Question

Im trying to do a select * from x where name LIKE y.I am very familiar with this code which is below.

SELECT * FROM ` arm` WHERE `u_fname` LIKE 'k%'

However, im trying to run this on a sever script and i am un sure where to put the “%” cause it results in errors. This is part of the sever code, this code works as it is with there ‘=’ but errors out when i use LIKE becuase im unsure how to use the “%” here

   elseif(empty($_POST["f_name"]) &&  !empty($_POST["l_name"])) 
{ 

 $l_name=$_POST["l_name"];
$stmt = $this->db->prepare('SELECT id, image, l_name, f_name, status  FROM table1  WHERE l_name =?');
           $stmt->bind_param("s",$l_name);
         $stmt->execute();
         $stmt->bind_result($id,$image,$l_name,$f_name,$status);   

I tried SELECT id, image, l_name, f_name, status FROM table1 WHERE l_name =?% and i have also tried putting $l_name=$_POST[“l_name%”]; but neither of them seem to work.

Thank you for help.

Answer 1

Your code isn’t using ‘LIKE’
Your code is: SELECT id, image, l_name, f_name, status FROM table1 WHERE l_name =?
Try

$stmt = $this->db->prepare('SELECT id, image, l_name, f_name, status  FROM table1
WHERE l_name LIKE ?'); 
$stmt->bind_param("s",$l_name.'%');

Or try

$stmt = $this->db->prepare('SELECT id, image, l_name, f_name, status  FROM table1
WHERE l_name LIKE CONCAT(?, "%")');
$stmt->bind_param("s",$l_name);