Number array

[Mutley]

I have a looping database query and want to create a variable for each one, there will only be 6 results from the query.

 

I tried:

$number = array(1,2,3,4,5,6);

...{
echo "$variable$number = $database_entry";
   }

 

But the array won't work, it's probably obviously wrong to you guys.

[fert]

$count=1;
while($row=mysql_fetch_array($result))
{
$var="result".$count;
$$var=$row['column'];
$count++;
}

I'm pretty sure that's what you want to do

[Mutley]

Doesn't appear to be working, I'm using a looping query list, like this:

 

    while (list($points) = mysql_fetch_row($res)) {
        echo "$points<br>";
    }

[Barand]

Why don't you put them into an array

<?php
$pointsArray = array();
while (list($points) = mysql_fetch_row($res)) {
        $pointsArray[] = $points;
}

// view array data

foreach ($pointsArray as $p) echo "$p<br>";
?>

[Mutley]

I receive lots of:

1

1

1

 

I need it to display the $points with a new variable for it, so it would show as:

 

$newvar1 = $points;

$newvar2 = $points;

$newvar3 = $points;

$newvar4 = $points;

 

But with a loop it would obvioulsy just be 1 of the above but I can't assign variables to it (as it would just loop the same number/variable).

[fert]

http://us2.php.net/manual/en/language.variables.variable.php

[Barand]

Then fert's earlier suggestion should work - read the link he provided

<?php
$count=1;
while (list($points) = mysql_fetch_row($res)) {
    $var = 'newvar'.$count;
    $$var = $points;
    ++$count;
}

echo $newvar1. '<br>';
echo $newvar2. '<br>';
echo $newvar3. '<br>';
echo $newvar4. '<br>';
echo $newvar5. '<br>';
echo $newvar6. '<br>';

// if you had a 1000, would you still insist on this method?
?>

[Mutley]

I got this to work:

 

<?php

$count=0;
$p1_slot = 'p1_slot';
$sql = "SELECT u.user_id, u.slot, p.name, p.points
            FROM loadout u
                INNER JOIN products p ON u.prod_id = p.prod_id WHERE u.user_id = '1'
            ORDER BY u.user_id, u.slot";
            
    $res = mysql_query($sql) or die (mysql_error()."<p>$sql</p>");
    while (list($user, $slot, $prod, $points) = mysql_fetch_row($res)) {
        echo "$$p1_slot$count = $points<br>";
	$count++;
    }

?>

 

$p1_slot0 = 3

$p1_slot1 = 5

$p1_slot2 = 4

$p1_slot3 = 4

$p1_slot4 = 10

$p1_slot5 = 3

 

But if I wanted to use these variables in a part of the script further on, how would I do that? As at the moment they don't show in the code, only when echoed.

[Mutley]

If I take away the echo to make them variables (not displayed as text) I get a unexpected variable message.

[per1os]

<?php

$count=0;
$p1_slot = 'p1_slot';
$sql = "SELECT u.user_id, u.slot, p.name, p.points
            FROM loadout u
                INNER JOIN products p ON u.prod_id = p.prod_id WHERE u.user_id = '1'
            ORDER BY u.user_id, u.slot";
            
    $res = mysql_query($sql) or die (mysql_error()."<p>$sql</p>");
    while (list($user, $slot, $prod, $points) = mysql_fetch_row($res)) {
          $newVar = "p1_slot" . $count;
          $$newVar = $points<br>";
	$count++;
    }

?>

 

I thought barand pointed this out already, but here it is for a second time.

[Mutley]

I got that to work.

 

Can you link to any resources or reply how I would do this without listing the query results, just using something like $row['field'] instead? I've never used the "INNER JOIN" SELECT method before, so not sure how it exactly works.