PHP Warning: Invalid argument supplied for foreach()

Question

Why am I getting this PHP Warning?

Invalid argument supplied for foreach()

Here is my code:

// look for text file for this keyword
if (empty($site["textdirectory"])) {
    $site["textdirectory"] = "text";
}
if (file_exists(ROOT_DIR.$site["textdirectory"].'/'.urlencode($q).'.txt')) {
    $keywordtext = 
     file_get_contents(ROOT_DIR.$site["textdirectory"].'/'.urlencode($q).'.txt');
}
else {
    $keywordtext = null;
}

$keywordsXML = getEbayKeywords($q);

foreach($keywordsXML->PopularSearchResult as $item) {
    $topicsString = $item->AlternativeSearches;
   $relatedString = $item->RelatedSearches;
   if (!empty($topicsString)) {
        $topics =  split(";",$topicsString);
    }
    if (!empty($relatedString)) {
        $related = split(";",$relatedString);
    }

}

$node = array();
$node['keywords'] = $q;

2

$xml = ebay_rss($node);

$ebayItems = array();
$totalItems = count($xml->channel->item);

$totalPages = $totalItems / $pageSize;


$i = 0;
foreach  ($xml->channel->item as $item) {
  $ebayRss = 
    $item->children('http://www.ebay.com/marketplace/search/v1/services');

    if ($i>=($pageSize*($page-1)) && $i<($pageSize*$page)) {
        $newItem = array();
        $newItem['title'] = $item->title;
        $newItem['link'] = buyLink($item->link, $q);
        $newItem['image'] = ebay_stripImage($item->description);
        $newItem['currentbid'] = ebay_convertPrice($item->description);
        $newItem['bidcount'] = $ebayRss->BidCount;
        $newItem['endtime'] = ebay_convertTime($ebayRss->ListingEndTime);
        $newItem['type'] = $ebayRss->ListingType;
        if (!empty($ebayRss->BuyItNowPrice)) {
            $newItem['bin'] = ebay_convertPrice($item->description);
        }
        array_push($ebayItems, $newItem);
    }
    $i++;
}


$pageNumbers = array();
for ($i=1; $i<=$totalPages; $i++) {
    array_push($pageNumbers, $i);
}

3

// get user guides
$guidesXML = getEbayGuides($q);
$guides = array();
foreach ($guidesXML->guide as $guideXML) {
    $guide = array();
    $guide['url'] = makeguideLink($guideXML->url, $q);
    $guide['title'] = $guideXML->title;
    $guide['desc'] = $guideXML->desc;
    array_push($guides,$guide);
}

What causes this warning?

Check out the {} button on the edit screen: It'll format the code for you like i've done now ;) — Nanne
– Nanne, Commented Jul 4, 2011 at 13:18
Because you supplied an invalid argument to foreach. Did you check what you feed into it? var_dump can help you. — Jon
– Jon, Commented Jul 4, 2011 at 13:18
Which one of the foreach produces the error? Your error message should give you a line number... — Michael Berkowski
– Michael Berkowski, Commented Jul 4, 2011 at 13:18
It has a PHP Warning: Invalid argument supplied for all three: foreach($keywordsXML->PopularSearchResult as $item) { foreach ($xml->channel->item as $item) { foreach ($guidesXML->guide as $guideXML) { — meme
– meme, Commented Jul 4, 2011 at 14:35

Christopher Wilson · Accepted Answer · 2012-11-06 10:47:52Z

90

You should check that what you are passing to foreach is an array by using the is_array function

If you are not sure it's going to be an array you can always check using the following PHP example code:

if (is_array($variable)) {

  foreach ($variable as $item) {
   //do something
  }
}

edited Nov 6, 2012 at 10:47

Christopher Wilson

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answered Jul 4, 2011 at 13:23

david

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3 Comments

meme Over a year ago

Why, I'm asking what would be causing me to get. Why am I getting a PHP Warning: Invalid argument supplied for foreach() on three of my foreach in my script.

John Magnolia Over a year ago

In most cases this will work but not always see: stackoverflow.com/a/2630032/560287

Heres2u Over a year ago

Excellent. I was having an issue for a custom Wordpress plugin where I parse an RSS feed. This feed doesn't always have content by default, and expires until there is more content. So the array is not always populated. This solved the "offset" warnings I was getting in that case.

Eric Leschinski · Accepted Answer · 2014-08-11 18:07:47Z

12

This means that you are doing a foreach on something that is not an array.

Check out all your foreach statements, and look if the thing before the as, to make sure it is actually an array. Use var_dump to dump it.

Then fix the one where it isn't an array.

How to reproduce this error:

<?php
$skipper = "abcd";
foreach ($skipper as $item){       //the warning happens on this line.
    print "ok";
}
?>

Make sure $skipper is an array.

edited Aug 11, 2014 at 18:07

Eric Leschinski

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answered Jul 4, 2011 at 13:20

Nanne

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Comments

Lightness Races in Orbit · Accepted Answer · 2011-07-04 13:19:54Z

7

Because, on whatever line the error is occurring at (you didn't tell us which that is), you're passing something to foreach that is not an array.

Look at what you're passing into foreach, determine what it is (with var_export), find out why it's not an array... and fix it.

Basic, basic debugging.

answered Jul 4, 2011 at 13:19

Lightness Races in Orbit

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Comments

Pang · Accepted Answer · 2018-01-16 07:54:47Z

3

Try this.

if(is_array($value) || is_object($value)){
    foreach($value as $item){
     //somecode
    }
}

edited Jan 16, 2018 at 7:54

Pang

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answered Jan 16, 2018 at 7:38

Mr Nobody

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