On Fri, Mar 6, 2009 at 4:07 PM, PJ <af.gour...@videotron.ca> wrote:
> haliphax wrote:
>> On Fri, Mar 6, 2009 at 3:00 PM, PJ <af.gour...@videotron.ca> wrote:
>>> I've been racking my little peanut-brain as well as big Google with
>>> little hope...
>>> I can retrieve the array from the multiple select dropdown box but I
>>> can't quite manage to insert the data which is just id numbers for a
>>> table.
>>> I've tried some while stuff but doesn't work. I know I don't have the
>>> right syntax on anything at this point. Here's an effort that doesn'
>>> work.
>>> Dont know if I should be using $categoriesIN[] type of stuff...
>>> but my tries were from web illustrations...
>>> the outcommented line returns the values of the array... but how to pass
>>> them to the query?
>>>
>>> foreach ($categoriesIN as $category) {
>>> Â Â //"$category<br>";
>>> $sql = "INSERT INTO book_categories ( book_id, category )
>>> Â Â VALUES( book.id WHERE title = $titleIN, $category )";
>>> Â Â Â Â $result = mysql_query($query, $db);;
>>> Â Â }
>>
>> Phil, you don't want to use an INSERT if you're just updating an
>> existing row. You can't do an INSERT and use the current row's values
>> (book.id/book_id/whatever) in your insert, since there IS no current
>> row. I'm assuming what you're trying to do is add a record to an
>> associative table that links Categories to Books, but only if that
>> Book title matches your $title string.
>>
>> Try this:
>>
>> foreach($categoriesIN as $category)
>> $sql = "update book_categories set category = '$category' where
>> book_id in (select id from book where title = '$titleIN')";
>>
>> If that's not what you're going for, and you just wanted to update
>> something's category, try this:
>>
>> foreach($categoriesIN as $category)
>> $sql = "update book_categories set category = '$category' where
>> book_id = '$somevalue'";
>>
>> Perhaps it would help us help you if you would explain what it is
>> you're trying to accomplish with this query?
> Hmmm, maybe I'm chasing my own tail...
> following the trail of execution is tough... :'(
> I'm trying to enter a new book into the database and that means I have
> to enter the info for the relational tables as well.
> I think I stumbled here a bit as what I need is the id that is inserted
> in the db before inserting this category stuff.
> As I look at the code, it looks like i have to find the
> "last-entered-id" for the book that is being entered.
Well, there are two ways you can do that with PHP/MySQL. First, you
can do it on the DB server. Second, you can do it in PHP after your
insert query has been performed.
MySQL:
SELECT LAST_INSERT_ID();
PHP:
mysql_query("insert into tablename(foo, bar) values('foo', 'bar');", $db);
echo mysql_insert_id();
If you're just trying to tie a book to existing categories, I would
insert the book and then separately insert the records in the
associative table (book_categories).
Pseudocode:
insert book into book table
grab last inserted id with mysql_insert_id()
for each category book is a part of
{
insert book (via last inserted id) and category IDs into book_categories
}
If you're trying to insert a book and a category and then link them
together, I would do this:
Pseudocode:
insert book into book table
grab last inserted book id with mysql_insert_id()
insert category into category table
grab last inserted category id with mysql_insert_id()
insert last book and last category IDs into book_categories
Hope this helps,
--
// Todd
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