Retrieving a column from mySQL database and passing to array

Question

I have previously constructed an array manually, e.g.:

<?php
$noupload = array('nick', 'cliff');
?>

but now I am trying to populate the array automatically from users in a MySQL database. So far I have this:

<?php
// Make a MySQL Connection
$debug=false;
$conn = mysql_connect("host","user","password"); // your MySQL connection data
$db = mysql_select_db("database");

$query = "SELECT * FROM users WHERE access LIKE '%listen%'"; 

$result = mysql_query($query) or die(mysql_error());

while($row = mysql_fetch_array($result)){
    $listen = "'". $row['login']. "', ";
}
?>

$listen constructs the user list in the way that I previously entered it manually (I tested this using echo), but I am not sure how to pass this to $noupload. I tried:

$noupload = array($listen);

but this didn't work. I think I'm close and would be grateful for some help over the final hurdle,

Thanks,

Nick

To confirm - you may get multiple results from your SQL query and you want to create a string array with the 'login' column from each result? — lampej
– lampej, Commented Apr 30, 2011 at 19:52
Also, why have a SELECT * if you're only accessing the login attribute, you can SELECT login FROM users WHERE access like "%listen%". If you don't need more data then the login field, don't ask the DB for more. — Oerd
– Oerd, Commented Apr 30, 2011 at 20:25

user732694 · Accepted Answer · 2011-04-30 20:02:56Z

1

You can declare $listen as an array

$listen = array();
while ($row = mysql_fetch_array($result)) {
    $listen[] = "'". $row['login']. "'";
}

edited Apr 30, 2011 at 20:02

answered Apr 30, 2011 at 19:51

user732694

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7 Comments

Nick Over a year ago

Thanks, that worked great! I will mark as the answer in a couple of minutes

Nick Over a year ago

I can see you have now edited the line $listen[] = "'". $row['login']. "'"; It worked before you edited it. Should I change it?

Oerd Over a year ago

As a side note, use $listen[] = "'{$row['login']}'"; It saves two unnecessary string concatenations. (. operator)

cOle2 Over a year ago

This makes an array but from your example I think you just want $listen[] = $row['login']; ie drop the quotes. It seems you were only adding the quotes to simulate the syntax of declaring an array.

Nick Over a year ago

Yes, the edited version above doesn't work, but the original did work, with the line $listen[] = $row['login'];.

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