问题描述
我想从数据库表中检索值并将它们显示在页面的 html 表中.我已经搜索过这个,但我找不到答案,虽然这肯定很容易(这应该是数据库的基础知识,哈哈).我想我搜索的术语具有误导性.数据库表名是 ticket,它现在有 6 个字段(submission_id、formID、IP、name、email 和 message),但应该有另一个字段,称为 ticket_number.我怎样才能让它在这样的 html 表中显示来自数据库的所有值:
I want to retrieve the values from a database table and show them in a html table in a page.I already searched for this but I couldn't find the answer, although this surely is something easy (this should be the basics of databases lol). I guess the terms I've searched are misleading.The database table name is tickets, it has 6 fields right now (submission_id, formID, IP, name, email and message) but should have another field called ticket_number.How can I get it to show all the values from the db in a html table like this:
<table border="1">
<tr>
<th>Submission ID</th>
<th>Form ID</th>
<th>IP</th>
<th>Name</th>
<th>E-mail</th>
<th>Message</th>
</tr>
<tr>
<td>123456789</td>
<td>12345</td>
<td>123.555.789</td>
<td>John Johnny</td>
<td>johnny@example.com</td>
<td>This is the message John sent you</td>
</tr>
</table>
然后是 'john' 下面的所有其他值.
And then all the other values below 'john'.
推荐答案
示例取自 W3Schools:PHP从 MySQL 中选择数据
Example taken from W3Schools: PHP Select Data from MySQL
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
这是一个学习的好地方!
It's a good place to learn from!
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