我在网上发现了一些随机代码,并尝试将其用于我的日历,但我不断收到此错误:
Notice: Undefined variable: nextHoliday
此错误指的是末尾的代码"RETURN $nextHoliday;
我相信 nextHoliday 是定义的,所以我尝试了一些事情,但没有任何效果.. 有人可以帮忙吗?
代码如下:
FUNCTION GetTimeStamp($MySqlDate)
{
$date_array = EXPLODE("-",$MySqlDate); // split the array
$var_year = $date_array[0];
$var_month = $date_array[1];
$var_day = $date_array[2];
$var_timestamp = MKTIME(0,0,0,$var_month,$var_day,$var_year);
RETURN($var_timestamp); // return it to the user
} // End function GetTimeStamp()
FUNCTION ordinalDay($ord, $day, $month, $year)
// ordinalDay returns date of the $ord $day of $month.
// For example ordinalDay(3, 'Sun', 5, 2001) returns the
// date of the 3rd Sunday of May (ie. Mother's Day).
//
// Note: $day must be the 3 char abbr. for the day, as
// given by date("D");
//
{
$firstOfMonth = GetTimeStamp("$year-$month-01");
$lastOfMonth = $firstOfMonth + DATE("t", $firstOfMonth) * 86400;
$dayOccurs = 0;
FOR ($i = $firstOfMonth; $i < $lastOfMonth ; $i += 86400)
{
IF (DATE("D", $i) == $day)
{
$dayOccurs++;
IF ($dayOccurs == $ord)
{ $ordDay = $i; }
}
}
RETURN $ordDay;
} // End function ordinalDay()
FUNCTION getNextHoliday()
// Looks through a lists of defined holidays and tells you which
// one is coming up next.
//
{
$year = DATE("Y");
CLASS holiday
{
VAR $name;
VAR $date;
VAR $catNum;
FUNCTION holiday($name, $date, $catNum)
// Contructor to define the details of each holiday as it is created.
{
$this->name = $name; // Official name of holiday
$this->date = $date; // UNIX timestamp of date
$this->catNum = $catNum; // category, we used for databases access
}
} // end class holiday
$holidays[] = NEW holiday("Groundhog Day", GetTimeStamp("$year-2-2"), "20");
$holidays[] = NEW holiday("Valentine's Day", GetTimeStamp("$year-2-14"), "14");
$holidays[] = NEW holiday("St. Patrick's Day", GetTimeStamp("$year-3-17"), "15");
$holidays[] = NEW holiday("Easter", EASTER_DATE($year), "16");
$holidays[] = NEW holiday("Mother's Day", ordinalDay(2, 'Sun', 5, $year), "3");
$holidays[] = NEW holiday("Father's Day", ordinalDay(3, 'Sun', 6, $year), "4");
$holidays[] = NEW holiday("Independence Day", GetTimeStamp("$year-7-4"), "17");
$holidays[] = NEW holiday("Christmas", GetTimeStamp("$year-12-25"), "13");
$numHolidays = COUNT($holidays);
FOR ($i = 0; $i < $numHolidays; $i++)
{
IF ( DATE("z") > DATE("z", $holidays[$i]->date) && DATE("z") <= DATE("z",
$holidays[$i+1]->date) )
{
$nextHoliday["name"] = $holidays[$i+1]->name;
$nextHoliday["dateStamp"] = $holidays[$i+1]->date;
$nextHoliday["dateText"] = DATE("F j, Y", $nextHoliday["dateStamp"]);
$nextHoliday["num"] = $holidays[$i+1]->catNum;
}
}
RETURN $nextHoliday;
} // end function GetNextHoliday
$nextHoliday = getNextHoliday();
ECHO $nextHoliday["name"]." (".$nextHoliday["dateText"].")";
您有两种选择之一。实例化数组值的空键:
$nextHoliday = array();
$nextHoliday['name'] = '';
$nextHoliday['dateStamp'] = '';
$nextHoliday['dateText'] = '';
$nextHoliday['num'] = '';
$numHolidays = COUNT($holidays);
for ($i = 0; $i < $numHolidays; $i++) {
// ... Blah
}
或者在每次数组查找之前使用 isset:
echo (isset($nextHoliday['name'] ? $nextHoliday['name'] : '') .
" (" .
(isset($nextHoliday) ? $nextHoliday["dateText"] : '' ) .
")";
没有什么比内联条件的三元运算符更好的了。
我们在一月份对此进行测试实际上很好,否则此错误稍后会咬您。问题是您使用的小于/大于来确定下一个假期是什么。这没有考虑到去年的最后一个假期。
要解决此问题,变量 $lastHoliday 需要是上一个假日的负表示:
$numHolidays = COUNT($holidays);
$nextHoliday = array('name' => '', 'dateStamp' => '', 'dateText' => '', 'num' => '');
for ($i = 0; $i < $numHolidays - 1; $i++) {
$today = DATE("z");
if ($i == 0) {
$lastHoliday = (365 - DATE("z", $holidays[$numHolidays - 1]->date)) * -1;
} else {
$lastHoliday = DATE("z", $holidays[$i]->date);
}
$futureHoliday = DATE("z", $holidays[$i+1]->date);
//print_r($today); echo "<br />";
//print_r($lastHoliday); echo "<br />";
//print_r($futureHoliday); echo "<br />";
if ($today > $lastHoliday && $today <= $futureHoliday ) {
$nextHoliday["name"] = $holidays[$i+1]->name;
$nextHoliday["dateStamp"] = $holidays[$i+1]->date;
$nextHoliday["dateText"] = DATE("F j, Y", $nextHoliday["dateStamp"]);
$nextHoliday["num"] = $holidays[$i+1]->catNum;
}
}
还可以考虑以小写而不是大写形式键入 PHP,因为它几乎是 PHP 中的通用标准。一种真正的支撑样式也不会受到伤害。
if
但未声明
在for
之前声明变量:
[...]
$nextHoliday = array();
FOR ($i = 0; $i < $numHolidays; $i++)
[...]