Prepare Statement Issue.

[chuflasky]

Hi all.

I'm being working on an insert query using prepare statement to store information retrieve from the user. this are the query that i'm using.

$q = " INSERT INTO author(author_name) VALUES(?)";
$stmt = mysqli_prepare($dbc, $q);
mysqli_stmt_bind_param($stmt,'s', $author);
$author_id = mysqli_insert_id($dbc);

//            //                 //         //         
$q = " INSERT INTO categories( category_name) VALUES(?)";
$stmt = mysqli_prepare($dbc, $q);
mysqli_stmt_bind_param($stmt, 's', $category);
$category_id = mysqli_insert_id($dbc);


///        ///                ///              ///
$q = "INSERT INTO titles(title, user_id, category_id) VALUES(?,?,?)";
$stmt = mysqli_prepare($dbc, $q);
mysqli_stmt_bind_param($stmt,'s,i,i', $title, $user_id, $category_id);                  //Line 289
mysqli_stmt_execute($stmt) or die ("Error: " . mysqli_stmt_error($stmt) );            
$title_id = mysqli_insert_id($dbc);

 

 

For me everything look fine, but i getting the following error.

 

Warning: mysqli_stmt_bind_param() [function.mysqli-stmt-bind-param]: Number of elements in type definition string doesn't match number of bind variables inC:\xampp\htdocs\index2\user_php\upload.php on line 289Error: No data supplied for parameters in prepared statement

Can any one tell me what i doing wrong here. I suspect that mysqli_insert_id is causing the problem because i might no generating a id yet since the statement get execute down further the script.

 

 

[Larry]

The problem is you're separating sii with commas, which is not the correct syntax. It should just be sii.

[chuflasky]

Thanks Larry, that would do it.