BenGoldberg Posted August 15, 2008 Share Posted August 15, 2008 Hi, I have a script that compares integers and ranks them from lowest to highest, and then submits those rankings to a mysql table. If two or more integers are the same, I have it so they're ranked the same and the next integer is ranked where it would appear if there was no tie, example... $a = 1 $b = 1 $c = 2 $a and $b would be ranked 1st because they're tied for the lowest value while $c would be ranked third, not second. Anyways, I created a relatively simple function to do this and in testing, it's outputting the values I expected it to. The only thing is that it's giving me undefined offset errors, but I don' see why. Here's the function... <?php /* FUNCTION TO ENTER RANKINGS INTO DATABASE */ function post_rankings() { $result = mysql_query("SELECT * FROM auf_records") or die(mysql_error()); $u = 17; $x = 18; $y = 19; while ($u < mysql_num_fields($result)) { /* RETRIEVE DATA FROM TABLE "AUF_RECORDS" */ $level = mysql_field_name($result, $u); $date = mysql_field_name($result, $x); $recrank = mysql_field_name($result, $y); $result2 = mysql_query("SELECT $level, $date, id FROM auf_records") or die(mysql_error()); while ($row = mysql_fetch_row($result2)) { if ($row[0] > 0) { $records[] = $row; } } if (isset($records)) { sort($records); $i = 0; $numrec = count($records[0]); while ($i < $numrec) { $uid = $records[$i][2]; // FIRST UNDEFINED OFFSET if ($i == 0) { $rank = $i + 1; mysql_query("UPDATE auf_records SET $recrank = '$rank' WHERE id = '$uid' ") or die(mysql_error()); } else { $a = $i - 1; if ($records[$i][0] == $records[$a][0]) { // SECONDS UNDEFINED OFFSET $prevuid = $records[$a][2]; // THIRD UNDEFINED OFFSET $result3 = mysql_query("SELECT $recrank FROM auf_records WHERE id = '$prevuid' ") or die(mysql_error()); $row2 = mysql_fetch_row($result3); $rank = $row2[0]; mysql_query("UPDATE auf_records SET $recrank = '$rank' WHERE id = '$uid' ") or die(mysql_error()); } else { $rank = $i + 1; mysql_query("UPDATE auf_records SET $recrank = '$rank' WHERE id = '$uid' ") or die(mysql_error()); } } $i++; } } $u = $u + 3; $x = $x + 3; $y = $y + 3; unset($records); } } ?> I'm getting Undefined offset: 1 for all three undefined offsets but only Undefined offset: 2 for the first two undefined offsets. If $records isn't set if (isset($records)) should take care of that, so why am I getting this error. I'm sure it's something simple, but I can't find it by myself. Any help would be greatly appreciated! Quote Link to comment Share on other sites More sharing options...
btherl Posted August 15, 2008 Share Posted August 15, 2008 Can you add this code after setting $records: var_dump($records); And show us what it displays. Quote Link to comment Share on other sites More sharing options...
BenGoldberg Posted August 15, 2008 Author Share Posted August 15, 2008 Well it's displaying a lot, but here's the first two arrays it displays... array(3) { [0]=> array(3) { [0]=> string(3) "106" [1]=> string(11) "08 15, 2008" [2]=> string(1) "1" } [1]=> array(3) { [0]=> string(3) "106" [1]=> string(11) "08 14, 2008" [2]=> string(1) "2" } [2]=> array(3) { [0]=> string(3) "108" [1]=> string(11) "08 14, 2008" [2]=> string(1) "3" } } array(1) { [0]=> array(3) { [0]=> string(3) "107" [1]=> string(11) "08 15, 2008" [2]=> string(1) "1" } } It then has 6 undefined offsets and then more arrays and errors... The values that var_dump are displaying are the ones I expected. Does this help? Quote Link to comment Share on other sites More sharing options...
btherl Posted August 15, 2008 Share Posted August 15, 2008 Ah I think I got it. Try counting $records instead of $records[0] to set $numrec Quote Link to comment Share on other sites More sharing options...
BenGoldberg Posted August 15, 2008 Author Share Posted August 15, 2008 Holy moly! Thank you so much, works perfectly now. Can I ask why this fixes the problem though? Why does $records[0] not count the values correctly? It seems like count($records) and count($records[0]) would output the same number no matter what. In any case, I can move on with my website now, so thanks again! Quote Link to comment Share on other sites More sharing options...
btherl Posted August 15, 2008 Share Posted August 15, 2008 Both $records and $records[0] are arrays. But $records is an array that lists rows of data from the database, whereas $records[0] (and [1], [2], etc) are arrays that list data in a single row. So count($records[0]) is always 3, same for the others. But count($records) will give you the number of results. In the first var_dump above, the number of items happens to be exactly the same as the number of items in each result. But in the second var_dump, there is one result, and that result contains 3 items. Counting the number of items ($records) will give you 1, but counting the number of data items in a result ($records[0]) will give you 3. Quote Link to comment Share on other sites More sharing options...
BenGoldberg Posted August 15, 2008 Author Share Posted August 15, 2008 Ah yes, I see now. Duh. Well I knew it was something silly like that. Quote Link to comment Share on other sites More sharing options...
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