Error putting data into database - php

Question

I have the following php code:
foreach($html->find('dl[class=movie-info]') as $info) {
for($i = 0; $i <= 20; $i++) {
$contenido = $info->find('dt',$i)->plaintext;
if($contenido == 'Año'){
$year = utf8_encode($info->find('dd',$i)->plaintext);}}}
(the code has more if functions)
And a mysql table where I put the content of the variables....
The problem is with the $year content, I need to fill it in a smallint(5) unsigned.
When I use
$con = mysqli_connect("localhost","root","tdguchiha","phpbb3");
mysqli_query($con,"INSERT INTO pablo (forum_id, calidad, titulo, caratula, sinopsis, pais, director, reparto, genero) VALUES ('$forum_id', '$calidad', '$titulo', '$img', '$sinopsis', '$pais', '$director', '$reparto', '$genero')");
mysqli_close($con);
All the content is inserted, but when i try to insert $year into año with type smallint(5) unsigned nothing happens, no row is created...
how can I convert $year to a number (it must be a number) to fill it in that column? or I need to change the column type?
PD: I am learning right now to "play" with mysql
thanks

Answer 1

There is no difference in mysql between a number and a string when doing queries. Make sure you include the column name in the list of all the columns:
mysqli_query($con,"INSERT INTO pablo (forum_id, calidad, titulo, caratula, sinopsis, pais, director, reparto, genero, year) VALUES ('$forum_id', '$calidad', '$titulo', '$img', '$sinopsis', '$pais', '$director', '$reparto', '$genero', '$year')");
^notice year the column ^now the actual value
Just replace year in the column section with the name of your mysql column.

Answer 2

Perhaps try casting it to an integer instead of using utf8_encode:
$year = intval($info->find('dd',$i)->plaintext);
Or, in addition to it:
$year = intval(utf8_encode($info->find('dd',$i)->plaintext));
Can you give us an example of the data?
EDIT: Damien is right, intval() shouldn't make a difference.
Try echoing out the contents instead, and making sure it's an actual number in the string to do some debugging:
if($contenido == 'Año'){
$year = utf8_encode($info->find('dd',$i)->plaintext);
echo $year.'<br />';}}}

Answer 3

Before posting a question to StackOverflow, identify any error-codes / error-messages and post them with your question:
/* check connection */
if ( mysqli_connect_errno() ) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
And:
if ( ! mysqli_query( $link, "SELECT ..." )) {
printf("Errorcode: %d\n", mysqli_errno($link));
}
More on the PHP doc site.

Answer 4

I think your problem is with value type and data you are trying to insert:
A smallint is between -32768 and 32767 signed, or 0 and 65535 unsigned.
The (5) represents the display width of the field - if you will try to put 90000 you will be rejected.
In your case it should not be a problem (years value I assume) but show us the data you are trying to insert.
I guess that you are parsing some content using simple_html_dom - make sure all data retrieved is
as you expect before inserting it to DB.
Also: Try changing the type to int(5) and now tell us....
It will be easier to see the query you use (with the year) and a sample of data.
For error loging and display use:
$execute = mysqli_query($con,"INSERT INTO pablo ( ....your query..... ")
or
printf("Errormessage: %s\n", mysqli_error($con));
Have Fun