mysql_fetch_array not creating right elements?

Question

I am getting the following error message:

Notice: Undefined index: timein in time.php on line 42 Notice: Undefined index: timeout in time.php on line 45

Here is my code:

if (isset($_POST['submit'])) {
    $name = $_POST['name'];

$querytimeout = mysql_query("
    SELECT timein 
    FROM studentInfo
    WHERE name = '$name' 
    ORDER BY time DESC
    LIMIT 1,1
    ")
    or die("Error querying database ".mysql_error());
    $querytimein = mysql_query("
    SELECT timeout
    FROM studentInfo
    WHERE name = '$name' 
    ORDER BY time DESC
    LIMIT 1
    ")
    or die("Error querying database ".mysql_error());
while($minutestimein = mysql_fetch_array($querytimein)){
    $ltimein = $minutestimein['timein'];
    }
        while($minutestimeout = mysql_fetch_array($querytimeout)){
    $ltimeout = $minutestimeout['timeout'];
  }
  $timegone = $ltimein - $ltimeout; 
  echo $timegone;
  }

The problem is that mysql_fetch_array() isn't making timein or timeout an element of the array $minutestimein.

Answer 1

If you compare your queries to the fields that they are looking up, they don't match.

In the first query you select only timeout yet you try to reference $minutestimein['timein']. You do the opposite in the second query.

Answer 2

Your two queries are redundant - they're fetching two different fields from the SAME record. Why not just run ONE query?

SELECT timein, timeout
FROM studentInfo
WHERE name = '$name'
ORDER BY time DESC
LIMIT 1

then

$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_assoc($result);
$diff = $row['timeout'] - $row['timeout'];

Plus, there is literally NO point in using a while() loop to fetch your results. You've already limited your queries to returning only ONE row anyways, making the entire looping infrastructure useless.