Question

数据未显示在图像的顶部。

<h1>Pernix RsPs<br />
  Official ItemDB!</h1>
<br />
<form action="search.php" method="POST">
<input type="text" name="search" /><br /><br />
<input type="Submit" value="Search ItemDB" />
</form>
 <table border="0" width="400">
  <img src="top.png"></table>
  <table border="1" width="400">
 <?php
    $search = $_POST["search"];
        MySQL_connect("localhost", "host", "pass");
        MySQL_select_db("dbname");
        $result = mysql_query("SELECT * FROM items WHERE name LIKE '%" . mysql_real_escape_string($search) . "%'  ORDER by name") or trigger_error(mysql_error().$sql);

while($row = mysql_fetch_array($result))
  {
  ?>

<?

    echo '<tr>';
    echo '<img src="post.png">';
    echo '<td>' . $row['img'] . '</td>';
    echo '<td>' . $row['name'] . '</td>';
    echo '<td>' . $row['price'] . '</td>';

    echo '</tr>';
        }
    }
?>  

</table>
</center>

Answer 1

只需在表格周围添加div，名称为theme，如下所示

<div class="theme">
<table border="1" width="400">
 <?php
    $search = $_POST["search"];
        MySQL_connect("localhost", "pernixr2_ge", "pass");
        MySQL_select_db("pernixr2_ge");
        $result = mysql_query("SELECT * FROM items WHERE name LIKE '%" . mysql_real_escape_string($search) . "%'  ORDER by name") or trigger_error(mysql_error().$sql);

while($row = mysql_fetch_array($result))
  {
  ?>

<?

    echo '<tr>';
    echo '<img src="post.png">';
    echo '<td>' . $row['img'] . '</td>';
    echo '<td>' . $row['name'] . '</td>';
    echo '<td>' . $row['price'] . '</td>';

    echo '</tr>';
        }
    }
?>  

</table>
</div>

在.css文件中添加此代码

.theme-table, .theme {
    height: 100%;
    width: 100%;  
}

.theme-table {
    border:1px solid black;
    table-layout: fixed;

}

.theme {
    background:url(http://www.psdgraphics.com/wp-content/uploads/2012/04/light-wooden-background.jpg);
    background-repeat:no-repeat;
}

注意：

将图片的网址替换为top.png或您想要的任何图片

更新：

由于mysql被折旧，我已经使用mysqli给出了代码，其内部有样式

<?php
$Connection = new mysqli("localhost", "root", "", "mysqli");
if ($Connection->connect_errno) 
{
    echo "Connection Failed".$Connection->connect_error;
    exit();
}
$query = "SELECT * from users";
if ($result = $Connection->query($query)) {
?>
<div class="theme">
<style type="text/css">
.theme {
    background:url(http://www.psdgraphics.com/wp-content/uploads/2012/04/light-wooden-background.jpg);
    background-repeat:no-repeat;
}

</style>
<table style="width:45%" border="1">
  <tr>
    <td>Jill</td>
    <td>Smith</td> 
    <td>50</td>
  </tr>
  <?php
    while ($row = $result->fetch_assoc()) 
    {
        echo "
                <tr>
                    <td>".$row["id"]."</td> 
                    <td>".$row["username"]."</td>
                    <td>".$row["username"]."</td>
                  </tr>
        ";
    }
    $result->free();
}
?>
</table>
</div>
<?php
$Connection->close();
?>

以下是Image

在图像PHP上显示数据

1 个答案: