Question

我想用php减去两个日期。 Dateone存储在数据库中，而datetwo是当前日期。现在，我有这种奇怪的情况： Dateone是23-03-13 第二天是02-04-13

使用不同的减法方法，给出不同的结果。

方法一 - 返回-21

$sqldate ="SELECT exam_date FROM exam_table";
$fetchdate = mysql_query($sqldate);
$rowdate = mysql_fetch_array($fetchdate);
//Fetch the date stored in the database
$dateone = $rowdate['exam_date'];
//Calculate the current date today
$datetwo =date('d-m-y');
//Calculate the diff between the two dates
$datediff = $datetwo-$dateone;

在这种情况下，$ datediff返回-21

方法二 - 返回-7639

$sqldate ="SELECT exam_date FROM exam_table";
$fetchdate = mysql_query($sqldate);
$rowdate = mysql_fetch_array($fetchdate);
//Fetch the date stored in the database
$dateone = $rowdate['exam_date'];
//Calculate the current date
$datetwo =date('d-m-y');
//Calculate the diff between the two dates
$datetime1 = strtotime("$dateone");
$datetime2 = strtotime("$datetwo");
//seconds between the two times
$secs = $datetime2 - $datetime1;
$days = $secs / 86400;

在这种情况下，$ days返回-7639

Answer 1

为什么不在SQL中全部使用？

"SELECT time_to_sec(datediff(`exam_date`, curdate())) as datedifference FROM exam_table";

仅在PHP中使用：

fetch_assoc { $datediff = $row['datedifference'];

Answer 2

试试这个

echo $dateone = '02-04-13';
echo $datetwo = '06-04-13';
echo $datediff = $datetwo-$dateone;

Answer 3

SQL查询将是理想的

SELECT DATEDIFF(exam_date,CURDATE()) AS DiffDate FROM exam_table

<强> PHP

$dateone = $rowdate['DiffDate'];

Answer 4

为什么不使用DateTime::diff？资料来源：http://php.net/manual/en/datetime.diff.php

$datediff = $datetwo->diff($dateone);
echo $datediff;

这种差异将在时代。您需要将其转换为所需的格式。

Answer 5

是的问题是因为您的日期格式不是标准差异。您需要通知当前的格式和日期。将其转换为标准版以获得正确的差异。

$datetime1 = DateTime::createFromFormat('d-m-Y', '23-03-13'); #In you case it is considering 13-03-2023
$datetime1->format('d-m-YY');
$datetime2 = DateTime::createFromFormat('d-m-Y', '02-04-13'); #In you case it is considering 13-04-2002
$datetime2->format('d-m-YY');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days'); #output +10 days

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注意： PHP版本应为＆gt; = 5.3.0。

Php减去两个日期，不同的日期返回错误的结果

5 个答案: