Adding key-value in PHP array?

Question

Inside a loop where I'm dealing with variables related to a product and a number of units, I'm trying to add these two to an array:

$pedido = array();

So,

    foreach($_POST as $post_key => $post_value){

        if ($post_value=="on") {
            $nombreProducto = mysql_fetch_assoc($mySQL->query("SELECT nombre from productos WHERE id_producto='$post_key'"));
            $cantidad = $_POST[$post_key."Number"];

            echo "<h1>".$nombreProducto['nombre']."</h1>"." Cantidad: ".$cantidad." <br><br>";
            $pedido["$nombreProducto"] = $cantidad;
        }
    }

It's right in:

$pedido["$nombreProducto"] = $cantidad;

Where I try to perform the adding, however the output of var_dump is like:

array(1) { ["Array"]=> string(1) "3" }

Not exactly what I wanted neither the format.

Than what do you want? Why are you putting a variable in quotes? Are you trying to $pedido[$nombreProducto['nombre']] = $cantidad;? You're SQL injecting yourself BTW.. ' OR 1=1 --=Whatever would return all products in your database. — h2ooooooo
– h2ooooooo, Commented Oct 25, 2013 at 10:55
You really shouldn't loop mysql queries, at least not on a live server. There usually is a better way, especially when you fetch! — Horse SMith
– Horse SMith, Commented Oct 25, 2013 at 10:56
I hope you read my full answer, because there was a lot of other issues with this code you should fix up in as well. — Horse SMith
– Horse SMith, Commented Oct 25, 2013 at 11:14

redelschaap · Accepted Answer · 2013-10-25 10:58:42Z

1

Use $pedido[$nombreProducto['nombre']] = $cantidad; instead of $pedido["$nombreProducto"] = $cantidad;

answered Oct 25, 2013 at 10:58

redelschaap

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2 Comments

diegoaguilar Over a year ago

Right! One more question, this is the var_dump output array(3) { ["Laptop Asus N56VM"]=> string(1) "3" ["Pantalla Samsung LED 40'"]=> string(1) "1" ["Tarjeta SD 32 GB ADATA Clase 10"]=> string(1) "2" } So that means the numbers storaged are actually strings, what if I want them to be numbers right away?

redelschaap Over a year ago

In most cases you can just use the numeric string (like the numbers are stored now) to do whatever you want (getting more info out of your database, echoing the numbers, inserting something into the database etc). But if you really need to use an integer, use $pedido[$nombreProducto['nombre']] = (int)$cantidad;

Sal00m · Accepted Answer · 2013-10-25 10:56:34Z

1

Remove quotes and

$pedido[$nombreProducto['nombre']] = $cantidad;

EDITED

It seems $nombreProducto is an array so you need to indicate the key field, so i changed to use the field "nombre"

If you see your var_dump it's an Array with the key "Array" this is why you are trying to convert the array to string and it return the word "Array"

answered Oct 25, 2013 at 10:56

Sal00m

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2 Comments

rorypicko Over a year ago

wrapping a varible in double quotes doesnt affect the output... $t = 'ddd'; echo $t; //ddd echo "$t" //ddd using double quotes tells php to check every word for a variable

Sal00m Over a year ago

You're right (both), i've edited the answer with some explanation, thanks

Niet the Dark Absol · Accepted Answer · 2013-10-25 10:57:16Z

1

You shouldn't be putting a variable by itself in quotes. Remove the quotes.

Also, since you were just accessing $nombreProducto['nombre'] on the previous line, it's fairly obvious that that variable is an array. You cannot use an array as a key, only integers and strings are allowed. So use something that identifies it, such as its ID number.

answered Oct 25, 2013 at 10:57

Niet the Dark Absol

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Comments

Horse SMith · Accepted Answer · 2013-10-25 11:09:14Z

0

The following makes no sense to do:

$pedido["$nombreProducto"] = $cantidad;

What are you trying to do here? I think what you want to do is this:

$pedido[$nombreProducto['nombre']] = $cantidad;

Also you may want to try to output the array like this:

print_r($pedido);

I recommend you re-write the mysql query so you don't need to loop it. That is for safety and efficiency reasons. I also recommend you use mysqli instead of mysql, because mysql is deprecated and unsafe to use. You don't check if $_POST[$post_key."Number"] is set, at least not in this code. I hope you sanitize/validate the input from $_POST before using it against the database?

answered Oct 25, 2013 at 11:09

Horse SMith

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