【发布时间】:2023-04-06 02:17:01
【问题描述】:
在将输入传递给变量时遇到问题。我的html是
<form id="checkform" action="DateTest.php" method = "post">
From Date:<label for="date"></label><br />
<input id="date" type = "date" name = "date" placeholder = "Date" autocomplete="off" /><br />
Until Date:<label for="date1"></label><br />
<input id="date1" type = "date" name = "date1" placeholder = "Date" autocomplete="off" /><br />
<input class="submit" title="submit" type = "submit" value = "Submit" name = "Submit"/><br />
</form>
</div>
我的php是
<?php
$fromDate = isset($_POST['date']) ? $_POST['date'] : 0;
$toDate = isset($_POST['date1']) ? $_POST['date1'] : 0;
var_dump($fromDate);
var_dump($toDate);
$dateMonthYearArr = array();
$fromDateTS = strtotime($fromDate);
$toDateTS = strtotime($toDate);
for ($currentDateTS = $fromDateTS; $currentDateTS <= $toDateTS; $currentDateTS += (60 * 60 * 24)) {
$currentDateStr = date("d-m-Y",$currentDateTS);
$dateMonthYearArr[] = $currentDateStr;
}
echo "<pre>";
print_r($dateMonthYearArr);
echo "</pre>";
?>
这是打印一个空数组,所以我在传输后转储了它,它显示$fromDate 按指示传递,但将$toDate 显示为一个空整数。
【问题讨论】:
-
如果你的
var_dump($_POST);是date1也不见了? -
是的,只显示日期。超级困惑!
标签: php date variables transfer