    $this->db->join("order_package_bespoke as bes", "bes.order_id = o.id", "left");
    $this->db->join("order_payment as op", "op.order_id = o.id", "left"); 
    $this->db->where("o.id" , $id);
    $query = $this->db->get();
    return $query->row();

我在“order\u package\u beskope”表中有3条记录，在其他所有表中有1条记录 我想通过订单id从所有表中获取价格

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我是否需要使用“组”_bydistinct或任何其他方法，需要您的建议。

更改第一个语句，如下所示

$this->db->select('o.*,seo.price as seo_price,domain.price ,SUM(bes.price) as bes_price, op.price as payment_price');
$this->db->from('order as o');
$this->db->join("order_package_seo as seo", "seo.order_id = o.id", "left");
$this->db->join("order_package_domain as domain", "domain.order_id = o.id", "left");
$this->db->join("order_package_bespoke as bes", "bes.order_id = o.id", "left");
$this->db->join("order_payment as op", "op.order_id = o.id", "left"); 
$this->db->where("o.id" , $id);
$this->db->group_by("o.id");
$query = $this->db->get();
return $query->row();

因为您提到表“order\u package\u beskope”有3条记录，所以只有表字段“bes\u price”需要总和。所有其他表只有一条记录，因此如果加上SUM，它将乘以3（订单\包装\定制记录的数量）

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因此，请删除“按o.id选择并添加分组”中其他表字段中的总和。好，那么它返回的是什么。它返回的是所有价格乘以3。i、 e域名的价格是123，但它的回报是369。如果你能帮助我，我可以解释更多。。嗨@IrfanAli，英语中没有像：update这样的词。我已经改正了。检查此项：您是否已尝试添加

$this->db->group_by（“o.id”）在
$this->db->where（“o.id”，$id）之后？