Re: [PHP] populate form input option dropdown box from existing data

Ford, Mike wrote:
> On 16 June 2009 20:48, PJ advised:
>
>> Now, I was happy to learn that it is simpler to populate the
>> insert new
>> books page dynamically from the db. Much shorter & neater.
>> It looks to me like the best solution for the edit page is
>> close to what
>> Yuri suggests.
>> Since the edit page is very similar to the insert new books page, I
>> merely need to populate the Select options box slightly differently.
>> This is the code to populate the insert page:
>> <select name="categoriesIN[]" multiple size="8">
>> <?php
>> $sql = "SELECT * FROM categories";
>> if ( ( $results = mysql_query($sql, $db) ) !== false ) {
>> while ( $row = mysql_fetch_assoc($results) ) {
>> echo "<option value=", $row['id'], ">", $row['category'],
>> "</option><br />"; }
>> }
>> </select>
>>
>> The problem nowis to find a way to add a conditional clause above that
>> will insert the option="selected" in the output.
>> The input for this comes from:
>> // do categories
>> $sql = "SELECT id, category FROM categories, book_categories
>> WHERE book_categories.bookID = $idIN &&
>> book_categories.categories_id = categories.id";;
>> if ( ( $results = mysql_query($sql, $db) ) !== false ) {
>> while ( $row = mysql_fetch_assoc($results) ) {
>> echo$row['id'], "<br />";
>> }
>> }
>>
>> This may return any number of category ids so the problem is to figure
>> out a way to pass the ids from the above code to the right ids in the
>> first code above. How & what do I search to match the two ids?
>
> Well, if I'm understanding your queries correctly, you need to compare
> the two sets of $row['id'] from the two queries above -- so your first
> query should be the second one above ("SELECT id, category FROM ..."),
> and you need to save the ids it returns for use in the loop which emits
> the <select>s. This can be done by replacing the "echo $row['id']" with
> "$selected_ids[] = $row['id']". Now you have an array of the selected
> ids which you can use in your in_array(). So your finished code is going
> to look something like this:
>
> <select name="categoriesIN[]" multiple size="8">
> <?php
> // do categories
> $selected_ids = array();
> $sql = "SELECT id, category FROM categories, book_categories
> WHERE book_categories.bookID = $idIN &&
> book_categories.categories_id = categories.id";
> if ( ( $results = mysql_query($sql, $db) ) !== false ) {
> while ( $row = mysql_fetch_assoc($results) ) {
> $selected_ids[] = $row['id'];
> }
> }
> $sql = "SELECT * FROM categories";
> if ( ( $results = mysql_query($sql, $db) ) !== false ) {
> while ( $row = mysql_fetch_assoc($results) ) {
> echo "<option value=", $row['id'],
> (in_array($row['id'], $selected_ids)?" selected":""),
> ">", $row['category'],
> "</option>\n";
> }
> }
> ?>
> </select>
>
> Hope this helps.
It does, indeed. This confirms my inexperienced conclusion that
in_array() does not work on associative arrays per se; it works on
simple arrays and I just don't have the experience to think of
extracting only the id fields.
I actually am using a slightly more complicated if else statement which
works.
Also, the other problem was the option selected definition required
Shawn's clarification
<select name='component-select' multiple ... which now highlights the
selected fields.
In all my searches (horrendously wasted time) I did not find any mention
of "component-select" either in php.net or w3c.org (I don't think my
queries on Google brought up anything from php.net) but w3c.org did and
I had looked at the page but somehow missed it.
I'm going to have to look at the way I search things. When you are
looking for something specific, other, even relevant, solutions seem to
get screened out.

Anyway, I learned quite a bit, here.
Thank you very, very much, gentlemen.
PJ

-- 
Hervé Kempf: "Pour sauver la planète, sortez du capitalisme."
-------------------------------------------------------------
Phil Jourdan --- [email protected]
http://www.ptahhotep.com
http://www.chiccantine.com/andypantry.php