Question

我试图用PHP编写代码来启用闰年友好的选择选项标签。例如，您选择一年，然后检查它是否是闰年，然后它会显示一个下拉菜单（选择选项标签），其中包含12个月的其他选择选项标签。 PHP代码嵌入在HTML中。下面是我失败的尝试（我是PHP的新手）：

$sql=mysql_query("SELECT friendName,createdDate FROM friends where userId='$userId'");
$json = array(); // create empty array
$i = 0; // start a counter
while($result=mysql_fetch_assoc($sql)){ // start iteration on result-set object
    // fetch values one-by-one and assing to them in the array
    $json[$i]['friendName'] = $result['friendName'];
    $json[$i]['createdDate'] = $result['createdDate'];
    $i++; // increase the counter so that values will assign to the new indexes every-time (prevent from over-writing)
}
if(count($json)>0){ // check array have some values or not
 // now you have all records in $json array and here you can do your next code whatever you want based on this resultant array
}else{
$json = array( "msg" => "No infomations Found");
}
header('Content-type: application/json'); //i don't know what purpose this line will serve, so check your self.

备注：

我知道以上是不完整的代码，但它不起作用。此外，我不知道如何从select标签定义<form method="post"> <select name="year"> <option value="" selected>Pick a year</option> <option value="2016">2016</option> <option value="2017">2017</option> <option value="2018">2018</option> <option value="2019">2019</option> <option value="2020">2020</option> </select> <select name="month"> <option value="" selected>Pick a month</option>  <?php for( $i = 1; $i <= 12; $i++ ): ?> <option value="<?php echo $i; ?>"><?php echo $i; ?></option> <?php endfor; ?> </select> <select name="day"> <option value="" selected>Pick a day</option>  <?php if ( $month == 1 || 3 || 5 || 7 || 8 || 10 || 12 ) { //Show dates until 31 for ( $i = 1; $i <= 31; $i++ ) { ?> <option value="<?php echo $i; ?>"><?php echo $i; ?></option> <?php } } elseif ( $month == 2 ) { //Leap year if ( $year != "" && $year % 4 == 0 && $year % 100 == 0 && $year % 400 == 0 ) { //Show dates until 29 for ( $i = 1; $i <= 29; $i++ ) { ?> <option value="<?php echo $i; ?>"><?php echo $i; ?></option> <?php } } //Regular year (Non-leap year) else { //Show dates until 28 for ( $i = 1; $i <= 28; $i++ ) { ?> <option value="<?php echo $i; ?>"><?php echo $i; ?></option> <?php } } } elseif ( $month == 4 || 6 || 9 || 11 ) { //Show dates until 30 for ( $i = 1; $i <= 30; $i++ ) { ?> <option value="<?php echo $i; ?>"><?php echo $i; ?></option> <?php } } ?> </select> <input type="submit" value="submit"> </form>或$year或$month。我应该这样做吗？如果是的话，我应该把它放在哪里？

$day

最好，我不想在几个月和几天内使用<?php if( $_SERVER["REQUEST_METHOD"] == "POST" ) { $year = $_POST['year']; $month = $_POST['month']; $day = $_POST['day']; //The rest of the php code? } ?>循环显示年份。我想这次手动做（抱歉，如果你建议我应该使用for循环，我感谢你的建议。）

我在2月闰年for阻止了$year =! ""，因为我想避开空的默认值。

我热烈欢迎您的建议和更正，请帮助我：）

提前致谢！

Answer 1

据我所知，闰年：

/**
 * In the Gregorian calendar 3 criteria must be taken into account to identify leap years:
 *  1. The year is evenly divisible by 4;
 *  2. If the year can be evenly divided by 100, it is NOT a leap year, unless;
 *  3. The year is also evenly divisible by 400. Then it is a leap year.
 *
 * @param $year integer
 * @return bool
 */
private static function isLeapYear($year)
{
    // the rule was applied after 1582
    if ($year > 1582) {
        if (($year % 400 == 0 && $year % 100 != 0) || ($year % 4 == 0))
        {
            //"It is a leap year"
            return true;
        }
        else
        {
            return false;
            //"It is not a leap year"
        }
    }
    return false;
}

Answer 2

这不起作用：

( $month == 1 || 3 || 5 || 7 || 8 || 10 || 12 )

这只会检查$month == 1，然后其他条件将被评估为true，因为整数被评估为true（并且PHP无法将变量与布尔值与此代码进行比较）。

看起来你正试图获得每月的天数。使用date()格式的DateTime::Format()可以轻松完成此操作：

t返回给定月份的天数：

$dt = new Datetime('2016-02-15');
print $dt->format('t');
// display "29"

Example

如果一年飞跃，

L会返回1，否则会返回0：

$dt = new Datetime('2016-02-15');
print $dt->format('L');
// display "1"
$dt = new Datetime('2015-02-15');
print $dt->format('L');
// display "0"

Example

您可能不关心输入并显示所有年份，即每月12个月和31天，并在数据提交后用checkdate()检查日期。

闰年友好年/月/日选择嵌入了PHP的选项标签

2 个答案: