Errors when using array_push -- "First argument should be an array"

Question

I have the following code:

<?php

function foo($bar) 
{
    global $products; 

    //$products = array();

    $query = 'SELECT p_name FROM 0_products WHERE p_category IN (' . $bar . ')';
    $results = mysql_query($query);

    while($row = mysql_fetch_array($results, MYSQL_ASSOC))
    {
        array_push($products, $row);
        echo 'name pushed, ';
    }
}

require('mysql_ipb_connect.php'); // connect to ipb mysql database

$products = array(); 
foo(5);

?>

When I run it I get the following output:

Warning: array_push() [function.array-push]: First argument should be an array in /home/rgcpanel/public_html/category/category.php on line 14
name pushed,
Warning: array_push() [function.array-push]: First argument should be an array in /home/rgcpanel/public_html/category/category.php on line 14
name pushed,
Warning: array_push() [function.array-push]: First argument should be an array in /home/rgcpanel/public_html/category/category.php on line 14
name pushed,

If I uncomment "$products = array();" then the output is correct:

name pushed, name pushed, name pushed,

Why is this happening? I declare the $products array outside of a function (so it's global), and then specify it as being a global inside the function. Something is not right, but I'm not sure what that is?

Thanks for your advice.

Test that although $products is declared elsewhere, that it is actually an array. var_dump($products), and that you declared it outside the function at global scope, rather than in another function, and finally that you declared it before calling foo(). — Michael Berkowski
– Michael Berkowski, Commented Mar 22, 2012 at 18:01
Michael, @CoryDee -- I didn't think of this until now, but the file this is in is actually being included from inside a function that's a part of the CMS I'm using, so while the $products is outside of a function in this file, it's actually being included into a function. Does that mean it's not considered as a global variable? — Nate
– Nate, Commented Mar 22, 2012 at 18:10
If it was defined in an include inside a function, it is not global unless you first declare global $products in the function, before it is initialized by the include. — Michael Berkowski
– Michael Berkowski, Commented Mar 22, 2012 at 18:13

Michael Berkowski · Accepted Answer · 2012-03-22 18:15:07Z

Per the comments, $products was initialized by an included file which was included inside a function. That defines its scope to the function, rather than globally. So you'll need to use global $products; before calling the include.

function func_that_defined_products() {
  global $products;
  include('file_that_defines_products.php');
}

// Now when called globally later, it will be at the correct scope.


function foo($bar) 
{
    global $products; 
    $query = 'SELECT p_name FROM 0_products WHERE p_category IN (' . $bar . ')';
    // etc...
}

In any case, I find it a little more readable to use $GLOBALS['products'] instead of the global keyword. And as always, wherever possible it is a preferred practice to pass the variable into a function rather than accessing it globally.

// If you can, do it this way
function foo($bar, $products) {
  // $products was a param, and so global is unnecessary
}

However in your case, if the CMS defines it you may lose the flexibility to do it that way...

Cory Dee · Accepted Answer · 2012-03-22 18:01:27Z

0

You haven't initialized the global variable as an array. To PHP, that variable is just null, which is NOT AN ARRAY.

answered Mar 22, 2012 at 18:01

Cory Dee

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1 Comment

Nate Over a year ago

"$products = array();" isn't initializing $products as an array?

MonkeyMonkey · Accepted Answer · 2012-03-22 18:06:47Z

0

Make sure you initialize variables before using them. Code like yours will result in side effects (working fine if there is a database result; failing if its empty).

Another point: Why don't you return the result instead of using the global operator? This is really bad style.

Next one: When creating SQL statements: escape your variables!

answered Mar 22, 2012 at 18:06

MonkeyMonkey

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Comments

Starx · Accepted Answer · 2012-03-22 18:09:58Z

0

Uncomment this part

//$products = array();

Initialise the $products as an array

answered Mar 22, 2012 at 18:09

Starx

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