Question

因此，此PHP代码将在Web页面上打印我数据库中的所有表数据。

<?php
$hostname = '127.0.0.1:3306';        
$dbname   = 'login'; // Your database name.
$username = 'root';             // Your database username.
$password = '';                 // Your database password. If your database has no password, leave it empty.

mysql_connect($hostname, $username, $password) or DIE('Connection to host is failed, perhaps the service is down!');
mysql_select_db($dbname) or DIE('Database name is not available!');
$query="SELECT * FROM markers";
$result=mysql_query($query);

$fields_num = mysql_num_fields($result);
echo "<table border='2'><tr>";
// printing table headers
for($i=0; $i<$fields_num; $i++)
{
    $field = mysql_fetch_field($result);
    echo "<td>{$field->name}</td></div>";
}
echo "</tr>\n";
// printing table rows

while($row = mysql_fetch_row($result))

{
    echo "<table>";
    echo "<tr>";

    echo "<td>$row[0]</td>";
    echo "<td>$row[1]</td>";
    echo "<td>$row[2]</td>";
    echo "<td>$row[3]</td>";
    echo "</tr>\n";
    echo "</table></div>";

}
?>

但基本上它正在做的是。它格式不正确。就是这样。

Unformatted Table

我想要做的是将此表准确地放在网页的中心，给它一些花哨的边框（弯曲的边缘，很棒的字体。）

Answer 1

您无需为要输出的每一行重新创建表格，因此请更改

while($row = mysql_fetch_row($result))

{
    echo "<table>";
    echo "<tr>";

    echo "<td>$row[0]</td>";
    echo "<td>$row[1]</td>";
    echo "<td>$row[2]</td>";
    echo "<td>$row[3]</td>";
    echo "</tr>\n";
    echo "</table></div>";

}

到

echo "<div><table>";
while($row = mysql_fetch_row($result))

{
    echo "<tr>";
    echo "<td>$row[0]</td>";
    echo "<td>$row[1]</td>";
    echo "<td>$row[2]</td>";
    echo "<td>$row[3]</td>";
    echo "</tr>\n";
}
echo "</table></div>";

Answer 2

您正在为每次迭代创建一个新表。将<table>标记移到循环外部。

Answer 3

更改

    while($row = mysql_fetch_row($result))

{
    echo "<table>";
    echo "<tr>";

    echo "<td>$row[0]</td>";
    echo "<td>$row[1]</td>";
    echo "<td>$row[2]</td>";
    echo "<td>$row[3]</td>";
    echo "</tr>\n";
    echo "</table></div>";

}

到

 echo "<table>";
 while($row = mysql_fetch_row($result))

{
    echo "<tr>";
    echo "<td>$row[0]</td>";
    echo "<td>$row[1]</td>";
    echo "<td>$row[2]</td>";
    echo "<td>$row[3]</td>";
    echo "</tr>\n";    
}
echo "</table>";

Answer 4

您不需要在循环中放置表头，因为标头始终是相同的。对于标题，您可以尝试：

<table border="1" align="left">
<tr>
<th>Employee ID</th>
<th>Designation ID</th>
<th>First Name</th>
</tr>
if(!result)
while($row = mysql_fetch_row($result))
{
   <tr>
           <td><?php $row[0];?></td>
   </tr>
 } 
</table>

Answer 5

使用

echo "<table>";
while($row = mysql_fetch_row($result))

{
    echo "<tr>";
    echo "<td>$row[0]</td>";
    echo "<td>$row[1]</td>";
    echo "<td>$row[2]</td>";
    echo "<td>$row[3]</td>";
    echo "</tr>\n";
}
echo "</table>";

然后在文档的头部，使用以下代码：

<style>
table {
margin: 0 auto;
width: 50%;
}
</style>

为了使表格正确居中，需要为其设置宽度。否则它将不起作用。

Answer 6

你需要2个循环（第一个内部的第二个），

$result = mysqli_query('your sql');
// begin table, only once
echo '<table>';
// use thead for the table header :)
echo '<thead>';
for($i=0; $i<$fields_num; $i++)
{
    $field = mysqli_fetch_field($result);
    echo '<th>'.$field->name.'</th>';
}
echo '</thead>';
echo '<tbody>';
// 1st loop, each rows
while($row = mysqli_fetch_assoc($result))
{
    // each rows start with  <tr> / ends with </tr>
    echo '<tr>';
    // 2nd loop, each field
    foreach($row as $field => $value)
    {
        // you can also add class name in each case of your table, to add custom style
        // for each field in your css
        echo '<td class="'.$field.'">'.$value.'</td>';
    }
    echo '</tr>';
}
echo '</tbody>';
echo '</table>';

PHP中的MySql表格式

6 个答案: