gpt4 book ai didi

javascript - 如何获取前一天(连续)

转载 作者:行者123 更新时间:2023-12-03 03:40:15 28 4
gpt4 key购买 nike

大家好,我想获取前一天或最后一天的信息,只有当我按下按钮时,它才会显示最后一天(星期六)的所有信息,如果我再次单击按钮,它将显示最后一天的信息(星期五)如果我再次点击它(星期四)谢谢你们帮助我

编辑:

generate_attendance.php

if(isset($_POST['submit1'])){
$prev_date= date('Y/m/d',strtotime("-1 days"));


$query=mysqli_query($dbcon,"select * from attendance where date_added '$prev_date'")or die(mysql_error());
while($row=mysqli_fetch_array($query)){

$attendance_id=$row['attendance_id'];


?>

<tr>
<td><?php echo $row['lastname'].', '.$row['firstname']; ?></td>
<td><?php echo $row['course']; ?></td>
<td><?php echo $row['type']; ?></td>
<td><?php echo $row['year_level']; ?></td> <td><?php echo $row['date_added']; ?</td>
</tr>
<?php

<div class="controls">


<button name="submit1" type="submit1" class="btn btn-success"><i class="icon-plus-sign icon-large"></i> Previous Day</button>
</div>
</div>

generate_attendance.php(完整代码)

<div class="container">
<div class="margin-top">
<div class="row">
<div class="alert alert-info">
<button type="button" class="close" data-dismiss="alert">&times;</button>
<strong><i class="icon-user icon-large"></i>&nbsp;Attendance Report</strong>
</div>

<div class="span12">
<center class="title">
<h1>Attendance List</h1>
</center>

<div class="pull-right">
<a href="" onclick="window.print()" class="btn btn-info"><i class="icon-print icon-large"></i> Print</a>
</div>
<form method="post">
<div class="span3">



<div class="control-group">
<label class="control-label" for="inputEmail"><!-- Attendance Report --></label>
<div class="controls">
<label class="control-label" for="inputEmail">From</label>
<div class="controls">
<input type="date" name="from_date" id="date1" alt="date" class="IP_calendar" title="d/m/Y">
</div>
</div>
</div>
<div class="control-group">
<label class="control-label" for="inputEmail">To Date</label>
<div class="controls">
<input type="date" name="to_date" id="date2" alt="date" class="IP_calendar" title="d/m/Y">
<!-- <input type="text" class="w8em format-d-m-y highlight-days-67 range-low-today" name="due_date" id="sd" maxlength="10" style="border: 3px double #CCCCCC;" required/> -->
</div>
</div>
<div class="control-group">
<div class="controls">

<button name="submit" type="submit" class="btn btn-success"><i class="icon-plus-sign icon-large"></i> Search</button>
</div>
</div>
<div class="controls">


<button name="submit1" type="submit1" class="btn btn-success"><i class="icon-minus-sign icon-large"></i> Previous Day</button>
</div>
</div>



<div class="span8">
<div class="alert alert-success"><strong>Attendance Report</strong></div>
<table cellpadding="0" cellspacing="0" border="0" class="table" id="example">

<thead>
<tr>



<th>Name</th>
<th>Program Code</th>
<th>Type</th>
<th>Year level</th>
<th>Date Log-in</th>

</tr>
</thead>
<tbody>
<?php
if(isset($_POST['submit'])){
$from_date=$_POST['from_date'];
$to_date=$_POST['to_date'];

$query=mysqli_query($dbcon,"select * from attendance where date_added between '$from_date' and '$to_date'")or die(mysql_error());
while($row=mysqli_fetch_array($query)){

$attendance_id=$row['attendance_id'];


?>

<tr>
<td><?php echo $row['lastname'].', '.$row['firstname']; ?></td>
<td><?php echo $row['course']; ?></td>
<td><?php echo $row['type']; ?></td>
<td><?php echo $row['year_level']; ?></td>
<td><?php echo $row['date_added']; ?></td>

</tr>
<?php
}}
?>

<?php


if(isset($_POST['submit1'])){



$prev_date= date('Y/m/d',strtotime("-1 days"));

$query=mysqli_query($dbcon,"select * from attendance where date_added between '$curr_date' and '$prev_date'")or die(mysql_error());
while($row=mysqli_fetch_array($query)){

$attendance_id=$row['attendance_id'];


?>

<tr>
<td><?php echo $row['lastname'].', '.$row['firstname']; ?></td>
<td><?php echo $row['course']; ?></td>
<td><?php echo $row['type']; ?></td>
<td><?php echo $row['year_level']; ?></td>
<td><?php echo $row['date_added']; ?></td>

</tr>
<?php
}}
?>

</tbody>
</table>

</form>
</div>
</div>

</div>
</div>
</div>

最佳答案

可能会传递一个查询参数,该参数指示应减去多少个日期,因此网址如下所示:

yourphpscript.php? days = 3

然后你可以在 php 中获取它,并更改日期构建:

$days = $_GET["days"];
if(!isset($days)){
$days = 1;
}
$days = intval($days);
$prev_date= date('Y/m/d',strtotime("-".$days." days"));

所以现在唯一缺少的就是更改下一个请求的 url,如下所示:

<form href="?days=<?php echo $days+1;?>" >

关于javascript - 如何获取前一天(连续),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/45658893/

28 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com