PHP display error - php

Question

I've created the following code but for some reason it is echoing Array instead of the result:
<?php
include("../config.php");
include("functions.php");
$count = "SELECT `monthly_slots` FROM users WHERE `username` = 'Bill'";
$count = mysql_query($count);
$data = mysql_fetch_assoc($count);
echo "$data";
?>
Any ideas?
I have no idea why it is outputting Array because there should only be one result from that query.

Answer 1

mysql_fetch_assoc() returns an array, you need to use print_r($data) instead, to dump out the array's contents.

Answer 2

try this
print_r($data);
This will output the contents of the array.
The return type of mysql_fetch_assoc is array. So you should use print_r to see the result.
Returns an associative array that corresponds to the fetched row and moves the internal data
pointer ahead. mysql_fetch_assoc() is equivalent to calling mysql_fetch_array() with MYSQL_ASSOC
for the optional second parameter. It only returns an associative array.

Answer 3

As the array manual page explains, when you output an array variable in string context (as echo does here) it will become just "Array".
To see the array contents use print_r or var_dump instead:
print_r($data);
Or better yet just access the content you wanted:
print($data["monthly_slots"]);

Answer 4

use print_r($data) to print the $data array, or $data['column_name'] for a perticular row. because mysql_fetch_assoc() aloways returns an array.

Answer 5

Solutions
You could simply print out the whole contents of the array with
print_r($data)
however I wouldn't usually recommend doing it like that considering you're only looking for a specific item soo..
You could also try referencing the output as the first item (assuming there is only one result) with
echo "$data[0]";.
you can also simply reference the specific item in the array you are looking for with
echo $data['monthly_slots'].
References:
php arrays
mysql_fetch_assoc
print_r

Answer 6

It is expected. You should try print_r($data)