Question

我在Mysql数据库中存储日期，我正在尝试以年，月，日格式格式化那些日子。我存储日期的变量称为$days

这就是我所拥有的：

$years = ($days / 365);
$years = floor($years);

$month = ($days % 365) / 30.5;
$month = floor($month);

$days = ($days % 365) % 30.5;

$display = "$years" . " years, " . "$month" . " months, and " . "$days" . " days";

然而，每当我有90天的价值时，它只显示为2个月，并且没有显示任何日期，只有2个月。我在网上找到了这个，直到最近才开始工作。

任何人都可以给我一些更好的方法，或者我的代码中有任何错误吗？

Answer 1

如下所示： -

<?php
  $days = '365';
  $start_date = new DateTime(date("Y/m/d"));
  $end_date = new DateTime(date("Y/m/d",strtotime("+$days days")));
  $dd = date_diff($start_date,$end_date);
  echo "$dd->y year(s) $dd->m month(s) $dd->d day(s)";
?>

输出： - https://eval.in/918854和https://eval.in/918858

Answer 2

由于没有指定特定日期，您可以这样做

 {
"msg_id": "387b8515-0c1d-42a9-aa80-e68b66b66c27",
"_text": "how many people between Tuesday and Friday",
"entities": {
  "metric": [ {
    "metadata": "{'code': 324}",
    "value": "metric_visitor",
    "confidence": 0.9231
  } ],
  "datetime": [ {
    "value": {
      "from": "2014-07-01T00:00:00.000-07:00",
      "to": "2014-07-02T00:00:00.000-07:00"
    },
    "confidence": 1
  }, {
    "value": {
      "from": "2014-07-04T00:00:00.000-07:00",
      "to": "2014-07-05T00:00:00.000-07:00"
    },
    "confidence": 1
  } ]
}
}

现场演示：https://eval.in/918861

输出将如下：<?php $days = 500; $years = intval($days / 365); $days = $days % 365; $months = intval($days / 30); $days = $days % 30; echo "$years years, $months months, $days days"; ?>

考虑到PHP中的天数，如何计算它的年，月和日？

2 个答案: