php - CodeIgniter 中的 SQL 查询

Question

我在 CI 模型中有一个查询。我需要使这个查询简短。我需要为每个字母表获取前五个构建器名称。有什么办法可以让它更短吗？

型号

public function getBuilders() {        
    $query1 = $this->db->select('builder_name')->from('builders_list')->where('builder_name like', 'a%')->limit('5')->get();
    $query2 = $this->db->select('builder_name')->from('builders_list')->where('builder_name like', 'b%')->limit('5')->get();
    $query3 = $this->db->select('builder_name')->from('builders_list')->where('builder_name like', 'c%')->limit('5')->get();
    $query4 = $this->db->select('builder_name')->from('builders_list')->where('builder_name like', 'd%')->limit('5')->get();
    $query5 = $this->db->select('builder_name')->from('builders_list')->where('builder_name like', 'e%')->limit('5')->get();
    $query6 = $this->db->select('builder_name')->from('builders_list')->where('builder_name like', 'f%')->limit('5')->get();
    $result1 = $query1->result();
    $result2 = $query2->result();
    $result3 = $query3->result();
    $result4 = $query4->result();
    $result5 = $query5->result();
    $result6 = $query6->result();
    return array($result1, $result2, $result3, $result4, $result5, $result6); 
}

Answer 1

您可以使用以下代码:

$this->db->select('builder_name')
         ->from('builders_list')
         ->like('name','a%')
         ->or_like('name','b%')
         ->or_like('name','c%')
         ->or_like('name','d%')
         ->or_like('name','e%')
         ->or_like('name','f%')
         ->limit(5)
         ->get();