如何在帶有綁定參數的代碼iginiter query（）中使用like子句？

Question

如何在帶有綁定參數的代碼iginiter query（）中使用like子句？

例如：當我使用

$query = 'SELECT mycol FROM mytable WHERE name LIKE %?';
$name = 'foo';
$db->this->query($query,array($name));
//the clause generated
//SELECT mycol FROM mytable WHERE id LIKE '%'foo'%' 
//I expected this
//SELECT mycol FROM mytable WHERE id LIKE '%foo'

我不將參數值放在查詢中並按如下方式使用：$ query ='從mytable那里選擇mycol名稱，例如'％foo';

我也不能使用$ this-> db-> like（）函數，因為我的查詢包括：INSERT IGNORE和INSERT INTO表SELECT col FROM table2;

請提出建議？ 謝謝，

Answer 1

只需使用CodeIgniter查詢生成器

$name = 'foo';

$query = $this
    ->db
    ->select('mycol')
    ->like('name', $name)
    ->get('mytable');

Answer 2

$query = 'SELECT mycol FROM mytable WHERE name LIKE ?';
$name = '%foo';
$this->db->query($query,array($name));

Answer 3

codeigniter會取代? 具有'params'值。 如果你寫這個

 $query = 'SELECT mycol FROM mytable WHERE name LIKE %?';
 $name = 'foo';
 //$db->this->query($query,array($name)); //you wrote this line wrong.
 //it should be like this
  $this->db->query($query,array($name));

它會產生

SELECT mycol FROM mytable WHERE name LIKE %'foo' //inverse comma after % ,actually before and after foo.

所以你的正確方法是

 $query = 'SELECT mycol FROM mytable WHERE name LIKE ?';
 $name = '%foo';    
 $this->db->query($query,array($name));

它將產生

SELECT mycol FROM mytable WHERE name LIKE '%foo'

注意
你寫錯了

$db->this->query($query,array($name));

正確的路

 $this->db->query($query,array($name));