Question

当我更新数据库设置表codeigniter上的行时。如果我只是更新一个项目，那么它会将每一行更改为已更新的那一行。如果我使用INSERT INTO它可以，但宁愿使用更新。

它应该只更新一个已更新的。

模型

<?php    
class Model_setting extends CI_Model {    
   public function editSetting($group, $data, $website_id = 0) {    
      //$this->db->query("DELETE FROM " . $this->db->dbprefix . "setting WHERE website_id  = '" . (int)$website_id . "' AND `group` = " . $this->db->escape($group) . " ");

      foreach ($data as $key => $value) {    
         if (!is_array($value)) {
            $this->db->query("UPDATE " . $this->db->dbprefix . "setting SET website_id = '" . (int)$website_id . "', 

            `group` = " . $this->db->escape($group) . ", 
            `key` = " . $this->db->escape($key) . ", 
            `value` = " . $this->db->escape($value) . "

            ");

         } else {

            $this->db->query("UPDATE " . $this->db->dbprefix . "setting SET 

            website_id = '" . (int)$website_id . "', 
            `group` = " . $this->db->escape($group) . ", 
            `key` = " . $this->db->escape($key) . ", 
            `value` = " . $this->db->escape(serialize($value)) . ", 
            serialized = '1'

            ");    
         }    
      }    
   }

   public function deleteSetting($group, $website_id = 0) {    
      $this->db->query("DELETE FROM " . $this->db->dbprefix . "setting WHERE 
         website_id = '" . (int)$website_id . "' AND 
         `group` = " . $this->db->escape($group) . "
      ");    
   }

   public function editSettingValue($group = '', $key = '', $value = '', $website_id = 0) {
      if (!is_array($value)) {

         $this->db->query("UPDATE " . $this->db->dbprefix . "setting SET 

            `value` = ". $this->db->escape($value) ." WHERE 
            `group` = " . $this->db->escape($group) . " AND 
            `key` = ". $this->db->escape($key) ."
            AND website_id = '" . (int)$website_id . "'

         ");

      } else {

         $this->db->query("UPDATE " . $this->db->dbprefix . "setting SET 

            `value` = ". $this->db->escape(serialize($value)) .", 
            serialized = '1' WHERE 
            `group` = " . $this->db->escape($group) ."
            AND `key` = " . $this->db->escape($key) . "
            AND website_id = '" . (int)$website_id . "'

         ");    
      }
   }       
}

Answer 1

您可以将WHERE子句与UPDATE查询一起使用来更新所选行，否则所有行都会受到影响。

$this->db->query("UPDATE " . $this->db->dbprefix . "setting SET website_id = '" . (int)$website_id . "', 

            `group` = " . $this->db->escape($group) . ", 
            `key` = " . $this->db->escape($key) . ", 
            `value` = " . $this->db->escape($value) . "
            WHERE `id`= $id ");
                    ^ replace with your condition

同样应该在else部分进行查询。

Answer 2

您错过了查询中的 WHERE 子句。也可以使用Codeigniter's Active Record代替原始SQL查询。

$data = array(
          'website_id' => $website_id,
          'group' => $group,
          'key' => $key,
          'value' => $value
       );

$this->db->where('id', $id);       // `id` is just for example
$this->db->update($this->db->dbprefix . 'setting', $data);

使用Active Records的优势：

1）使用Active Record功能的一个主要好处是它允许您创建独立于数据库的应用程序，因为查询语法是由每个数据库适配器生成的。
2）它提供了更简化的界面 3）它还允许更安全的查询，因为系统会自动转义值。

模型不更新正确的Codeigniter

2 个答案: