MySQL in_array function - Invalid second argument

Question

I'm trying to determine whether or not the user has voted on a certain item and changing the image for the vote icon accordingly.

$user_likes = mysql_query("SELECT * from ratings WHERE userid = '$uid' AND aggregate = 1");
while ($row_likes = mysql_fetch_array($user_likes)) {
$row_likes['imageid'] = $likes_array;
}

Now in my content, I have a PHP foreach that returns a menu item for each submission (that is what users can vote on), the $image['idnum'] is the ID value for that individual image. The 'ratings' table seen in the query above is where I store votes like this: user id, image id, aggregate (in this case 1, to represent a like). Here is the in_array function that is giving me trouble:

if (in_array($image['idnum'], $likes_array)) {
    echo 'vote_triangle.png';
}

I'm getting the following error message:

Warning: in_array() [function.in-array]: Wrong datatype for second argument in (filename) on line 33

You are not passing an array as the second parameter.

alex
– alex

2011-12-28 04:39:35 +00:00
Commented Dec 28, 2011 at 4:39 — alex
– alex, Commented Dec 28, 2011 at 4:39

No Results Found · Accepted Answer · 2011-12-28 04:41:34Z

2

I think you have this part backwards:

while ($row_likes = mysql_fetch_array($user_likes)) {
    $row_likes['imageid'] = $likes_array; // ????
}

Should be:

// Initialize the array
$likes_array = array();

while ($row_likes = mysql_fetch_array($user_likes)) {
     $likes_array[] = $row_likes['imageid'];
}

answered Dec 28, 2011 at 4:41

No Results Found

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5 Comments

Mark Lyons Over a year ago

Awesome! I had a feeling it would be something messed up with my array. I'll select your answer as soon as it lets me.

No Results Found Over a year ago

Do you fully understand it? The first version makes no sense, looks like it was basically a typo or "dumb" mistake.

Mark Lyons Over a year ago

I'm a bit embarrassed, but MySQL arrays confuse the hell out of me. I do better understand why I needed the [] to determine that it's an array and then use it as such in the function. I just thought that the variable would have stored $row_likes['imageid'] as an array.

No Results Found Over a year ago

$row_likes is reset in every iteration in the while loop, and $likes_array is undefined. I used $likes_array = array(); in case we don't get any results from the database, you'd still have a valid (empty) array.

Mark Lyons Over a year ago

I didn't even notice the array() but that definitely makes more sense. Thanks again!

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