Explode array and assign as variables

Question

I have a table in the following structure...

ID     USER_ID      INTEREST
1      290          soccer
2      290          tennis
3      290          badminton
4      560          netball

I want to grab all values from the table where the user id is equal to the session user id and assign them as variables.

I have the following which works but displays the data very peculiarly..

$interestsquery  = "SELECT * FROM user_interests WHERE user_id = " . $usersClass->userID();
$result = mysql_query($interestsquery);

while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
 $interests[] = $row['interest'];
 $interest1 = $interests[1];
 $interest2 = $interests[2];
 $interest3 = $interests[3];

 print $interest1 . " - " . $interest2 . " - " . $interest3;

}

The above however outputs something along the lines of...

- - tennis - - tennis - badminton -

Can anybody see where I'm going wrong?

What is the structure of your table and what are you trying to do? — Cam
– Cam, Commented Jul 7, 2011 at 8:20
I want to grab all values from the table where the user id is equal to the session user id and assign them as variables. — Liam
– Liam, Commented Jul 7, 2011 at 8:21

Edoardo Pirovano · Accepted Answer · 2011-07-07 08:25:06Z

2

This should do what you need:

$interestsquery  = "SELECT `interest` FROM `user_interests` WHERE `user_id` = " . $usersClass->userID();
$result = mysql_query($interestsquery);

$interests = array();
while(list($interest) = mysql_fetch_array($result))
    $interests[] = $interest;

print explode($interests, ' - ');

answered Jul 7, 2011 at 8:25

Edoardo Pirovano

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2 Comments

Andy Over a year ago

ah, you should just do this then: echo '<pre>' . print_r($interests,true) . '</pre>';

Liam Over a year ago

Thanks Andy, Is it possible however to assign the results as a variable? So... $interest1 = Soccer; $Interest2 = Tennis; etc etc...

Eineki · Accepted Answer · 2011-07-07 08:43:16Z

Think what happens to your loop at the first iteration:

$interest is an empty array
you fetch the first value and put it into $interest[0],
you fill $interest1 with the value that lies into $interest[1] (it is empty)
same for $interest2 and $interest3
you print ""." - ".""." - ".""

in the second run:

$interest is [0=>soccer]
you fetch the second value and put it into $interest[1],
you fill $interest1 with the value that lies into $interest[1] tennis
same for $interest2 and $interest3 (that are still empty)
you print "tennis"." - ".""." - ".""

and so on.

You need print the result when you exit the while loop (and the code is still flawed as it don't get the value into the index 0 of the array).

An alternative should be:

$interestsquery  = "SELECT `interest` FROM `user_interests` WHERE `user_id` = " 
                 . $usersClass->userID();
$result = mysql_query($interestsquery);

$interests = array();
// you fetch just a field, fetch_row will be sufficent
while($interest = mysql_fetch_row($result)) {
    array_push($interests, $interest[0]);
}
echo implode(' - ', $interests);

Subdigger · Accepted Answer · 2011-07-07 08:44:27Z

0

$interests = array();

while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
  $interests[] = $row['interest'];
}

$int_count = count($interests);
$i=0;

foreach ($interests as $interest) {
    $var = 'interest' . ($i + 1);
    $$var = $interest;
    $i++;
} 

print $interest1 . " - " . $interest2 . " - " . $interest3;

answered Jul 7, 2011 at 8:44

Subdigger

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