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[SOLVED] Getting undefined offset error, but don't see why...


BenGoldberg

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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!

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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?

 

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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!

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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.

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