在PostgreSQL上错误ILIKE

问题描述 投票:-1回答:2

我有一个使用PostgreSQL的项目和新的。以下是我所面临的错误。我也在谷歌搜索但无法找到解决方案。

致命错误:在第162行的D:\ htdocs \ poes \ system \ application \ rapyd \ classes \ datafilter.php中调用未定义的方法CI_DB_postgre_driver :: ilike()

我正在使用codeigniter框架。此代码来自datafilter.php $this->db->ilike($name, $value);在我的控制器上,

$query1 = "SELECT DISTINCT users.* FROM users LEFT JOIN login ON users.user_id=login.login_user_id;";
$filter->db->query($query1);
$filter->name = new inputField("Name :", "user_name");
$filter->name->clause = "ilike";
php postgresql codeigniter-2
2个回答
0
投票

我只是解决了这个问题,需要在system / database / DB_active_rec.php中添加这个函数才能使用ilike

function ilike($field, $match = '', $side = 'both')
{
    return $this->_ilike($field, $match, 'AND ', $side);
}

function _ilike($field, $match = '', $type = 'AND ', $side = 'both', $not = '')
{
if ( ! is_array($field))
{
$field = array($field => $match);
}

  foreach ($field as $k => $v)
  {
$k = $this->_protect_identifiers($k);

$prefix = (count($this->ar_like) == 0) ? '' : $type;

$v = $this->escape_like_str($v);

if ($side == 'before')
{
$like_statement = $prefix." $k $not ILIKE '%{$v}'";
  }
  elseif ($side == ‘after’)
  {
$like_statement = $prefix." $k $not ILIKE '{$v}%'";
  }
  else
  {
$like_statement = $prefix." $k $not ILIKE '%{$v}%'";
  }

  // some platforms require an escape sequence definition for ILIKE wildcards
  if ($this->_like_escape_str != ‘’)
  {
$like_statement = $like_statement.sprintf($this->_like_escape_str, $this->_like_escape_chr);
}

  $this->ar_like[] = $like_statement;
  if ($this->ar_caching === TRUE)
  {
$this->ar_cache_ilike[] = $like_statement;
$this->ar_cache_exists[] = 'ilike';
}

  }
  return $this;
}

来自https://ellislab.com/forums/viewthread/87725/#808269的消息来源


0
投票

像这样

$param = "%".urldecode($this->uri->segment(4))."%";
            if(!empty($this->uri->segment(4))){
                $this->db->where("name ilike",$param); 
             }
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