如何动态地将datediff插入网页以保持实时更新(现在)
how to dynamically insert datediff into webpage to keep updated in real time (now)
我可以在 phpmyadmin 中请求显示今天日期和用户选择的日期之间的天数:
SELECT DATEDIFF(CURDATE(), date_instructed
) 从 date_tester
开始的天数
.. 但我认为这不会更新所有 table,并且只会在 table 中创建新记录时更新。
因此,任何人都可以建议使用用户输入的日期动态 update/count 列中天数的最佳方法。
它甚至应该放在一栏中吗?或者我应该使用 php 来回显网页上每一行的结果。
如果我应该用脚本操纵“date_instructed”,然后将其单独输入 html table,他们的 link 会告诉我如何操作要做到这一点?
非常感谢
<?php
include 'dbh.php';
?>
<table class="table table-sm table table-bordered
table table-hover table table-striped "> <!--, table table-inverse-->
<thead class="thead-inverse">
<tr class="warning">
<th>First Name</th>
<th>Last</th>
<th>Date Instructed</th>
<th>Days On Market</th>
</tr>
</thead>
<?php
$query = "SELECT first_name, last_name, date_instructed , days_onmarket
FROM date_tester";
$result = mysqli_query($conn, $query);
while ($date_tester = mysqli_fetch_assoc($result)) {
echo "<tr>";
echo "<td>" . $date_tester['first_name'] . "</td>";
echo "<td>" . $date_tester['last_name'] . "</td>";
echo "<td>" . $date_tester['date_instructed'] . "</td>";
}
?>
</tbody>
</table>
</div>
据我了解,这就是您所需要的:
$query = "SELECT first_name, last_name, date_instructed , days_onmarket, DATEDIFF(CURDATE(), 'date_instructed') AS DAYS FROM date_tester";
$result = mysqli_query($conn, $query);
while ($date_tester = mysqli_fetch_assoc($result)) {
echo "<tr>";
echo "<td>" . $date_tester['first_name'] . "</td>";
echo "<td>" . $date_tester['last_name'] . "</td>";
echo "<td>" . $date_tester['date_instructed'] . "</td>";
echo "<td>" . $date_tester['DAYS'] . "</td>";
}
我可以在 phpmyadmin 中请求显示今天日期和用户选择的日期之间的天数:
SELECT DATEDIFF(CURDATE(), date_instructed
) 从 date_tester
.. 但我认为这不会更新所有 table,并且只会在 table 中创建新记录时更新。
因此,任何人都可以建议使用用户输入的日期动态 update/count 列中天数的最佳方法。
它甚至应该放在一栏中吗?或者我应该使用 php 来回显网页上每一行的结果。
如果我应该用脚本操纵“date_instructed”,然后将其单独输入 html table,他们的 link 会告诉我如何操作要做到这一点?
非常感谢
<?php
include 'dbh.php';
?>
<table class="table table-sm table table-bordered
table table-hover table table-striped "> <!--, table table-inverse-->
<thead class="thead-inverse">
<tr class="warning">
<th>First Name</th>
<th>Last</th>
<th>Date Instructed</th>
<th>Days On Market</th>
</tr>
</thead>
<?php
$query = "SELECT first_name, last_name, date_instructed , days_onmarket
FROM date_tester";
$result = mysqli_query($conn, $query);
while ($date_tester = mysqli_fetch_assoc($result)) {
echo "<tr>";
echo "<td>" . $date_tester['first_name'] . "</td>";
echo "<td>" . $date_tester['last_name'] . "</td>";
echo "<td>" . $date_tester['date_instructed'] . "</td>";
}
?>
</tbody>
</table>
</div>
据我了解,这就是您所需要的:
$query = "SELECT first_name, last_name, date_instructed , days_onmarket, DATEDIFF(CURDATE(), 'date_instructed') AS DAYS FROM date_tester";
$result = mysqli_query($conn, $query);
while ($date_tester = mysqli_fetch_assoc($result)) {
echo "<tr>";
echo "<td>" . $date_tester['first_name'] . "</td>";
echo "<td>" . $date_tester['last_name'] . "</td>";
echo "<td>" . $date_tester['date_instructed'] . "</td>";
echo "<td>" . $date_tester['DAYS'] . "</td>";
}