Question

我有两个不同的tables testimonial和pagetitles。从Pagetitles获取database时。我需要比较前4个字符，如果两个匹配，那么我应该得到数据。

function getpagetitle($id)
{

    $this->db->select('P.*,T.testimonial_name');        
    $this->db->from('pagetitle AS P');      
    $this->db->join('testimonials AS T','SUBSTR(T.testimonial_name, 4) = SUBSTR(P.page_title, 4)','INNER');
    $this->db->where(array('P.page_title'=>$id));       
    $q=$this->db->get();        
    //var_dump($this->db->last_query());
    //print_r($q->num_rows());
    if($q->num_rows()>0)
      {
    $output = $q->result();

   return $output[0];
        }
    else
    {
    return false;
    }
}

数据库表

推荐：

+----------------+-------------------+-------------+
| testimonial_id |  testimonial_name | client_name |
+----------------+-------------------+-------------+
| 1              |  testimonial      | abc         |
| 2              |  testimonial      | def         |
+----------------+-------------------+-------------+

PAGETITLE

+--------------+-------------+
| pagetitle_id | pagetitle   |   
+--------------+-------------+
|  1           | testimonial |
|  2           | career      |
+--------------+-------------+

Answer 1

我认为你必须在mysql中使用if then statement，因此，如果我们想要更改您的查询，它必须是这样的：

select P.*,T.testimonial_name from pagetitle AS P , testimonials as T (
 select IF 'SUBSTR(T.testimonial_name, 4) = SUBSTR(P.page_title, 4)'  
 where P.page_title>5
)

我不测试它，但我认为一定是真的

Answer 2

如果您还没有任何解决方案，请尝试以下代码，

$testimonials = $this->db->query("SELECT `P`.*, `T`.`testimonial_name` 
FROM (`pagetitle` P) INNER JOIN `testimonials` T ON 
((SUBSTR(`T`.`testimonial_name`, 1, 4)) = (SUBSTR(`P`.`page_title`, 1, 4))) 
WHERE `P`.`page_title` = SUBSTR('{$id}', 1, 4)")->result_array();
print_r($testimonials);
/* Output of $testimonials will be like below:
Array
(
    [0] => Array
        (
            [fieldname_1] => val
            [fieldname_2] => val
        )

    [1] => Array
        (
            [fieldname_1] => val
            [fieldname_2] => val
        )
)
*/
if(count($testimonials) > 0) {
//$output = $testimonials->result();
return $testimonials[0];
} else {
return false;
}

比较表中的前4个字符和从codeigniter中的数据库中获取数据

2 个答案: