Invalid argument supplied for foreach() - but script still works

Question

The suggested questions have not quite covered my issue. I'm intrigued by the answer here, Invalid argument supplied for foreach! but don't quite see how it applies in my case.

I have the following code, which does work as I would expect; the code builds an array of email addresses by matching an id given as hidden form posted from the previous page, finding the 'email' column in a mysql database.

However, I also get the error: Invalid argument supplied for foreach() on line 50

Any help appreciated.

function findEmail($id) {
$mysqli = new mysqli(DBHOST,DBUSER,DBPASS,DB);
if ($mysqli->connect_errno) {
    error_log("Cannot connect to MySQL: " . $mysqli->connect_error);
    return false;
}

$query = "SELECT email FROM `Students` WHERE id = '$id'";
$result = $mysqli->query($query);
while($row = $result->fetch_array()) {
    $rows[] = $row;
}
$email = "";
foreach($rows as $row) {      //This is line 50
    $email = ($row['email']);
}
return $email;
}

$userId = array();
$userEmail = array();

foreach($_POST as $key => $value) {
array_push($userId, $value);
array_push($userEmail, findEmail($value));
}

you loop twice onece in the while then the foreach, its pointless — , Apr 16 '14 at 21:21
You also seem to just assign `$email` in that `foreach` loop, so at the end you are not adding to a variable but over-riding it every time. — ek9, Apr 16 '14 at 21:29
The id is unique so there should only ever be one email address. — Russell, Apr 16 '14 at 21:37

score 0 · Answer 1 · answered Apr 16 '14 at 21:24

0

The error isn't fatal, you can make it be... however the error is that $rows is not defined -- your query must be returning empty.

answered Apr 16 '14 at 21:24

awm

2,723
2
18
26

score 0 · Answer 2 · answered Apr 16 '14 at 21:25

0

You might have an element in the $rows array that is NULL, if you var_dump($rows) do you have any NULLS?

answered Apr 16 '14 at 21:25

rmcfrazier

444
4
8

Yes - in fact despite only having hidden elements in the form the value "Submit" is included in $_POST, which is causing the error. – Russell Apr 16 '14 at 22:07

score 0 · Answer 3 · answered Apr 16 '14 at 21:25

You need to initialize the vaiable $rows, else as stated on http://www.php.net/manual/en/control-structures.foreach.php the foreach "will issue an error when you try to use it on a variable with a different data type or an uninitialized variable"

So before line 50 put this:

$rows=null;

score 0 · Answer 4 · answered Apr 16 '14 at 21:26

the function could be optermised like so:

function findEmail($id) {
$mysqli = new mysqli(DBHOST,DBUSER,DBPASS,DB);
if ($mysqli->connect_errno) {
    error_log("Cannot connect to MySQL: " . $mysqli->connect_error);
    return false;
}

$query = "SELECT email FROM `Students` WHERE id = '$id'";
$result = $mysqli->query($query);  

  $row = $result->fetch_array($query);

  return $row[0];
}

score 0 · Answer 5 · answered Apr 16 '14 at 22:14

The value of "submit" was being sent to the function. Adding the second line below solved the problem.

foreach($_POST as $key => $value) {
    if($key == "submit") continue;
    array_push($userId, $value);
    array_push($userEmail, findEmail($value));
}

Invalid argument supplied for foreach() - but script still works

5 Answers5