MySQL => PHP Associative Array...Understanding the undefined variable error

Question

I am trying to store data in an array, and then output a key of that array whenever I want. I have a table of staff data and am successfully using print_r to print the array, but cannot use "echo $staff_data;" to do anything. I feel like I am very close, but it is slipping by me.

I have two functions:

function staff_data() {
    $staff_data = array();
    $query = mysql_query("SELECT * FROM `staff`");
    while ($row = mysql_fetch_assoc($query)) {
    $staff_data[] = $row;
    }
}

function output_staff($staff_data) {
return '<ul><li>' . implode('</li><li>', $staff_data) . '</li></ul>';
}

I am not receiving the correct output. Currently I get these errors:

Notice: Undefined variable: staff_data in C:\xampp\htdocs\projects\etk\internal\staff.php >on line 11

Warning: implode(): Invalid arguments passed in C:\xampp\htdocs...line 3

which is pointing to my output_staff function. Can someone point me in the right direction?

Where are you calling output_staff()? It looks like you aren't passing an array to that function. — scrowler
– scrowler, Commented Nov 15, 2013 at 22:38
When you call output_staff(), are you passing in a variable like so? output_staff($myArray); — user399666
– user399666, Commented Nov 15, 2013 at 22:39
i think in the first functions is missing a return, return $staff_data; — rray
– rray, Commented Nov 15, 2013 at 22:39
Yes, am writing this in my actual page: echo output_staff($staff_data); — idelest
– idelest, Commented Nov 16, 2013 at 14:39

Alma Do · Accepted Answer · 2013-11-15 22:39:36Z

2

And how are your functions related? Your staff_data() seems to be returning nothing (i.e. null). And your output_staff() returns string, so you need to output it. So that should be:

function staff_data() 
{
    $staff_data = array();
    $query = mysql_query("SELECT * FROM `staff`");
    while ($row = mysql_fetch_assoc($query)) {
       $staff_data[] = $row;
    }
    return $staff_data;
}
//code 

$data = staff_data();
echo output_staff($data);

answered Nov 15, 2013 at 22:39

Alma Do

37.4k10 gold badges81 silver badges106 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

idelest Over a year ago

Fair play, I wasn't returning anything. I added the return and used a variable to store the array returned from staff_data() but I get this array to string conversion error, which I have seen before. I thought that was what my implode was taking care of. I guess at this point I don't understand the error message completely

idelest Over a year ago

I would like to add that using print_r on your $data variable totally works, as we would expect. I am looking up this string to conversion error, and I am thinking that the problem lies in my output_staff() function that is trying to output a string, when really its handling an array

Collectives™ on Stack Overflow

MySQL => PHP Associative Array...Understanding the undefined variable error

1 Answer 1

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related