我试图从两个联接表中获取数据,从一个表中我收集列表标题,从第二个联接表中我收集列表数据.

型号代码：

function view_searching_type_items()

{

$this->db->select("searching_type.searchtype_name, search_type_item.st_item_id_pk, search_type_item.searchtype_itemname");

$this->db->from('searching_type');

$this->db->join('search_type_item', 'search_type_item.st_id_fk = searching_type.st_id_pk');

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

return $query->result_array();

}

控制器：

$url1 = 'http://localhost/pakistanmenu/index.php/Restaurant/ViewSearchingTypeItems';

$curl_hand = curl_init($url1);

curl_setopt($curl_hand, CURLOPT_TIMEOUT, 5);

curl_setopt($curl_hand, CURLOPT_CONNECTTIMEOUT, 5);

curl_setopt($curl_hand, CURLOPT_RETURNTRANSFER, true);

$data1 = curl_exec($curl_hand);// json_decode($data)

curl_close($curl_hand);

$result['searchingtype'] = (array) json_decode($data1,true);

$this->load->view('header');

$this->load->view('index',$result);

$this->load->view('footer');

查看代码：

<?php echo $searchingtyp['searchtype_name'];?>

请检查下面的图像,我已经显示了此代码的输出,并且我还画了我需要的输出,谢谢,请帮助我.谢谢

解决方法:

只需存储您的姓氏,并且仅在新名称与其不同时才打印它.发生这种情况时,我们还将创建新列表,以便每个组具有不同的列表.

// Define our test variable

$lastName = null;

foreach ($searchingtype as $searchingtyp):

?>

if ($lastName != $searchingtyp['searchtype_name']):

// It's not the same name

if ($lastName != null):

// Close the previous list (except from the first iteration).

?>

endif;

?>

?>

• <?php echo $searchingtyp['searchtype_name'];?>

// Store the current name so we can check again in the next iteration

$lastName = $searchingtyp['searchtype_name'];

endif;

?>

// The rest of your code.. just move the

标签：php,codeigniter,loops

来源： https://codeday.me/bug/20191012/1903106.html