Invalid argument supplied for foreach() - Yet output is showing - php

Question

I am working to show editable fields based on query results. I know the query is functioning properly, and it is returning an array. The array is populating the form fields properly, however, I am getting the "Invalid argument supplied for foreach()" warning. I am new at this, and at a loss as to what is happening. I appreciate any suggestions.
Here is the code:
// Grab the profile data from the database
$query8 = "SELECT * FROM EDUCATION WHERE ID_NUM = '" . $_SESSION['IDNUM'] . "' ORDER BY RECORD";
$data = mysqli_query($dbc, $query8);
echo '<pre>' . print_r($data, true) . '</pre>';
$rowcount = 1;
while ($row = mysqli_fetch_assoc($data))
{
if (is_array($row))
{
echo '<p> It is an Array</p>';
}
foreach($row as &$item)
{
$record = $row['RECORD'];
$school = $row['SCHOOL'];
$type = $row['TYPE'];
$degree = $row['DEGREE'];
$major = $row['MAJOR'];
$grad = $row['GRAD'];
?>
<form enctype="multipart/form-data" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<fieldset>
<legend>Education History </legend>
<?php
echo '<input type="hidden" id="record" name="record" value="' . $record . '">';
echo 'Rowcount' . $rowcount. '</br>';
// Insert Listbox here
$queryschool = "SELECT * FROM SCHOOL";
$list = mysqli_query($dbc, $queryschool);
if($list)
{
echo 'School Type? ';
echo '<select name="school_code">';
while($row = mysqli_fetch_assoc($list))
{
echo "<option value={$row['CODE']}>{$row['TYPE']}" ;
echo '</option>';
}
echo '</select>';
}
echo '<br />';
echo '<label for="school">School Name:</label>';
echo '<input type="text" id="school" name="school" size="40" maxlength="40" value="' . ( (!empty($school)) ? $school : "") . '" /><br />';
// Insert Listbox here
$querydegree = "SELECT * FROM DEGREE";
$list = mysqli_query($dbc, $querydegree);
if($list)
{
echo 'Degree Type? ';
echo '<select name="degree_code">';
while($row = mysqli_fetch_assoc($list))
{
echo "<option value={$row['CODE']}>{$row['DEGREE']}";
echo '</option>';
}
echo '</select>';
}
echo '<br />';
echo '<label for="major">Field of study:</label>';
echo '<input type="text" id="major" name="major" size="40" maxlength="40" value="' . ( (!empty($major)) ? $major : "") . '" /><br />';
echo '<label for="grad">Did you graduate?:</label>';
echo '<input type="radio" id="grad" name="grad" value="Y" ' . ($grad == "Y" ? 'checked="checked"':'') . '/>Yes ';
echo '<input type="radio" id="grad" name="grad" value="N" ' . ($grad == "N" ? 'checked="checked"':'') . '/>No<br />';
?>
</fieldset>
<?php
$rowcount++;
}
}
;
echo '<label for="another">Do you need to enter more educational experience?:</label>';
echo '<input type="radio" id="another" name="another" value="Y" ' . ($another == "Y" ? 'checked="checked"':'') . '/>Yes ';
echo '<input type="radio" id="another" name="another" value="N" ' . ($another == "N" ? 'checked="checked"':'') . '/>No<br />';
?>
<input type="submit" value="Save Profile" name="submit" />
</form>

Answer 1

foreach ($row as &$item)
replace this with:
foreach ($row as $item)
And then for each variable you should probably change
$record = $row['RECORD'];
to
$record = $item['RECORD'];

Answer 2

foreach($row as &$item) should be
foreach($row as $item)
there is no need to use the foreach here you can just do this like like
while ($row = mysqli_fetch_assoc($data))
{
$record = $row['RECORD'];
$school = $row['SCHOOL'];
$type = $row['TYPE'];
$degree = $row['DEGREE'];
$major = $row['MAJOR'];
$grad = $row['GRAD'];
}

Answer 3

You're not changing the row item, so don't pass by reference to the foreach. Also, shouldn't you be using $item instead of $row? Do this:
foreach($row as $item)
{
$record = $item['RECORD'];
$school = $item['SCHOOL'];
....
Don't do This:
foreach($row as &$item)
{
$record = $row['RECORD'];
$school = $row['SCHOOL'];
....