php 使用 MySQL 表数据填充 HTML 表
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Populate HTML Table with MySQL Table Data
提问by Tobias Karl
I'm really new to HTML/PHP and I want to build a simple website that returns the data from my MySQL database in form of a table..
我真的是 HTML/PHP 的新手，我想建立一个简单的网站，以表格的形式从我的 MySQL 数据库返回数据。
My code looks like this for now:
我的代码现在看起来像这样：
<?php
$server = mysql_connect("server", "username", "password");
$db = mysql_select_db("ni354077_2sql1", $server);
$query = mysql_query("SELECT * FROM Jobs");
?>
<table>
<tr>
<td>AuftragsID</td>
<td>Startort</td>
<td>Zielort</td>
<td>Gewicht</td>
<td>Fracht</td>
</tr>
<?php
while ($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>".$row[AuftragsID]."</td>";
echo "<td>".$row[Startort]."</td>";
echo "<td>".$row[Zielort]."</td>";
echo "<td>".$row[Gewicht]."</td>";
echo "<td>".$row[Fracht]."</td>";
echo "</tr>";
}
?>
</table>
However it doesn't seem to be working properly. My table is full of "echo (..)" here is a Picutre of what it looks like:
但是，它似乎无法正常工作。我的桌子上满是“回声（..）”，这是它的外观图片：
Am I missing something ? I'm coming from C# WinForms and am confused with that.
我错过了什么吗？我来自 C# WinForms 并且对此感到困惑。
回答by Vivek Singh
use this u have missed single quotes around php values, put like below
使用这个你错过了 php 值周围的单引号，如下所示
<?php
while ($row = mysql_fetch_array($query)) {?>
<tr>
<td><?php echo $row['AuftragsID'];?></td>
<td><?php echo $row['Startort'];?></td>
<td><?php echo $row['Zielort'];?></td>
<td><?php echo $row['Gewicht'];?></td>
<td><?php echo $row['Fracht'];?></td>
</tr>
<?php } ?>
回答by devnull
First of all, you mentioned that you're new to PHP.
首先，您提到您是 PHP 新手。
According to PHP mYSQL Documentation.
根据 PHP mySQL 文档。
WarningThis extension is deprecated as of PHP 5.5.0, and will be removed in the future. Instead, the MySQLi or PDO_MySQL extension should be used.
警告此扩展自 PHP 5.5.0 起已弃用，将来会被删除。相反，应使用 MySQLi 或 PDO_MySQL 扩展。
In other words. mySql will not be available and you better start using mySqli or PDO (PHP DATA OBJECTS).
换句话说。mySql 将不可用，您最好开始使用 mySqli 或 PDO（PHP 数据对象）。
MYSQLI: http://php.net/manual/en/book.mysqli.php
MYSQLI：http://php.net/manual/en/book.mysqli.php
PDO: http://php.net/manual/en/book.pdo.php
PDO：http: //php.net/manual/en/book.pdo.php
The error you're getting because you're getting the key of array without quotes
你得到的错误是因为你得到了不带引号的数组的键
Right Syntax:
正确的语法：
$arrName['keyName'];
So you're solution will be like this
所以你的解决方案将是这样的
<?php
while ($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row['AuftragsID'] . "</td>";
echo "<td>" . $row['Startort'] . "</td>";
echo "<td>" . $row['Zielort'] . "</td>";
echo "<td>" . $row['Gewicht'] . "</td>";
echo "<td>" . $row['Fracht'] . "</td>";
echo "</tr>";
}
?>
回答by sakshi
Try This Code:
<?php
$server = mysql_connect("server", "username", "password");
$db = mysql_select_db("ni354077_2sql1", $server);
$query = mysql_query("SELECT * FROM Jobs");
?>
<table>
<tr>
<td>AuftragsID</td>
<td>Startort</td>
<td>Zielort</td>
<td>Gewicht</td>
<td>Fracht</td>
</tr>
<?php
while ($row = mysql_fetch_array($query)) {
?>
<tr>
<td><?php echo $row[AuftragsID]; ?></td>
<td><?php echo $row[Startort]; ?></td>
<td><?php echo $row[Zielort]; ?></td>
<td><?php echo $row[Gewicht]; ?></td>
<td><?php echo $row[Fracht]; ?></td>
</tr>
<?php
}
?>
</table>
回答by Scoffy
first use table row then use the loop inside PHP then populate your table with the data from the database
首先使用表格行，然后使用 PHP 中的循环，然后使用数据库中的数据填充表格
<tr>
<?php
while ($row = mysql_fetch_array($query))
{
?>
<td><?php echo $row["AuftragsID"]?></td>
<td><?php echo $row["Startort"] ?></td>
<td><?php echo $row[Zielort]?></td>
<td><?php echo $row[Gewicht]?></td>
<td> <?php echo $row[Fracht]?></td>
</tr>
<?php }?>
回答by WebDeveloper
change the loop to
将循环更改为
while ($row = mysql_fetch_array($query)) {
echo "<tr>";
echo "<td>".$row['AuftragsID']."</td>";
echo "<td>".$row['Startort']."</td>";
echo "<td>".$row['Zielort']."</td>";
echo "<td>".$row['Gewicht']."</td>";
echo "<td>".$row['Fracht']."</td>";
echo "</tr>";
}
