[英]Make first option value null in select menu coming from MySQL database
[英]Value from MYSQL is coming null
我正在使用爆破功能從Codeignitor中獲取代碼。 這是代碼
$this->load->database();
$session_data = $this->session->userdata('logged_in');
$user_id = $session_data['user_id'];
$this->db->select('skill_id');
$this->db->from('user_info_skillset');
$this->db->where('user_id',$user_id);
$query = $this->db->get();
foreach($query->result() as $row)
{
$skill_id[] = $row->skill_id;
}
$test = implode(',',$skill_id);
// echo '<pre />';
// print_r($test); exit;
$this->db->select('project_id');
$this->db->from('project');
$this->db->where_in('required_skills',$test);
$query1 = $this->db->get();
echo '<pre />';
print_r($query1); exit;
return $query1->result();
問題是我無法獲取數據
echo '<pre />';
print_r($query1); exit;
return $query1->result();
當我嘗試在mysql工作台中手動輸入此數據庫查詢時,它正在工作,但是通過代碼顯示空值。 代碼中缺少什么嗎? 請指導我。 以下是我的輸出。
產量
CI_DB_mysql_result Object
(
[conn_id] => Resource id #34
[result_id] => Resource id #38
[result_array] => Array
(
)
[result_object] => Array
(
)
[custom_result_object] => Array
(
)
[current_row] => 0
[num_rows] => 0
[row_data] =>
)
您應該在查詢中使用FIND_IN_SET
子句:
$test = implode(',',$skill_id);
$where = " FIND_IN_SET(required_skills,'".$test."') ";
$this->db->select('project_id');
$this->db->from('project');
$this->db->where( $where, false );
//$this->db->where_in('required_skills',$test);
$query1 = $this->db->get();
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