Question

According to this question I succeeded to create upload image, but now I need to store the image_id into another table called articles. I do not know if this is correct, but I tried to select the image_id from table image like this

$select_image=mysql_query("select image_id from image where image_name = $fileName") or die(mysql_error());

and fetch the result to my article insert query like this

$fetch=mysql_fetch_array($select_image);

$query=mysql_query("insert into articles (article_name,article_category,article_subcategory,article_body,article_summary,article_tags,article_photo,article_timedate) values ('$article_title','$CategoryID','$ProductID','$article_body','$article_summary','$fetch[image_id]','$time')") or die ('Error, Query Faild'.mysql_error());

Is this correct ? The mysql_error keeps saying

Unknown column ‘Penguins.jpg’ in ‘where clause’

Answer 1

you can select the last inserted id by calling

SELECT LAST_INSERT_ID()

instead of

select image_id from image where image_name = $fileName