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copied!<p>From my comments:</p> <p><code>"$arrayname"</code> is a string, not an array. <code>call_user_func_array</code> will pass each element in <code>$arr_results</code> as argument to <code>array_intersect</code>. <code>array_intersect</code> expects arrays as arguments, but each item in <code>$arr_results</code> is a string, not an array. </p> <p>All you have to do is create an array of arrays instead of array names:</p> <pre><code>$arr_results = array(); foreach($arr_words as $word) { $ids = array(); while ($row = mysql_fetch_assoc($search)) { $ids[] = $row['id']; } $arr_results[] = $ids; } $matches = call_user_func_array('array_intersect',$arr_results); </code></pre> <blockquote> <p>I turn <code>$arrayname</code> into an array with <code>$$arrayname = array();</code></p> </blockquote> <p>Right, you create a variable, lets say <code>arr_0</code> which will point to array. But there is still a difference between the variable name <code>arr_0</code> and the <strong>string</strong> containing the variable name <code>"arr_0"</code>. You create an array of strings, and that just won't work. PHP does not know that the string contains a name of a variable. For example, consider this:</p> <pre><code>$arr = "arr_0"; echo $arr[0]; </code></pre> <p>Based on your logic, it should output the first element of the array, but it does not, because <code>$arr</code> is a string, not an array, although it contains the name of a variable. </p> <p>You'd have to use <code>eval</code>, but you really should not. You could also use variable variables again:</p> <pre><code>array_push($arr_results, $$arrayname); </code></pre> <p>that would work as well, but as I said, variable variables are confusing and in 99% of the cases, you are better of with an array.</p>

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