Question

嗨我是前端世界的新手，我做了一个小的SQL数据库，里面装满了一些数据，我可以将数据从数据库中获取为我的PHP代码作为关联数组，但问题是我不能将它们填入我的表中，它们出现了桌子上方和外面。在任何帮助下，我都会感恩。

<?php
include("page.html");
include("databaseConnection.php");

$sth = $conn->prepare("SELECT ID, userName from passwords");
$sth->execute();

$result[] = $sth->fetchAll();

if(isset($_POST["showData"])){

echo "<table style='border: 2px solid black'>
<tr>
<th style='border: 1px solid red'>ID</th>
<th style='border: 1px solid black'>User Name</th>
</tr> <tbody>";

foreach ($result as $key => $value) {
    foreach ($value as $key1 => $value1) {
        echo "<tr>
        <td>".print_r($value1['ID'])."</td>
        <td>".print_r($value1['userName'])."</td>
        </tr>";
    }

}
echo "</tbody></table>";


}
?>

enter image description of the result here

Answer 1

您在字符串连接中使用print_r。

foreach ($value as $key1 => $value1) {
    echo "<tr>
    <td>".print_r($value1['ID'])."</td>
    <td>".print_r($value1['userName'])."</td>
    </tr>";
}

删除print_r语句，值将显示在表格内。

Answer 2

快速而肮脏的解决方案，未经测试： -




<?php
include("page.html");
include("databaseConnection.php");

$sth = $conn->prepare("SELECT ID, userName from passwords");
$sth->execute();

$result[] = $sth->fetchAll();

if(isset($_POST["showData"])){

echo "<table style='border: 2px solid black'>
<tr>
<th style='border: 1px solid red'>ID</th>
<th style='border: 1px solid black'>User Name</th>
</tr> <tbody>";
foreach ($result as $key => $value) {
    foreach ($value as $key1 => $value1) {
?>
<tr>
<td><? echo $value1['ID']; ?></td>
<td><? echo $value1['userName']; ?></td>
</tr>
<?php
    }

}
echo "</tbody></table>";
}
?>

Answer 3

你的tbody标签应该是你所有tr元素的父级和你在thead中的标题。检查以下结构：

<table>
 <thead>
  <tr>
     <th>Month</th>
     <th>Savings</th>
  </tr>
 </thead>
 <tbody>
  <tr>
     <td>January</td>
     <td>$100</td>
  </tr>
  <tr>
     <td>February</td>
     <td>$80</td>
  </tr>
 </tbody>
 <tfoot>
  <tr>
     <td>Sum</td>
     <td>$180</td>
  </tr>
 </tfoot>
</table>

如何使用php创建一个用数据库数据填充的普通HTML表？

3 个答案: