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PHP试图比较两个日期,以显示费用_Php_Date - Fatal编程技术网

PHP试图比较两个日期,以显示费用

PHP试图比较两个日期,以显示费用,php,date,Php,Date,我正在为一个客户编辑一些代码,我花了很长时间才弄明白这一点。我试图比较两个日期,并根据差异设定一定金额的费用,但我在比较日期时遇到问题。第一个日期由客户选择,第二个日期设置为今天日期,如果客户到达日期与今天相同或在1天内,则费用应为25;例如,今天是01/25,如果到达日期是01/25或01/26,则费用应为25,如果到达日期是从今天起2天或更长时间,则费用为0 客户选择的日期格式为:m/d/Y 选择人: <tr style="border-bottom: 1px solid #ddddd

我正在为一个客户编辑一些代码,我花了很长时间才弄明白这一点。我试图比较两个日期,并根据差异设定一定金额的费用,但我在比较日期时遇到问题。第一个日期由客户选择,第二个日期设置为今天日期,如果客户到达日期与今天相同或在1天内,则费用应为25;例如,今天是01/25,如果到达日期是01/25或01/26,则费用应为25,如果到达日期是从今天起2天或更长时间,则费用为0

客户选择的日期格式为:m/d/Y 选择人:

<tr style="border-bottom: 1px solid #dddddd;"><td style="text-align:right;">Arrival Date:</td><td><input id="arrivaldate" name="arrivaldate" value="<?php if (strlen($arrivaldate)>4) { echo date("m/d/Y", strtotime($arrivaldate)); } else { echo date("m/d/Y"); } ?>" placeholder="click here for arrival date" readonly></td></tr>

到达日期:您试过类似的方法吗

$date1 = new DateTime($today);
$date2 = new DateTime($choosen_date);

// Should have the days difference between your 2 dates
$diff = $date2->diff($date1)->format("%a");
然后应该很容易设置您的费用

$date1 = new DateTime($today);
$date2 = new DateTime($choosen_date);

// Should have the days difference between your 2 dates
$diff = $date2->diff($date1)->format("%a");