發佈的數據值時,每當我試圖在PHP中我碰到下面的錯誤後的數據值:錯誤在PHP
Fatal error: Uncaught exception 'Exception' with message 'DateTime::__construct() [datetime.--construct]: Failed to parse time string (this wednesday G:i A) at position 16 (:): Unexpected character' in C:\xampp\htdocs\keypad\end.php:14 Stack trace: #0 C:\xampp\htdocs\keypad\end.php(14): DateTime->__construct('this wednesday ...') #1 {main} thrown in C:\xampp\htdocs\keypad\end.php on line 14
這裏是我使用的代碼:
<?php require_once('Connections/rent.php'); ?>
<?php
// post
$id = $_GET['id'];
$user = $_GET['user'];
$week = $_GET['week'];
$pRent = $_GET['pRent'];
$status = $_GET['status'];
$rentPaid = $_GET['rentPaid'];
$result = $_GET['result'];
$pDate = date("d/m/Y G:i A");
$lDate = new DateTime('this wednesday G:i A');
$cDate = new DateTime('next tuesday G:i A');
// update
mysql_query("UPDATE rent SET dNo = '$dNo', pdate = '$pDate', pRent = '$result' WHERE rent.id = $id");
// if pRent = 0 then set status to clear, colour to #3C0
if ($pRent == 0) {
mysql_query("UPDATE rent SET status = 'clear', colour = '#3C0' WHERE rent.id = $id");
}
// if pRent =! 0 but just paid then set status to paid/not clear, colour to #FF0
elseif ($pRent =! 0) {
mysql_query("UPDATE rent SET status = 'paid', colour = '#3C0' WHERE rent.id = $id");
}
// if status = on holiday, then set status to clear, colour to #3C0
elseif ($status == 'awaiting') {
mysql_query("UPDATE rent SET status = 'awaiting', colour = '#09F' WHERE rent.id = $id");
}
// check all drivers who have/have not paid rent in the past week
elseif ($lDate < $pDate && $cDate > $pDate) {
// date is within desired range
mysql_query("UPDATE rent SET status = 'paid', colour = '#3C0'");
} else {
// date is not within desired range
mysql_query("UPDATE rent SET status = 'awaiting', colour = '#09F' WHERE rent.status =! 'not working'");
}
?>
任何建議?
我認爲錯誤信息已經足夠清楚了。 `'今天星期三G:我A'`不是`DateTime()`理解的東西。 – deceze 2010-12-09 07:12:14