I don't think you're suppose to end your queries with a semicolon. Try:
$sql = "SELECT * FROM photographs WHERE
photo_filename LIKE '%2%' LIMIT 0, :q";
On Mon, Dec 12, 2011 at 1:49 PM, Stephen <stephen-d@xxxxxxxxxx> wrote:
> So I am getting this SQL error:
>
> Error selecting photographs: SQLSTATE[42000]: Syntax error or access
> violation: 1064 You have an error in your SQL syntax; check the manual that
> corresponds to your MySQL server version for the right syntax to use near
> ''4'' at line 2
>
> My code is:
>
> function updatephotos($dbh, $x) {
>
> echo $x['number'] . "<br />"; <<<<< this echo is 4
>
> $sql = "SELECT * FROM photographs WHERE
> photo_filename LIKE '%2%' LIMIT 0, :q;";
>
> $stmt = $dbh->prepare($sql);
>
> try {
>
> $stmt->bindValue( ':q', $x['number'], PDO::PARAM_INT );
> $stmt->execute();
>
> } catch (PDOException $e) {
> return 'Error selecting photographs: ' . $e->getMessage();
> }
>
> while ( list( $id, $name, $alt, $caption) = $stmt->fetch(PDO::FETCH_NUM)) {
> echo $name . "<br />";
> }
>
>
> return "test worked" ;
> }
>
> If I hard code the SQL as:
>
> $sql = "SELECT * FROM photographs WHERE
> photo_filename LIKE '%2%' LIMIT 0, 4";
>
> all works well.
>
> Can anyone see what is wrong?
>
> How can I see the prepared SQL statement before it is executed?
>
> Thanks
> Stephen
>
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