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PHP

Shon Levi
Shon Levi
6,036 Points
Posted by Shon Levi
Shon Levi
6,036 Points

Can't prepare with mysqli!

hey you all,

I tried to make some prepare query but it won't working:

$user_id = 1;
$result = $db->prepare("SELECT * FROM users WHERE id = ?");
$result->bind_param('i', $user_id);
$result->execute();

When I var_dump it that's what I get:

object(mysqli_stmt)[2]
  public 'affected_rows' => int -1
  public 'insert_id' => int 0
  public 'num_rows' => int 0
  public 'param_count' => int 1
  public 'field_count' => int 3
  public 'errno' => int 0
  public 'error' => string '' (length=0)
  public 'error_list' => 
    array (size=0)
      empty
  public 'sqlstate' => string '00000' (length=5)
public 'id' => int 1

What can be the problem?! (When I make it with query and put manually 1 it working...)

1 Answer

jcorum
jcorum
71,830 Points
jcorum
jcorum
71,830 Points

Could it have anything to do with the id = ? in your SQL query:

$result = $db->prepare("SELECT * FROM users WHERE id = ?");

I know the DBMS I use would complain.