如何使用CodeIgniter 3阻止select语句中的自动修复。
我想要的查询:SELECT REPLACE(item, ',', '') AS item
(没有空格)
我尝试了:$this->db->select("replace(item, ',', '') AS item", FALSE);
结果:SELECT REPLACE(item, ', ', '') AS item
(CI在我的查询中添加了空格)
我看了this documentation但没有工作
这是我的真实查询
$this->db->select("a.unit_id");
$this->db->select("replace(c.topik, ',', '') AS topik", FALSE); //this my problem (added space which shouldnt)
$this->db->select("COUNT(*) AS jumlah");
$this->db->from('opub_kliping a');
$this->db->join('opub_media b','a.media_id = b.organisasi_id');
$this->db->join('opub_tag c','a.id = c.kliping_id');
if(!empty($data['keyword']))$this->db->where("(a.judul LIKE '%".$data['keyword']."%' OR a.text LIKE '%".$data['keyword']."')");
if(!empty($data['tgl1']))$this->db->where("a.tanggal >=",$data['tgl1']);
if(!empty($data['tgl2']))$this->db->where("a.tanggal <=",$data['tgl2']);
if(!empty($data['unit']))$this->db->where("a.unit_id ",$data['unit']);
if(!empty($data['media']))$this->db->where("a.media_id ",$data['media']);
if(!empty($data['jenis']))$this->db->where("a.jenis_id ",$data['jenis']);
if(!empty($data['tone']))$this->db->where("a.karakter ",$data['tone']);
if(!empty($data['topik']))
{
$qTopik = array();
$expTopik = explode(',', $data['topik']);
foreach($expTopik as $t)
{
if(!empty($t))
{
$qTopik[] = "c.topik LIKE '%$t%'";
}
}
if(!empty($qTopik))
{
$qTopiks = implode(",", $qTopik);
$this->db->where($qTopiks);
}
}
$this->db->where('a.unit_id !=', 0);
$this->db->where('c.topik !=', NULL);
$this->db->where('c.topik !=', '');
$this->db->group_by("a.unit_id");
$this->db->group_by("replace(c.topik, ',', '')", FALSE); //this produce correct query (without adding space)
$this->db->order_by('a.unit_id', 'ASC');
$this->db->order_by('jumlah', 'DESC');
$q= $this->db->get();
感谢先进的
答案 0 :(得分:0)
更改查询的4行
$this->db->select("a.unit_id,replace(c.topik, ',', '') AS topik,COUNT(*) AS jumlah")->from('opub_kliping a');
$this->db->join('opub_media b','a.media_id = b.organisasi_id');
答案 1 :(得分:0)
我在尝试了一些实验后发现了这个
for ( i = j = 0 ; s[i] != '\0' ; i++)
{
if (s[i] != q[k])
s[++j] = s[i];
}
似乎很可能是黑客。但它的工作对我来说 希望能帮助别人