Question

Please review this code:

    $current_sql = mysql_query("SELECT * FROM `product_categories` WHERE `product`='{$_GET['id']}'");
    $all_sql = mysql_query("SELECT * FROM `categories`");
    $in = mysql_fetch_array($current_sql);

    while($ca = mysql_fetch_array($all_sql)){

        echo("<label><input type=\"checkbox\" name=\"categories[]\" value=\"".$ca['ID']."\" class=\"cat_check\"");
        if($edit){
            if(in_array($ca['ID'], $in)){
                echo(" checked=\"checked\"");   
            }
        }
        echo("> " . $ca['category'] . "</label><br/>\n");

    }

This is the output:

<label><input type="checkbox" name="categories[]" value="1" class="cat_check" checked="checked"> PS3 Games</label><br/> 
<label><input type="checkbox" name="categories[]" value="16" class="cat_check"> Category 2</label><br/> 
<label><input type="checkbox" name="categories[]" value="17" class="cat_check" checked="checked"> Category 3</label><br/> 
<label><input type="checkbox" name="categories[]" value="18" class="cat_check"> Category 4</label><br/> 
<label><input type="checkbox" name="categories[]" value="19" class="cat_check"> Category 5</label><br/>

The problem is only checkbox value=”17″ is supposed to be checked, there are 2 MySQL databases here are are the structures:

categories:

ID    |   name
1         PS3 Games
16        Category 3
17        Category 4
18        Category 5
19        Category 6

product_categories

category     |    product
1                 20
17                1

Please help!!

UPDATE:

Sorry forgot to add:

$_GET['id'] = 1

Answer 1

Instead

$in = mysql_fetch_array($current_sql);

do

while ($prod_cat = mysql_fetch_array($current_sql)) {
  $in[] = $prod_cat['cat_id']; // where cat_id is the category ID in product_categories table
}

Basically the problem is that mysql_fetch_array get you back 2 arrays and I guess that you have something like array([0] = 1, [1] =17, etc ...). This is causing both categories to be checked. If you print_r the $in in your code, you will see what I mean.