PHP-检查日期范围

Question

I have a table on database called period . The fields in this table are: code | name | status | open_date | close_date code | name | status | open_date | close_date code | name | status | open_date | close_date . The date format is Y/m/d .

This is what I want: if today's date is out of date range in the database, the site won't display a form and a table column.

This is my code:

<?php
$sql5 = "select close_date from period"
         . " where status='1'";
$result5 = mysqli_query($link, $sql5);
if (!$result5) {
    die("<h3>SQL Error</h3>" . $sql5);
}
$row5 = mysqli_fetch_array($result5);

$date = date("Y/m/d");
?>

<?php
if($date >= $row3['open_date'] && $date <= $row5['close_date']) {
?>
<form action="manageInputPerwalian.php" method="POST">
    <table align="center">
        <tr>
                <td>MK Code:</td>
                    <td><input type="text" name="mkCode"/></td>
        </tr>     
        <tr>
                <td>KP:</td>
                <td><input type="text" name="kp"/></td>
        </tr>
            <tr>
                    <td></td>
                    <td><input type="submit" value="SUBMIT"/></td>
            </tr>
</table>
</form>
<?php } ?>


<table border="1" style="width:100%">
<tr>
    <th>MK CODE</th>
            <th>MK NAME</th>
            <th>CLASS</th>
            <?php if($date >= $row3['open_date'] && $date <= $row5['close_date']) { ?>
                <th>CANCEL</th>
            <?php } ?>
    </tr>
</table>

But, either today's date is out of range or in the range, the form and the table column won't show up. What's wrong? Please explain your answer. Thanks

Answer 1

use strtotime() to convert the dates to timestamp and compare, as,

$openDateStr = strtotime($row3['open_date']);
$closeDateStr = strtotime($row5['close_date']);
$todayStr = strtotime(date("Y/m/d"));
//compare
if($todayStr >= $openDateStr && $todayStr <= $closeDateStr) {
..

Answer 2

Everything you did right. Some small mistakes only.

1. There is no $row3 variable. Only $row5 only. So change row3 to row5.
2. In select query you selected only close_date. you need to select open_date also.
3. Next use strtotime() functions to compare.

Answer 3

There's nothing really worng with your code. Just a few miner fixes. $row3 should probably be $row5 in your code.

// Create a DateTime object
$now = new DateTime();

$stmt = "select close_date
from period
where status='1'
";

$result5 = mysqli_query($link, $sql5);

if (!$result5) {
    die("<h3>SQL Error</h3>" . $sql5);
}
$row5 = mysqli_fetch_array($result5);

$date = date("Y/m/d");

And then, your HTML

<!-- Note that you were using $row3 here -->
<?php if ($now->format('Y-m-d') >= $row5['open_date'] && $now->format('Y-m-d') <= $row5['close_date']): ?>
<form action="manageInputPerwalian.php" method="POST">
    <table align="center">
        <tr>
                <td>MK Code:</td>
                    <td><input type="text" name="mkCode"/></td>
        </tr>
        <tr>
            <td>KP:</td>
            <td><input type="text" name="kp"/></td>
        </tr>
        <tr>
            <td></td>
            <td><input type="submit" value="SUBMIT"/></td>
        </tr>
</table>
</form>
<?php èndif; ?>


<table border="1" style="width:100%">
<tr>
    <th>MK CODE</th>
            <th>MK NAME</th>
            <th>CLASS</th>
            <?php if ($now->format('Y-m-d') >= $row5['open_date'] && $now->format('Y-m-d') <= $row5['close_date']): ?>
                <th>CANCEL</th>
            <?php endif; ?>
    </tr>
</table>