我不是一个耐心的编码员。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,我怎样才能在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>
答案 0 :(得分:1)
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); ?>
答案 1 :(得分:0)
您无法将变量从PHP代码“转移”到HTML代码,因为后者不是编程语言,它只是描述没有变量或过程的网页的结构化标记语言。
相反,您可以显示PHP变量的当前值,然后HTML标记将包含该变量的实际值。例如:
<table>
<tr>
<td><?php echo $name; ?></td>
</tr>
</table>