php - 将参数从 Controller 传递到模型以进行 mysql 查询的优化方法

Question

我正在 condeigniter(新手)中从事一个项目。我的项目包括使用不同过滤器(如类别、子类别、关键字、最低-最高价格等)列出产品。我已经在自定义 php 中执行了数百次这样的操作，其中我为所有条件创建变量 $where where 子句。例如

$where = '';
if($category != '')
{
  $where .= " category = '$category'";
}

现在在codeigniter中我有多个参数，我可以通过uri段进入我的 Controller 并调用model函数，传递所有参数并在那里运行 mysql 查询。问题是我正在检查每个参数是否为空或设置并相应地运行多个查询。例如。

if($category != '' && $subCategory != '')
{
  //whole query here
}
else  if($category == '' && $subCategory != '')
{
  //whole query here
}

//and so on

我需要的是任何优化方法来执行此操作，因为我有连接(多个参数)。

我的示例条件和查询

if($subCat == '' && $vendor == '')
    {
        $cat = str_replace("-"," ",str_replace("-and-"," & ",$cat));
        $this->db->select('listings.* ');
        $this->db->from('listings');
        $this->db->join('categories','categories.id = listings.category');
        $this->db->where('categories.name',$cat);
        $this->db->limit($perpage,$page);
        $query = $this->db->get();
        return $query->result();
    }
    else if($subCat == '' && $vendor != '')
    {
        $cat = str_replace("-"," ",str_replace("-and-"," & ",$cat));
        $this->db->select('listings.* ');
        $this->db->from('listings');
        $this->db->join('categories','categories.id = listings.category');
        $this->db->join('vendors','vendors.id = listings.programme');
        $this->db->where('categories.name',$cat);
        $this->db->where('vendors.name',$vendor);
        $this->db->limit($perpage,$page);
        $query = $this->db->get();
        return $query->result();
    }
    else if($subCat != '' && $vendor == '')
    {
        $cat = str_replace("-"," ",str_replace("-and-"," & ",$cat));
        $this->db->select('listings.* ');
        $this->db->from('listings');
        $this->db->join('categories','categories.id = listings.category');
        $this->db->join('subcategories','subcategories.id = listings.subcategory');
        $this->db->where('categories.name',$cat);
        $this->db->where('subcategories.name',$subCat);
        $this->db->limit($perpage,$page);
        $query = $this->db->get();
        return $query->result();
    }
    else
    {
        $cat = str_replace("-"," ",str_replace("-and-"," & ",$cat));
        $subCat = str_replace("-"," ",str_replace("-and-"," & ",$subCat));
        $vendor = str_replace("-"," ",str_replace("-and-"," & ",$vendor));

        $this->db->select('listings.* ');
        $this->db->from('listings');
        $this->db->join('categories','categories.id = listings.category');
        $this->db->join('subcategories','subcategories.id = listings.subcategory');
        $this->db->join('vendors','vendors.id = listings.programme');
        $this->db->where('categories.name',$cat);
        $this->db->where('subcategories.name',$subCat);
        $this->db->where('vendors.name',$vendor);
        $this->db->limit($perpage,$page);
        $query = $this->db->get();  
        return $query->result();
    }

Answer 1

您只需编写一次查询，而不是每次都编写。例如，您可能有您所描述的几种情况。例如，这是我所做的事情:

$this->db->select('resource.id,city.name as city_name,city.provinceID,resource.Rate, skills.name')->from('resource')
->join('city','resource.cityID = city.id')
->join('resourceToSkillsMap','resource.id = resourceToSkillsMap.resourceID')
->join('skills','skills.id =  resourceToSkillsMap.skillsID');

if($category != '' && $subCategory != '') #condition
{
    $this->db->where('condition');
}
else  if($category == '' && $subCategory != '') #condition
{
    $this->db->where('condition');
}

$rs = $this->db->group_by('resource.id')
->limit($limit,$offset)
->get();

查看查询如何仅编写一次，但where 条件在if else 内给出。所以你的查询将以这种方式建立。执行查询后，只需 echo $this->db->last_query(); 即可回显查询并根据需要进行调整。