Echo'ing a piece of the phpMyAdmin database in wordpress - php

Question

I have a wordpress toggle function that toggles a status depending on the value in the wp_database. The user basically has the option to report sick and to report healthy. When the user reports sick i want a piece of tekst to echo out the date stored in the database. So if for example i reported myself sick on the 1st of December 2014 i want to echo out that date. The code i have so far is listed below.
$date = $wpdb->get_results( "SELECT sick FROM ziekbeter WHERE person =
$user_ID AND healthy IS NULL" );
$status="You reported sick on DATE.";
Image being the 'ziekbeter' table in my database. I know for a fact that the array it echo's only contains a single date.
Lets say that we're person 2 and we take a look at my code above.
SELECT sick FROM ziekbeter WHERE person = $user_ID AND healthy IS NULL
This would select the 'sick' row from table 'ziekbeter' (the whole table) but only the value of 'sick' where person = $user_ID (which is the user thats currently logged in to the system) and when the 'healthy' field is empty. This value (which if we're person 2 is 2014-11-13) will get put into $date. Now the only problem i face is that i need to echo this date on my website. Is there any way to do this?
The second image is of my website's front-end. You see the big red button which says 'report healthy'. If the user clicks that butten the date which they enter in the date field on the right gets put in the database (in the healthy field). (NOTE: All of this code works, im just looking for a way to output my $date on my page).
EDIT
If i var_dump the array i get the following code:
array(1) { [0]=> object(stdClass)#2315 (1) { ["sick"]=> string(10)
"2014-11-03" } }

Answer 1

What
print_r($array);
tells you?
Did you try
echo $array["sick"];
?
or maybe
echo $array->sick ?
$array being the name of the array containing the result of your MySQL query (the one you var_dump'ed).

Answer 2

Ok turns out the answer was rather simple.
$date2 = $wpdb->get_results( "SELECT sick FROM ziekbeter WHERE person = $user_ID AND healthy IS NULL" );
foreach ($date2 as $row) {
echo $row->sick;
}
I simply had to declare he had to echo the sick field again. The original SELECT did select the table at first but didnt remember that value when stored in a variable. I added a foreach row in the date variable and echo'ed the sick field. Since there is only 1 value ever in the table it has my desired result. Thanks all!