PHP error question

I keep getting the following error below. How can I correct this error?

Warning: Invalid argument supplied for foreach() on line 25.

Line 25 is.

foreach ($parent_comment_id as $id => $comment) {

Here is the PHP & MySQL code.

function make_comments($parent_comment_id = 0, $comment_id = 0) {
global $user_id;
global $comment_order;

foreach ($parent_comment_id as $id => $comment) {

    if($comment['user_id'] == $user_id && $comment['parent_comment_id'] == 0){
        $comment_id = $comment['comment_id'];
        echo '<div>' . $comment['comment'] . '</div>';

        if($comment_id  == $comment['parent_comment_id']){
            if($comment['user_id'] == $user_id && $comment['parent_comment_id'] >= 1) {
                $comment_id = $comment['comment_id'];
                echo '<div>' . $comment['comment'] . '</div>';
            } else if($comment['parent_comment_id'] >= 1) {
                $comment_id = $comment['comment_id'];
                echo '<div>' . $comment['comment'] . '</div>';
            }
        }           

    } else if($comment['user_id'] != $user_id && $comment['parent_comment_id'] == 0) {
        $comment_id = $comment['comment_id'];
        echo '<div>' . $comment['comment'] . '</div>';

        if($comment_id  == $comment['parent_comment_id']){
            if($comment['user_id'] == $user_id && $comment['parent_comment_id'] >= 1) {
                $comment_id = $comment['comment_id'];
                echo '<div>' . $comment['comment'] . '</div>';
            } else if($comment['parent_comment_id'] >= 1) {
                $comment_id = $comment['comment_id'];
                echo '<div>' .开发者_运维问答 $comment['comment'] . '</div>';
            }
        }
    }       
        make_comments($comment_order[$id]);
    }
}

$dbc = mysqli_query($mysqli, "SELECT articles_comments.comment_id, articles_comments.parent_comment_id, articles_comments.comment, articles_comments.user_id FROM articles_comments LEFT JOIN users ON articles_comments.user_id = users.user_id WHERE article_id = '" . $article_id . "' ORDER BY articles_comments.parent_comment_id ASC");
if (!$dbc) {
    print mysqli_error();
} 

$comment_order = array();

while (list($comment_id, $parent_comment_id, $comment_text, $comment_user, $comment_id) = mysqli_fetch_array($dbc)) {
$comment_order[$parent_comment_id][$comment_id] =  array('parent_comment_id' => $parent_comment_id, 'comment_id' => $comment_id, 'comment' => $comment_text, 'user_id' => $comment_user, 'comment_id' => $comment_id);
}

make_comments($comment_order[0], $comment_id);

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In your foreach statement, $parent_comment_id must be an array expression. It's probably not when the execution reach that point, hence you are getting that runtime error.

---

Check the code where you call your function.

From what I can see, $parent_comment_id is a variable containing a number, and you're trying to iterate over it using a foreach statement.

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$parent_comment_id must be an array. Currently it's a value inside an array.

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I've never used MySql, but I've used foreach with arrays a lot. I don't know exactly what the variable $parent_comment_id looks like, but if it's a string perhaps you can use the function explode to change it to an array.

---

Your code is not good object oriented.

Besides you mess different types in single var. You should have this definition of function:

function make_comments(array $parent_comment_id, $comment_id=0) {

If you have array element ensure it is array. Do not mess integer and array in same variable!

---

This should be a simple fix to your code. It checks is $parent_comment_id is an array before executing the foreach loop.

if(is_array($parent_comment_id))
foreach ($parent_comment_id as $id => $comment) {
   //your for each code goes here
}

You can even add an else after the foreach.