Problem with the implode function.

[pthurmond]

Hey everyone. I am having a weird error with the implode function. I have used it dozens of times, but this time it doesn't want to work.

Here is the error:
[quote]
Warning: implode(): Bad arguments. in /home/content/p/a/l/paldata/html/TeamB/user_volunteer_edit.php on line 90
[/quote]

Here is the code:
[code]
$query = "SELECT Activity_ID FROM Activity_List WHERE Volunteer_ID = '$vid'";
$result = mysql_query($query) or die('Query failed: ' . mysql_error());

if($result)
{
$i = 0;

while($row = mysql_fetch_array($result, MYSQL_BOTH))
{
$acts[$i] = $row[0];
$i++;
}

$acts = implode(",", $acts);
}
[/code]


I have tried it with single quotations first with no luck and originally I was referring to the array with $acts[] in the while loop.
Any ideas?

Thanks,
Patrick

[chiprivers]

What exactly are you trying to do here? Why put the values into an array if you are going to take them out again into a string? Why not just build a string in the first place?

[pthurmond]

Because I am having trouble passing an array to a function. Of course you have a point, but this is slightly easier to code.
None the less, I would like to know why this is giving me issues. If nothing else but to learn from my mistakes.

[JasonLewis]

well from what i can gather $acts has nothing in it. try doing print_r($acts) and see whats in there.