insert multiple records into one table (one form)

[GJPeacock]

Hi all,

Please excuse my limited technical knowledge of PHP, so far i have only
created PHP sites using Dreamweaver, i find know that i need features
outside of it's capabilities.

I am trying to create a 'related articles' selection page. The
administrator, after creating an article, can select older articles to
include on the final displayed page.

As the number of older articles increases, so will the number of
possible relataed articles, so i am making a dynamic checkbox as the
selection. This is all covered in Dreamweavers features, but, what i
then need to be able to do is create a new record in the
'related_articles' table for each article selected when the form is
submitted.

I found this post below, which is more or less exactly what i want to
do, i just wasn't sure that this was php code, looks like asp. Would
someone be kind enough to show me how to change it to php code, or even
write the code for me? ANY help at all would be much appreciated.

http://groups.google.ca/group/macrom...41afa2366e40e2

Thanks,

Gavin

[NC]

GJ*******@gmail.com wrote:

I am trying to create a 'related articles' selection page. The
administrator, after creating an article, can select older articles to
include on the final displayed page.
Is this the best way to do it? You could generate a list of related
articles dynamically, based on article's publication date and topic
(which, of course, means that you need to keep track of topics in your
database).
As the number of older articles increases, so will the number of
possible relataed articles, so i am making a dynamic checkbox as the
selection. This is all covered in Dreamweavers features, but, what i
then need to be able to do is create a new record in the
'related_articles' table for each article selected when the form is
submitted.
So what seems to be the problem? You receieve several ID numbers or
URLs via POST, and run one or more INSERT queries with this data.
I found this post below, which is more or less exactly what i want to
do, i just wasn't sure that this was php code, looks like asp. Would
someone be kind enough to show me how to change it to php code,
or even write the code for me?

Well, so far you haven't even mentioned what database you are using...

Cheers,
NC

[GJPeacock]

Hi,

Maybe i should re-phrase my question. I'm using php with a MySQL
database, and i would like to know how to insert a new record into a
table for each checked item on one form.

The reason i want to create a relation in this way, is because i want
to be able to choose which article appears as a related link regardless
of it's category/date/etc.

So what seems to be the problem? You receieve several ID numbers or
URLs via POST, and run one or more INSERT queries with this data.

No problem with this, it's just this is what i'm not sure of how to do.

Thanks, Gavin

[NC]

GJ*******@gmail.com wrote:

Maybe i should re-phrase my question. I'm using php with a MySQL
database, and i would like to know how to insert a new record into a
table for each checked item on one form.

OK, let's say you have a form:

<form method="POST" action="insert.php">
<input type="checkbox" name="id[]" value="32">Article #32<br>
<input type="checkbox" name="id[]" value="38">Article #38<br>
<input type="checkbox" name="id[]" value="45">Article #45<br>
<input type="checkbox" name="id[]" value="59">Article #59<br>
<input type="Submit">
</form>

Now, let's assume the user checked articles #32 and #59. These values
will be available to insert.php as fields in $_POST['id'], which in
this case will be an array. So in insert.php you can write:

$query = 'INSERT INTO related_articles (id) VALUES (' .
implode('), (', $_POST['id']) . ')';
$result = mysql_query($query)
or die('Could not execute INSERT query');

That's it, really...

Cheers,
NC

[GJPeacock]

Thanks for the reply NC, the trouble is i don't want the field to be an
array, i want each item that is selected to be inserted into a new row.
So if #32 and #59 were selected, #32 would be inserted into a row, then
a new row would be created to insert #59 into... and so on if there
were more articles selected.

[Jerry Stuckle]

NC wrote:

GJ*******@gmail.com wrote:

Maybe i should re-phrase my question. I'm using php with a MySQL
database, and i would like to know how to insert a new record into a
table for each checked item on one form.

OK, let's say you have a form:

<form method="POST" action="insert.php">
<input type="checkbox" name="id[]" value="32">Article #32<br>
<input type="checkbox" name="id[]" value="38">Article #38<br>
<input type="checkbox" name="id[]" value="45">Article #45<br>
<input type="checkbox" name="id[]" value="59">Article #59<br>
<input type="Submit">
</form>

Now, let's assume the user checked articles #32 and #59. These values
will be available to insert.php as fields in $_POST['id'], which in
this case will be an array. So in insert.php you can write:

$query = 'INSERT INTO related_articles (id) VALUES (' .
implode('), (', $_POST['id']) . ')';
$result = mysql_query($query)
or die('Could not execute INSERT query');

That's it, really...

Cheers,
NC

It also violates first normal form for relational databases.

Rather, something like (for simplicity, no error checking shown):

foreach ($_POST['id']) as $id) {
mysql_query("INSERT INTO related_table (article_id, related_id) " .
"VALUES ($origId, $id)");

$origId is the original article id, $_POST['id'] is the id(s) of the related
article(s).

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================

[NC]

GJ*******@gmail.com wrote:

Thanks for the reply NC, the trouble is i don't want the field to be an
array,
Then tell us a little more about your form, so we can give you advice
appropriate to your situation... Also, why don't you want the field to
be an array?
i want each item that is selected to be inserted into a new row.

This is exactly what this piece of code does. It creates one record
for each selected item. Obviously, the data in the example are
incomplete (the ID of the article to which selected articles are
related is missing, which is probably what prompted Jerry Stuckle's
comment about it violating first normal form for relational databases),
but it was only meant to show you the sequence of steps... Well, let's
try it again, this time with more data...

Let's say you have a form:

<form method="POST" action="insert.php">
<input type="checkbox" name="id[]" value="32">Article #32<br>
<input type="checkbox" name="id[]" value="38">Article #38<br>
<input type="checkbox" name="id[]" value="45">Article #45<br>
<input type="checkbox" name="id[]" value="59">Article #59<br>
<input type="hidden" name="referer" value="123">
<!-- This is the ID of the "referring" article -->
<input type="Submit">
</form>

Then, let's assume the user checked articles #32 and #59. These
values will be available to insert.php as fields in $_POST['id'], which

in this case will be an array. So in insert.php you can write:

$ref = $_POST['referer'];
$query = 'INSERT INTO related_articles (id, referer) VALUES (' .
implode(", $ref), (", $_POST['id']) . ", $ref)";
$result = mysql_query($query)
or die('Could not execute INSERT query');

Here's the query that is going to be executed:

INSERT INTO related_articles (id, referer)
VALUES (32, 123), (59, 123);

This query will create two records in related_articles, one linking
article #32 to article #123, the other linking article #59 to article
#123.

The advantage over Jerry Stuckle's solution is that all necessary
records are created in one query, although I must admit that this
advantage is probably trivial.

Cheers,
NC

[GJPeacock]

Thanks NC

This is exactly what i wanted, sorry if it took a few posts to get
here. Much appreciated.

Gavin