Invalid argument supplied for foreach()

kirk5630072

I would like to know how to fix this error any help would be appreciated:
Warning: Invalid argument supplied for foreach() in B:\webserver\xampp\htdocs\computers\printinvoice.php on line 39

here is the code and surrounding code

<?
// form is subitted
if (!empty($_POST)) {
$q = "SELECT FROM invoice WHERE id = '$invoice'"; // get the current entry
$r = mysql_query($q) or die(mysql_error()); // exec
$row = mysql_fetch_assoc($r); // fetch the row
foreach($row as $key => $val) {$data[$key] = $val;
};
//does the subtotal calculations that display on the invoice
$hwsubtotal1 = ($row['qty1'] $row['unitcost1']);
$hwsubtotal2 = ($row['qty2'] $row['unitcost2']);
$hwsubtotal3 = ($row['qty3'] $row['unitcost3']);
$hwsubtotal4 = ($row['qty4'] $row['unitcost4']);
$hwsubtotal5 = ($row['qty5'] $row['unitcost5']);
$hwsubtotal6 = ($row['qty6'] * $row['unitcost6']);
//subtotal all the harware types
$hwsub = ($hwsubtotal1 + $hwsubtotal2 + $hwsubtotal3 + $hwsubtotal4 + $hwsubtotal5 + $hwsubtotal6);
..........
............
............

Running php 5.2.4

Thanks, Kirk

zabmilenko

print_r($row); just before foreach and see the contents. My guess is that $row is false or not an array

kirk5630072

how do i make $row an array im kinda new to all this stuff so the code will run correctly

NogDog

If your query does not match any rows and therefore returns an empty result set, then your call to mysql_fetch_assoc() will return boolean false, and therefore that result will not be an array. So you may want to add a check that the query actually returned something, either by checking the result of the mysql_fetch_assoc() to see if it is false, or doing a mysql_num_rows() right after the query to find out how many result rows were returned by the query.

kirk5630072

"SELECT * FROM invoice WHERE id = '$invoice'" thats the SQL it returns a result in phpmyadmin but not in code any ideas on how to rewrite the sql to work

kirk5630072

"SELECT * FROM invoice WHERE id = '$invoice'"

thats what i got for a sql command and it doesnt work with the index variable however if i change $invoice to the id number it works any ideas on how to properly include that invoice in the sql query.

zabmilenko

Ahhh... try this:

$invoice = mysql_real_escape_string($_POST['invoice']);

Just before the query... is it getting better?

rsmith

How are you getting the information from the form? Are you using GET or POST? Can you show the rest of the script?

bradgrafelman

Where is $invoice ever defined?

Looks like you're depending on register_globals being turned On (although you do check f $_POST isn't empty...), which was long ago identified as security threat. Instead, you should be using the superglobal arrays explained in the manual here: [man]variables.predefined[/man].