大家好我想要获取上一个日期或最后一天的信息，只有当我按下按钮时它会在最后一天（星期六）显示所有信息，如果我再次点击该按钮，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）感谢帮助我们
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">×</button>
<strong><i class="icon-user icon-large"></i> 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>
答案 0 :(得分：1)
可能会传递一个查询参数，该参数指示应该减去多少日期，因此网址如下所示：
yourphpscript.php? days = 3
然后你可以在php中获得它，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;?>" >
答案 1 :(得分：0)
您可以在POST数据中将当前显示的日期发送回服务器。
<input type="hidden" name="currently_displayed" value="<?php echo $prev_date ?>">
<button name="submit1" type="submit1" class="btn btn-success"><i class="icon-plus-sign icon-large"></i> Previous Day</button>
然后在你的PHP中：
if (isset($_POST['currently_displayed'])) {
// Convert as unix-timestamp
$currentlyDisplayed = strtotime($_POST['currently_displayed']);
} else {
// Not posted, assuming user wants to see yesterday (thus $currentlyDisplayed should be today/now)
$currentlyDisplayed = time();
}
// strtotime( $formatString, $relativeTo );
$prev_date= date('Y/m/d',strtotime("-1 days", $currentlyDisplayed));
答案 2 :(得分：0)
我建议您使用javascript / JQuery AJAX请求，您可以立即更改它，并且您可以跟踪您前几天的点击次数！你不再需要ifset POST submit1。
在按钮上＆＃34; Submit1＆＃34; ，具有发送numberPrevDays值的AJAX请求的onclick函数。确保跟踪此次点击（哎呀，你甚至可以在第二天使用另一个按钮）。有几种方法可以做到这一点。在发送到AJAX请求之前，要么只有一个将继续递增（或递减）的全局变量
有一个单独的文件＆amp;函数php，它接收Prev天数。此文件将处理数据并输出响应json您的SQL mysqli_fetch_array数据
使用该输出响应，只需清空并使用新数据重新加载表
抱歉，我没有写实际代码