我已经浏览了网络以寻找答案,并查看了Stack Overflow,但尚未找到解决我问题的可行解决方案。首先发布Stack Overflow,请耐心等待我!
我正在用PHP编写一个基本的考勤程序,每个成员都可以选择现在,迟到或缺席。该页面当前从数据库生成活动成员列表,并且每个页面都有一组单选按钮,其中包含可能的选项(显示,延迟或不存在)。
我遇到的问题是如何将值传递到数据库中以获取动态成员列表。由于成员是动态的,我如何将这些成员添加到数据库中?
修改
根据要求,以下是页面上的一些示例代码,它们根据成员生成单选按钮,但我的问题源于如何将这些值传递到数据库中。
<div class="row">
<div class="col-lg-12">
<div class="panel panel-default">
<div class="panel-heading">
Members
</div>
<!-- /.panel-heading -->
<div class="panel-body">
<div class="table-responsive">
<table class="table table-striped table-bordered table-hover">
<form action = "attendanceProcess.php" method="post">
<thead>
<tr>
<th>Name</th>
<th>Present</th>
<th>Late</th>
<th>Absent</th>
</tr>
</thead>
<tbody>
<?php
$sql = "SELECT ID, firstName, lastName, status FROM members WHERE status='1' OR status='2' OR status='3'";
$result = mysql_query($sql);
if (mysql_num_rows($result) > 0) {
while($row = mysql_fetch_assoc($result)) {
if($row['status'] == 1) {
$status = "<span class=\"label label-success pull-right\">" . statusIDtoStatusName($row['status']) . "</span>";
}
if($row['status'] == 2) {
$status = "<span class=\"label label-primary pull-right\">" . statusIDtoStatusName($row['status']) . "</span>";
}
if($row['status'] == 3) {
$status = "<span class=\"label label-warning pull-right\">" . statusIDtoStatusName($row['status']) . "</span>";
}
// Present = 1
// Late = 2
// Absent = 3
echo "<tr class=\"odd gradeX\"><td>" . $row['firstName'] . " " . $row['lastName'] . " " . $status . "</td><td><input type=\"radio\" name=\"". $row['ID'] ."\" value=\"1\"></td><td><input type=\"radio\" name=\"". $row['ID'] ."\" value=\"2\"></td><td><input type=\"radio\" name=\"". $row['ID'] ."\" value=\"3\"></td></tr>";
}
} else {
echo "0 results";
}
?>
</tbody>
</table>
<input type="submit" value="Submit"><br />
</form>
</div>
<!-- /.table-responsive -->
</div>
<!-- /.panel-body -->
</div>
<!-- /.panel -->
</div>
<!-- /.col-lg-12 -->
</div>
答案 0 :(得分:-1)
会员存储在一个表中?
UPDATE `TABLE` SET `status` = "$status" WHERE `member_id` = "$id"