PHP Multiple Dropdown Box Form Submit To MySQL (Part 2) - php

Question

This is a continuation of the discussion at
PHP Multiple Dropdown Box Form Submit To MySQL
which ended with the words: "Once you have the variables, it is trivial to create new rows." No doubt that's generally true, but apparently not for this learner... :-D
Given the following form:
<form action="form.php" method="POST">
<select name="colors[]" multiple="yes" size="2">
<option>Red</option>
<option>Blue</option>
</select>
<input type="submit" value="Go!">
</form>
how do I create new rows? The following script
foreach($_POST['colors[]'] as $color)
{
$id = mysqli_real_escape_string($link, $color);
$sql = "INSERT INTO colors SET id = '$id'";
}
raises the error
Warning: Invalid argument supplied for foreach() in form.php on line ...
whereas the following
$colors = $_POST['colors[]'];
for ($i = 0; $i < count($colors); $i++)
{
$color = $colors[$i];
$sql = "INSERT INTO colors SET id = '$color'";
}
raises no errors but does no row creation.
What triviality am I missing here?

Answer 1

Use:
foreach($_POST['colors'] as $color)
No need to specify [] here, php knows that it is an array.

Answer 2

Inside of your foreach loop I did not see that you are executing the query. You need to execute mysql_query with your INSERT statement.
foreach($_POST['colors'] as $color) {
$id = mysqli_real_escape_string($link, $color);
$sql = "INSERT INTO colors SET id = '$id'";
if (!mysql_query($sql)) { // if the query encountered a problem.
die('Invalid query: ' . mysql_error());
}
}