PHP DB - assigning variables after one-to-many query

Author

assigning variables after one-to-many query

Rachel Rodriguez

2004-04-22, 10:30 pm

Hi!

I have a one-to-many relationship between two tables
(table1 and table3) with a "linking" table between
them (table2):

table1: table2
+---+--------+ +---+--------+
|id | f_name | |id | emailID|
+---+--------+ +---+--------+
| 1 | bill | | 1 | 1 |
| 2 | john | | 1 | 4 |
| 3 | diana | | 1 | 3 |
| 4 | victor | | 2 | 2 |
| 5 | renata | | 4 | 5 |
+---+--------+ +---+--------+

table3
+--------+-----------------+
|emailID | email |
+--------+-----------------+
| 1 | bill@hotmail.com|
| 2 | diana@yahoo.com |
| 3 | bill@yahoo.com |
| 4 | bill@excite.com |
| 5 | vic@hotmail.com |
+--------+-----------------+

I would like to write a query that matches table1.id
with records
from table3.emailID via the linking table (table2) and
then
assign each match to a variable.

Here is what I have:

$query = "SELECT t3.email
FROM table3 AS t3
LEFT JOIN table2 AS t2
ON (t3.emailID = t2.emailID)
LEFT JOIN table1 AS t1
ON (t2.id = t1.id)
WHERE t1.id = 1";

$result = @myql_query($query, $db_connection);

$num = mysql_num_rows($result);

$email1 = "";
$email2 = "";
$email3 = "";

if ($num > 0)
{
while ($row = mysql_fetch_array($result))
{
// do something here to assign
// $email1 = $row[email] and
// $email2 = $row[email], etc.
}
}

Of course, the problem I am having is as I am going
through the "while" loop, I am assigning $email1,
$email2, and so on the

same e-mail address.

I would like to get: $email1 = bill@hotmail.com,
$email2 = bill@excite.com, and $email3 =
bill@yahoo.com.

I would prefer to do that rather than what I have seen
in most examples which is similar to the following:

// snippet of code

while ($row = mysql_fetch_array($result))
{
echo"Bill's email is $row[email].";
}

// end of snippet

Any assistance is greatly appreciated.

Thanks,
Rachel

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Torsten Roehr

2004-04-23, 4:30 am

"Rachel Rodriguez" <email2rachel2003@yahoo.com> wrote in message
news:20040423015539.12050.qmail@web41805.mail.yahoo.com...
> Hi!
>
> I have a one-to-many relationship between two tables
> (table1 and table3) with a "linking" table between
> them (table2):
>
> table1: table2
> +---+--------+ +---+--------+
> |id | f_name | |id | emailID|
> +---+--------+ +---+--------+
> | 1 | bill | | 1 | 1 |
> | 2 | john | | 1 | 4 |
> | 3 | diana | | 1 | 3 |
> | 4 | victor | | 2 | 2 |
> | 5 | renata | | 4 | 5 |
> +---+--------+ +---+--------+
>
> table3
> +--------+-----------------+
> |emailID | email |
> +--------+-----------------+
> | 1 | bill@hotmail.com|
> | 2 | diana@yahoo.com |
> | 3 | bill@yahoo.com |
> | 4 | bill@excite.com |
> | 5 | vic@hotmail.com |
> +--------+-----------------+
>

Hi Rachel,

if you know that you will always have a one to many relationship and never a
many to many relationship you don't need table2 at all. Just add column 'id'
to table three as the foreign key. This should make your life easier.

Regards, Torsten

> I would like to write a query that matches table1.id
> with records
> from table3.emailID via the linking table (table2) and
> then
> assign each match to a variable.
>
> Here is what I have:
>
> $query = "SELECT t3.email
> FROM table3 AS t3
> LEFT JOIN table2 AS t2
> ON (t3.emailID = t2.emailID)
> LEFT JOIN table1 AS t1
> ON (t2.id = t1.id)
> WHERE t1.id = 1";
>
> $result = @myql_query($query, $db_connection);
>
> $num = mysql_num_rows($result);
>
> $email1 = "";
> $email2 = "";
> $email3 = "";
>
> if ($num > 0)
> {
> while ($row = mysql_fetch_array($result))
> {
> // do something here to assign
> // $email1 = $row[email] and
> // $email2 = $row[email], etc.
> }
> }
>
> Of course, the problem I am having is as I am going
> through the "while" loop, I am assigning $email1,
> $email2, and so on the
>
> same e-mail address.
>
> I would like to get: $email1 = bill@hotmail.com,
> $email2 = bill@excite.com, and $email3 =
> bill@yahoo.com.
>
> I would prefer to do that rather than what I have seen
> in most examples which is similar to the following:
>
> // snippet of code
>
> while ($row = mysql_fetch_array($result))
> {
> echo"Bill's email is $row[email].";
> }
>
> // end of snippet
>
> Any assistance is greatly appreciated.
>
> Thanks,
> Rachel
>
>
>
>
> __________________________________
> Do you Yahoo!?
> Yahoo! Photos: High-quality 4x6 digital prints for 25¢
> http://photos.yahoo.com/ph/print_splash

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