我是PHP和MySQL的新手,我做了一些类似于手册和指南的内容,但我无法获取结果。当我在浏览器中运行它时,标题显示在页面上,而不是数据库内容。这是我的代码:
<?php
$hostname='localhost';
$username='root';
$password='12345678';
$db=mysqli_connect($hostame, $username, $password);
$selectdb=mysqli_select_db('journal', $db);
$result=mysqli_query("SELECT id,nn,npd,date FROM journal ORDER by id ASC"); ?>
<table border="1">
<thead>
<tr>
<th rowspan="2">Header 1</th> <!-- id -->
<th rowspan="2">Header 2</th> <!-- nn -->
<th rowspan="2">Header 3</th> <!-- npd -->
<th rowspan="2">Header 4</th> <!-- date -->
</tr>
</thead>
<tbody>
<?php while($array=mysqli_fetch_assoc($result))
{ ?>
<tr>
<td><?php echo $array['id']; ?></td>
<td><?php echo $array['nn']; ?></td>
<td><?php echo $array['npd']; ?></td>
<td><?php echo $array['date']; ?></td>
</tr>
<?php } ?>
</tbody>
</table>
<?php mysqli_close($db); ?>
答案 0 :(得分:0)
您没有将连接参数传递给错误的mysqli_query。
$db = mysqli_connect($hostname, $username, $password, $database_name);
$sql = "SELECT id,nn,npd,date FROM journal ORDER by id ASC";
$result = mysqli_query($db, $sql);
答案 1 :(得分:0)
将第一个参数添加到$result=mysqli_query($con,$sql_query);
作为连接
onChangeTab
答案 2 :(得分:0)
尝试此代码,看看是否对您有所帮助
$hostname='localhost';
$username='root';
$password='12345678';
$db=mysqli_connect($hostame, $username, $password,'journal');
$result=mysqli_query($db,"SELECT id,nn,npd,date FROM journal ORDER by id ASC");
?>
<table border="1">
<thead>
<tr>
<th rowspan="2">Header 1</th> <!-- id -->
<th rowspan="2">Header 2</th> <!-- nn -->
<th rowspan="2">Header 3</th> <!-- npd -->
<th rowspan="2">Header 4</th> <!-- date -->
</tr>
</thead>
<tbody>
<?php while($array=mysqli_fetch_assoc($result))
{ ?>
<tr>
<td><?php echo $array['id']; ?></td>
<td><?php echo $array['nn']; ?></td>
<td><?php echo $array['npd']; ?></td>
<td><?php echo $array['date']; ?></td>
</tr>
<?php } ?>
</tbody>
</table>
<?php mysqli_close($db); ?>