Codeigniter count_all_results 有

Question

我使用 Codeigniter 的查询生成器编写了一个查询 class。该查询使用别名和 having 方法。当我对此查询调用 count_all_results 方法时，发生异常。检查日志，我发现查询已删除 'having' 子句。有没有办法在调用 count_all_results 时保留这些子句？感谢您的帮助。

编辑：我首先认为问题是基于知识而不是基于代码的，因此没有共享代码，但在这里。如果需要更多，请告诉我。

这是控制器中模型的调用。

$where_array = array(
    $parent_key.' is not NULL' => null
);
$search_post = $request_data['search'];
if (isset($request_data['filter'])) {
    $filter_array = $request_data['filter'];
    foreach ($filter_array as $filter_pair) {
        if (isset($filter_pair['escape'])) {
            $where_array[$filter_pair['filterBy']] = null;
        } else {
            if ($filter_pair['filterBy'] == 'table3_id') {
            $where_array['table3.'.$filter_pair['filterBy']] = isset($filter_pair['filterId']) ?
                $filter_pair['filterId'] : null;
            } else {
                $where_array[$table.'.'.$filter_pair['filterBy']] = isset($filter_pair['filterId']) ?
                    $filter_pair['filterId'] : null;
            }
        }
    }
}
$like_array = array();
foreach ($request_data['columns'] as $key => $column) {
    if (!empty($column['search']['value'])) {
        $like_array[$column['data']] = $column['search']['value'];
    }
}
$totalFiltered = $this->$model_name->modelSearchCount($search, $where_array, $like_array);

这是模型方法。

public function modelSearchCount($search, $where_array = null, $like_array = null)
{
    $this->joinLookups(null, $search);
    if ($where_array) {
        $this->db->where($where_array);
    }
    if ($like_array) {
        foreach($like_array as $key => $value) {
            $this->db->having($key." LIKE '%". $value. "%'");
        }
    }
    return $this->db->from($this->table)->count_all_results();
}

protected function joinLookups($display_config = null, $search = null)
{
    $select_array = null;
    $join_array = array();
    $search_column_array = $search ? array() : null;
    $i = 'a';

    $config = $display_config ? $display_config : $this->getIndexConfig();
    foreach ($config as $column) {
        if (array_key_exists($column['field'], $this->lookups)) {
            $guest_model_name = $this->lookups[$column['field']];
            $this->load->model($guest_model_name);
            $join_string =$this->table.'.'.$column['field'].'='.$i.'.'.
                $this->$guest_model_name->getKey();
            $guest_display = $this->$guest_model_name->getDisplay();
            if ($search) {
                $search_column_array[] = $i.'.'.$guest_display;
            }
            $join_array[$this->$guest_model_name->getTable().' as '.$i] = $join_string;
            $select_array[] = $i.'.'.
                $guest_display;
        } else {
            $select_array[] = $this->table.'.'.$column['field'];
            if ($search) {
                $search_column_array[] = $this->table.'.'.$column['field'];
            }
        }
        $i++;
    }
    $select_array[] = $this->table.'.'.$this->key;
    foreach ($join_array as $key => $value) {
        $this->db->join($key, $value, 'LEFT');
    }
    $this->db->join('table2', $this->table.'.table2_id=table2.table2_id', 'LEFT')
        ->join('table3', 'table2.table3_id=table3.table3_id', 'LEFT')
        ->join('table4', $this->table.'.table4_id=table4_id', 'LEFT')
        ->join('table5', 'table4.table5_id=table5.table5_id', 'LEFT');
    $this->db->select(implode($select_array, ', '));
    if ($search) {
        foreach (explode(' ', $search) as $term) {
            $this->db->group_start();
                $this->db->or_like($this->table.'.'.$this->key, $term);
            foreach ($search_column_array as $search_column) {
                $this->db->or_like($search_column, $term);
            }
            $this->db->group_end();
        }
    }
    $this->db->select('table2_date, '. $this->table.'.table2_id, table4_id, '. 'table5.table5_description');
}

Answer 1

为了回答这个问题，count_all_results() 通过将您的选择替换为 SELECT COUNT(*) FROM table[=14= 来转换原始查询].不会选择别名列，并且 having 子句不会识别该列。这就是 count_all_results() 不适用于 having 的原因。

Answer 2

因为 count_all_results() 基本上运行一个 Select count(*) 并且不计算结果集中的行（基本上使查询对您的目的无用）您可以使用其他 Codeigniter 方法来获取结果集和行数。

尝试运行将查询转换为变量：

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

从那时起，您几乎可以做任何事情。除了 return 使用 $query->result(); 计算结果外，您还可以使用以下方法将行数放入另一个变量：

 $rownum = $query->num_rows();

然后您可以 return 将其放入您的控制器中，甚至 return $query 对象然后运行控制器上的 num_rows() 方法

Codeigniter count_all_results 有

Codeigniter count_all_results with having

codeigniter

codeigniter-query-builder