Question

我尝试在codeigniter模型中运行查询。它工作但是当我回显模型函数查询时如下所示。

＆＃13;

SELECT * FROM `table1` WHERE `id` = '17'
SELECT * FROM `table1` WHERE `id` = '20'
SELECT * FROM `table1` WHERE `id` = '21'
SELECT * FROM `table1` WHERE `id` = '22'
SELECT * FROM `table1` WHERE `id` = '23'

＆＃13;

我的模型功能如下：

＆＃13;

function get_quick_navi_menu($q_code)
{
$this->db->select("*");
$this->db->where('q_id',$q_code);
$this->db->from("table0");
$q = $this->db->get();
//echo $this->db->last_query();
$final = array();
if ($q->num_rows() > 0) 
{
foreach ($q->result() as $row) {
$this->db->select("*");
$this->db->from("table1");
$this->db->where("id",$row->id);
$q = $this->db->get();
echo $this->db->last_query();
if ($q->num_rows() > 0) {
$row->children = $q->result();
}
array_push($final, $row);
}
}

＆＃13;

我想运行如下的查询

＆＃13;

SELECT * FROM `table1` WHERE `id` = '17,18,19..'

＆＃13;

＆＃13; 表结构

＆＃13;

Table0

id   q_id  value1

1      2       4

2      2       5

3      2       6

Table1

t1_id   id  value1   value2

 1      1      2         2
 
 2      2      5         6
 
 3      3      8         12

＆＃13;

＆＃13; 视图

＆＃13;

<ul class="dropdown-menu" aria-labelledby="dropdownMenu1" style="margin-left: 1px; opacity: .9;">
						
						<?php foreach ($menus as $menu) { ?>	
                         <li class="dropdown-submenu"><a href="#" class="pan-btn" data-look="<?php echo $menu->sceneid;?>"><?php echo $menu->title;?></a>  
                           <ul class="dropdown-menu">        <?php
            if (isset($menu->children)) {
                foreach ($menu->children as $child) {?>
                
                <li><a href="#" class="pan-btn" data-look="<?php echo $child->menu_scene;?>"><?php echo $child->menu_item;?></a></li> <?php
                }
            }
            ?></ul></li><?php } ?>
 						</ul>

＆＃13;

＆＃13; 控制器

＆＃13;

$menus = $this->Home_model->get_quick_navi_menu($q_code);
$data = array('menus' => $menus);

＆＃13;

必需的输出

根据table0中的id从table1中选择value1和value2。如何解决这个请帮助。

Answer 1

在控制器中使用此功能

public function getTableData()
    {
        $this->db->select('GROUP_CONCAT(id) as id');
        $tbl0 = $this->db->get('table0')->row_array();
        if($tbl0) {
                $ids = explode(',', $tbl0['id']);
                $this->db->where_in('id', $ids);
                $tbl1 = $this->db->get('table1')->result_array();
                echo "<pre>"; print_r($tbl1);

        }
    }

Answer 2

希望这会对您有所帮助;

在此处获取数组名称中的所有ids并在循环外使用where_in

public function get_quick_navi_menu($q_code)
{
    $this->db->select("*");
    $this->db->where('q_id',$q_code);
    $this->db->from("table0");
    $q = $this->db->get();
    $final = array();
    if ($q->num_rows() > 0) 
    {
        foreach ($q->result() as $key => $row) 
        {
            $ids[$key] = $row->id;
            $data[$key] = $row;
        }

        $this->db->select("*");
        $this->db->from("table1");
        $this->db->where_in("id",$ids);
        $q = $this->db->get();
        //echo $this->db->last_query();
        if ($q->num_rows() > 0) 
        {
            if ( ! empty($data))
            {
                foreach ($data as $key => $item) 
                {
                    $item->children = $q->result();
                    $final[] = $item;
                }
            }

        }
    }
    return $final;
    /*print_r($final);*/
}

在您的控制器类中：

确保您已在控制器或autoload.php中加载模型和数据库

 $q_code = 'q_code_value';
 $data['menus'] =$this->Home_model->get_quick_navi_menu($q_code);
 /* pass the $data in the view like this*/
 $this->load->view('your_view_file_path',$data);

在您看来：

<div><?php print_r($records);?></div>

更多信息：https://www.codeigniter.com/user_guide/database/query_builder.html#looking-for-specific-data

Codeigniter模型函数和查询不起作用

2 个答案: