Question

我想减去2个日期，并在几分钟内得到结果，如下面的代码给出了我想要的答案。

$to_time = strtotime("2017-03-27 17:31:40");  
$from_time = strtotime("2017-03-27 18:32:40");  
echo "sunolo1: ".round(abs($to_time - $from_time) / 60,2). " minute";

但是当我尝试使用php从mysql中检索动态日期时它不起作用它返回0.（我的表中的日期是时间戳）

 $d = new DateTime("now", new DateTimeZone("Europe/Athens"));

    $dateM = $d->format("Y-m-j H:i:s");
    $result = mysqli_prepare($con, "SELECT date FROM mytable WHERE id= ? ");

      mysqli_stmt_bind_param($result, 'i', $ids);

      mysqli_stmt_execute($result);

      mysqli_stmt_bind_result($result, $ddd);

  while(mysqli_stmt_fetch($result)){
    $sunolo_krathshs = round(abs($ddd - $dateM) / 60,2);
    echo "sunoloo: ".$sunolo_krathshs;
  }

Answer 1

您需要将$ ddd的值解析为DateTime对象，因为最简单的方法是比较DateTime对象。

$date = new DateTime();
$ddd = $date->setTimestamp($ddd);

$sunolo_krathshs = round(abs($ddd - $d) / 60,2);

Answer 2

请检查一下。

＆＃13;

<?php

$datetime1 = new DateTime("2017-03-27 17:31:40");
$datetime2 = new DateTime("2017-03-27 18:32:40");
$interval = $datetime1->diff($datetime2);
$hours   = $interval->format('%h'); 
$minutes = $interval->format('%i');
echo 'Diff. in minutes is: '.($hours * 60 + $minutes); 

?>

＆＃13;

Answer 3

在传递日期值时使用strtotime函数进行计算。像这样：

$sunolo_krathshs = round(abs(strtotime($dateM) - strtotime($ddd)) / 60,2);

Answer 4

使用DateTime类可以非常简单

$result = mysqli_prepare($con, "SELECT date FROM mytable WHERE id= ? ");

mysqli_stmt_bind_param($result, 'i', $ids);
mysqli_stmt_execute($result);
mysqli_stmt_bind_result($result, $ddd);

$to_time = new DateTime("now", new DateTimeZone("Europe/Athens"));
while(mysqli_stmt_fetch($result)){
    // probably want UTC as date times are stored on the db in UTC
    // but you may need to experiment with timezones here
    $from_date = DateTime::createFromFormat('Y-m-d H:i:s', $ddd, new DateTimeZone("UTC"));   
    echo round(abs($to_date->getTimestamp() - $from_date->getTimestamp()) / 60,2). " minute";
}

Answer 5

变量$ ids没有分配值操作...

从mysql获取日期时的奇怪结果

5 个答案: