我想创建一个HTML表，但应从mysql数据库中的值生成行。问题是我希望有一个布尔框，用户可以在其中标记它，然后按一个按钮以更新数据库中的表。我该怎么做？
到目前为止的代码：
<?php
session_start();
require("connectToEvent_log.php");
$connectToEvent = connect2db();
$uid = '2'; // for the filnal version: @$_SESSION['uid'];
$view_event = "SELECT * FROM event_log WHERE uid = $uid";
$view_event_query = mysqli_query($connectToEvent, $view_event);
$row = mysqli_num_rows($view_event_query);
$print = mysqli_fetch_array($view_event_query);
?>
<html>
<head>
<title>Events</title>
</head>
<body>
<form action="viewEvents.php" method="POST">
<table border = '1'>
<tr>
<?php
while($row != 0){
echo "<td>".$print['hours']."</td>";
echo "<td>".$print['date']."</td>";
}
?>
</tr>
</table>
<form/>
</form>
</body>
</html>
最佳答案
您可以轻松地遍历mysqli_fetch_array函数的结果以创建表行。创建复选框标记很容易完成，我假设表具有主键id，并且存储复选框值（0或1）的列称为checkbox。
<?php
session_start();
require("connectToEvent_log.php");
$connectToEvent = connect2db();
$uid = '2'; // for the filnal version: @$_SESSION['uid'];
$view_event = "SELECT * FROM event_log WHERE uid = $uid";
$view_event_query = mysqli_query($connectToEvent, $view_event);
$num_rows = mysqli_num_rows($view_event_query);
$rows = mysqli_fetch_array($view_event_query);
?>
<html>
<head>
<title>Events</title>
</head>
<body>
<form action="viewEvents.php" method="POST">
<table border="1">
<thead>
<tr>
<td>Id</td>
<td>Date</td>
<td>Hours</td>
<td>Checkbox</td>
</tr>
</thead>
<tbody>
<?php
for ($i = 0; $i < count($num_rows); $i++) {
?>
<tr>
<td><?php print $rows[$i]["eid"]; ?></td>
<td><?php print $rows[$i]["date"]; ?></td>
<td><?php print $rows[$i]["hours"]; ?></td>
<td><input type="checkbox" name="row[<?php $rows[$i]["eid"]?>][checkbox]" value="1" <?php if ($rows[$i]["accepted"]) print ' checked="checked"'; ?>/></td>
</tr>
<?php
}
?>
</tbody>
</table>
<input type="submit" />
</form>
</body>
</html>
关于php - 带内联PHP的HTML表，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/30262506/