Question

我有一些用Codeigniter框架编写的PHP代码，它不返回任何内容（空集）。看看它并告诉我它有什么问题。

function sbsn($serial){
    $this->db->select('asset_types.name as type_name,asset_brands.name as brand_name');
    $this->db->from('asset_types,asset_brands,assets');
    $this->db->where('assets.type_code','asset_types.code');
    $this->db->where('assets.brand_code','asset_brands.code');
    $this->db->where('serial_no',$serial); 
    $result = $this->db->get();
    return $result;
}

Answer 1

嗨，使用db get你只得到结果集而不是记录来获得结果你必须得到以下内容

$this->db->get()->result(); // will return an array of objects
$this->db->get()->result_array(); //will return result in pure array
$this->db->get()->row() // just single row as object
$this->db->get()->row_array() // just single row as array

you can use $result to perform the same above things
$result->result();
$result->row();

有关详细信息，请参阅生成结果user guide

Answer 2

变化：

$result = $this->db->get();

到

$result = $this->db->get()->row_array();  //to get single row
//OR
$result = $this->db->get()->result_array();  //to get multiple rows

或尝试使用JOIN，如

$this->db->select('asset_types.name as type_name,asset_brands.name as brand_name');
$this->db->from('assets');
$this->db->join('asset_types', 'asset_types.code = assets.type_code', 'left');
$this->db->join('asset_brands', 'asset_brands.code = assets.brand_code', 'left');
$this->db->where('assets.serial_no',$serial); 
$result = $this->db->get()->result_array();
return $result;

Answer 3

通过以下方式加入表格可以轻松解决问题。

$this->db->select('at.name as type_name,ab.name as brand_name');
$this->db->from('asset_types as at,asset_brands as ab');
$this->db->join('assets as a', 'a.type_code = at.code and a.brand_code as ab.code');
$this->db->where('a.serial_no',$serial); 
$result = $this->db->get()->result_array();
return $result;

在Codeigniter中查询结果为空

3 个答案: