我不是一个狂热的程序员。我现在正在学习 php/htmp/mysql/css :)
所以，我用 php 进行 mysql 查询，现在我想在 html 表中显示查询结果。但我不知道，如何在 php 和 html 代码之间传递变量。
我的 php 文件现在看起来像这样:
<?php
$con=mysqli_connect("my_db_host","my_db_username","my_db_pass","my_db_name");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM hlstats_Servers
WHERE serverId='2'");
while($row = mysqli_fetch_array($result))
{
$name=$row['name'];
$ip=$row['address'] . ":" . $row['port'];
$player=$row['act_players'] . "/" . $row[max_players];
$map=$row['act_map'];
}
mysqli_close($con);
?>
<html>
<head>
</head>
<body>
<table>
<tr>
<td>$name</td>
</tr>
</table>
<table>
<tr>
<td>Ip cím:</td>
<td>$ip</td>
</tr>
<tr>
<td>Jelenlegi pálya:</td>
<td>$map</td>
</tr>
<tr>
<td>Játékosok</td>
<td>$player</td>
</tr>
</table>
</body>
</html>
最佳答案
Sok sikert(祝你好运):)
<?php
$con = mysqli_connect("my_db_host", "my_db_username", "my_db_pass", "my_db_name");
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query(
$con,
"SELECT * FROM hlstats_Servers
WHERE serverId='2'"
);
?>
<html>
<head></head>
<body>
<table>
<?php while ($row = mysqli_fetch_array($result)): ?>
<?php $name = $row['name']; ?>
<?php $ip = $row['address'] . ":" . $row['port']; ?>
<?php $player = $row['act_players'] . "/" . $row['max_players']; ?>
<?php $map = $row['act_map']; ?>
<tr>
<td colspan="2"><?php echo $name; ?></td>
</tr>
<tr>
<td>Ip cím:</td>
<td><?php echo $ip; ?></td>
</tr>
<tr>
<td>Jelenlegi pálya:</td>
<td><?php echo $map; ?></td>
</tr>
<tr>
<td>Játékosok</td>
<td><?php echo $player; ?></td>
</tr>
<?php endwhile; ?>
</table>
</body>
</html>
<?php mysqli_close($con); ?>
关于PHP数据库查询显示在html中，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20307865/