2011-04-22 62 views
0

這裏的一切正在工作,除了更新命令...插入工作正常,但唉,我卡住了。更新命令語法

$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++; 
     } 
+0

切忌在對/而行再次使用'count'。放入一個變量並使用變量本身。 – Pentium10 2011-04-22 19:39:49

+0

@ Pentium10 - 爲什麼不呢?除了不必要的微觀優化之外,還有一個原因嗎 – 2011-04-22 19:47:13

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用於低延遲響應時間。 – Pentium10 2011-04-22 19:55:59

回答

2

試着寫更新這樣的:

"UPDATE schedule SET schedule.jakes = (SELECT schedule_default.jakes FROM schedule_default WHERE schedule.date = '$newDatesArray[$i]' AND schedule_default.ID = '5')"; 
1

多臺UPDATE語法是:

UPDATE schedule, schedule_default 
SET schedule.jakes = schedule_default.jakes 
WHERE schedule.date = '$newDatesArray[$i]' 
    AND schedule_default.ID = '5' 
+0

這樣做!非常感謝你。 – Andrew 2011-04-22 19:57:17