UPDATE command syntax

Question

Everything here is working except the UPDATE command... The INSERT works fine but alas, I'm stuck.

$queryLastDateArray = "SELECT date FROM schedule ORDER BY ID DESC LIMIT 1";
$lastDateArray = mysql_query($queryLastDateArray);
while($row = mysql_fetch_array($lastDateArray))
                        {
                        $lastDate = $row['date'];
                        }                       
$lastDatePlusOne = date("Y-m-d", strtotime("+1 day", strtotime($lastDate)));

$newDatesArray = GetDays($lastDatePlusOne, $_POST[date]);
$i = 0;
while($i < count($newDatesArray))
        {
        if ((date('D', strtotime($newDatesArray[$i]))) == 'Fri')
                {
                $insDate = "INSERT INTO schedule (date) VALUES ('$newDatesArray[$i]')";
                $result = mysql_query($insDate);
                $insEmp = "UPDATE schedule SET schedule.jakes = schedule_default.jakes FROM schedule, schedule_default WHERE schedule.date = '$newDatesArray[$i]' AND schedule_default.ID = '5'";
                $result2 = mysql_query($insEmp);
                }
        $i++;
        }

Never use again count in the for/while line. Put in a variable and use the variable itself. — Pentium10
– Pentium10, Commented Apr 22, 2011 at 19:39
@Pentium10 - Why not? Is there a reason besides unnecessary micro-optimization? — John Kugelman
– John Kugelman, Commented Apr 22, 2011 at 19:47

dcp · Accepted Answer · 2011-04-22 19:39:22Z

2

Try writing the update like this:

 "UPDATE schedule SET schedule.jakes = (SELECT schedule_default.jakes FROM  schedule_default WHERE schedule.date = '$newDatesArray[$i]' AND schedule_default.ID = '5')";

answered Apr 22, 2011 at 19:39

dcp

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Oswald · Accepted Answer · 2011-04-22 19:41:20Z

1

Multiple table UPDATE syntax is:

UPDATE schedule, schedule_default
SET schedule.jakes = schedule_default.jakes
WHERE schedule.date = '$newDatesArray[$i]'
  AND schedule_default.ID = '5'

answered Apr 22, 2011 at 19:41

Oswald

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