循环日期函数php

Question

我有桌名＆＃34; guest_room＆＃34;在数据库中，guest_room的字段是＆＃34;到达＆＃34;并且＆＃34;离开＆＃34;，到达的价值是＆＃34; 2016-12-27＆＃34;离开的价值是＆＃34; 2016-12-31＆＃34;。

在我的php文件中我想显示从到达到出发日期的数据，这里是我的代码：

$date = $g[arrival];
$end_date = $g[departure];
while (strtotime($date) <= strtotime($end_date)) {                                       
$date = date ("Y-m-d", strtotime("+1 day", strtotime($date)));
echo "$date<br>";
}

但在我的结果中显示的日期来自＆＃34; 2016-12-28＆＃34;直到＆＃34; 2017-01-01＆＃34; 我想要的是，我想显示来自＆＃34; 2016-12-27＆＃34;直到＆＃34; 2016-12-31＆＃34;

我知道我不应该使用mysql_但这不是重点，请帮帮我

Answer 1

试试这段代码，

$date =  "2016-12-27" ;
$end_date = "2016-12-31";
while (strtotime($date) <= strtotime($end_date)) {
    echo "$date<br>";
    $date = date ("Y-m-d", strtotime("+1 day", strtotime($date)));
}

Answer 2

试试这个：

$date = $g[arrival];
$end_date = $g[departure];
while (strtotime($date) < strtotime($end_date)) {                                       
$date = date ("Y-m-d", strtotime("+1 day", strtotime($date)));
echo "$date<br>";
}

Answer 3

/ **  *生成两个日期之间的字符串日期数组  *  * @param string $ start开始日期  * @param string $ end结束日期  * @param string $ format输出格式（默认值：Y-m-d）  *  * @return数组  * /

function getDatesFromRange($start, $end, $format = 'Y-m-d') {
    $array = array();
    $interval = new DateInterval('P1D');

    $realEnd = new DateTime($end);
    $realEnd->add($interval);

    $period = new DatePeriod(new DateTime($start), $interval, $realEnd);

    foreach($period as $date) { 
        $array[] = $date->format($format); 
    }

    return $array;
}

然后，您可以按预期调用该函数：

getDatesFromRange('2010-10-01', '2010-10-05');

Answer 4

首先简单地回显日期然后递增日期

while (strtotime($date) <= strtotime($end_date)){
    echo "$date<br>";
    $date = date ("Y-m-d", strtotime("+1 day", strtotime($date)));
}

Answer 5

只需将增量放在循环体的末尾：

$date = $g[arrival];
$end_date = $g[departure];
while (strtotime($date) <= strtotime($end_date)) {                                       
    // Switched these lines so the increment is at the end of the body
    echo "$date<br>";
    $date = date ("Y-m-d", strtotime("+1 day", strtotime($date)));
}

Answer 6

$date =  "2016-12-27";
$end_date = "2016-12-31";
while (strtotime("+1 day", strtotime($date)) <= strtotime($end_date)) {  
    echo "$date<br>";
    $date = date ("Y-m-d", strtotime("+1 day", strtotime($date)));

}
echo "$date<br>";

查看截图