I write a website and use php. When ı write
$cat = mysql_query("
Select m.Image_link AS link
, m.Name
, m.ID
, c.Name AS catName
from movie m
join has_category mc
on m.ID = mc.movie_id
join category c
on c.ID = mc.category_id
where c.Name = '$category'
ORDER
BY ID desc"
);
this query everything is okay but when ı insert limit, ı take error.
@$sayfa=$_GET['s'];
$kacar=12;
$toplansayfa=ceil(mysql_num_rows($cat)/$kacar);
$baslangic=($sayfa * $kacar) - $kacar;
$bul_cat=mysql_query("
Select m.Image_link AS link
, m.Name
, m.ID
, c.Name AS catName
from movie m
join has_category Mc
on m.ID = mc.movie_id
join category c
on c.ID = mc.category_id
where c.Name = '$category'
ORDER
BY ID desc
limit $baslangic, $kacar
");
echo mysql_error();->
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-12, 12' at line 1
Where is my mistake..thanks for your interest.
As with many SQL programming errors, it helps a great deal to look at the actual text of the query you tried to run.
So, try this:
$query = "Select movie.Image_link AS link, movie.Name, movie.ID, category.Name AS catName from (movie inner join has_category on movie.ID=has_category.movie_id inner join category on category.ID=has_category.category_id) where category.Name='$category' ORDER BY ID desc limit $baslangic, $kacar";
$bul_cat = mysql_query($query); /*deprecated API! */
if (!$bul_cat) {
echo "query failed: " . $query;
echo mysql_error(); /* deprecated API! */
}
You will almost certainly spot the problem right away when you display the actual query you tried to run. I'm sure Paul Siegel's diagnosis is correct... your query says LIMIT -12,12
. You Can't Do That™.