doubi4435 2018-06-08 11:58
浏览 43
已采纳

如何在PHP中创建表并对其进行样式设置

I have managed to take information from my database of the website, however I have no idea how to present the results in a table and how to style it. I tried putting <table> outside of the whole PHP, clearly did not work. I tried echoing a <table> tag before the result echo and a closing </table> tag after it, but that did not do it. This is the code I am working with:

<?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "onlib";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    //Takes all the results from the table with genre 5.
    $sql = "SELECT name, description, content FROM books WHERE genre='5'";
    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
            echo "<span style='color:white;'>"."<br> Name: ".$row["name"]."<br> Description: ".$row["description"]."<br> Content: ".$row["content"] ."<br>"."</p>";
        }
    } else {
        echo "0 results";
    }

    $conn->close();
?>

I am still new in PHP, trying to understand how the whole thing works. Thanks in advance!

  • 写回答

5条回答 默认 最新

  • 「已注销」 2018-06-08 12:24
    关注
    <?php

function tableV1 ($row) {
    echo '<tr>';
    echo '<td>' . $row['name'] . '</td>';
    echo '<td>' . $row['description'] . '</td>';
    echo '<td>' . $row['content'] . '</td>';
    echo '</tr>';
}

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "onlib";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } 
?>

    Always do Database Connection first, before Outputting anything, that way, you can create custom error message to show instead of a failed database conntection or no content at all.

    <style type="text/css">
table {}
tbody {}
td {}
th {}
thead {}
tr {}
</style>

    Style is used inside the <head></head> to style the table, CSS it's called.

    <table>
    <thead>
        <th>Name</th>
        <th>Description</th>
        <th>Content</th>
    </thead>
    <tbody>
<?php
// Takes all the results from the table with genre 5.
$sql = "SELECT name, description, content FROM books WHERE genre='5'";

$result = $conn->query($sql);

if ($result->num_rows > 0) {

    // Output data of each row
    while($row = $result->fetch_assoc()) {
        tableV1($row);
    }

} else {
    echo '<tr><td colspan="3">0 results</td></tr>';
}
?>
    </tbody>
</table>

    Output contents from database.

    <?php

$conn->close();

?>

    Close database connection in the end. All together:

    <?php

function tableV1 ($row) {
    echo '<tr>';
    echo '<td>' . $row['name'] . '</td>';
    echo '<td>' . $row['description'] . '</td>';
    echo '<td>' . $row['content'] . '</td>';
    echo '</tr>';
}

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "onlib";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } 
?>
<!DOCTYPE html>
<html>
    <head>
        <meta charset="utf-8" />
        <meta http-equiv="X-UA-Compatible" content="IE=edge">
        <title>Page Title</title>
        <meta name="viewport" content="width=device-width, initial-scale=1">
        <style type="text/css">
        table {}
        tbody {}
        td {}
        th {}
        thead {}
        tr {}
        </style>
    </head>
    <body>
        <table>
            <thead>
                <th>Name</th>
                <th>Description</th>
                <th>Content</th>
            </thead>
            <tbody>
<?php
// Takes all the results from the table with genre 5.
$sql = "SELECT name, description, content FROM books WHERE genre='5'";

$result = $conn->query($sql);

if ($result->num_rows > 0) {

    // Output data of each row
    while($row = $result->fetch_assoc()) {
        tableV1($row);
    }

} else {
    echo '<tr><td colspan="3">0 results</td></tr>';
}
?>
            </tbody>
        </table>
    </body>
</html>
<?php

$conn->close();

?>
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(4条)

报告相同问题？

悬赏问题

  • ¥15 表达式必须是可修改的左值
  • ¥15 如何绘制动力学系统的相图
  • ¥15 对接wps接口实现获取元数据
  • ¥20 给自己本科IT专业毕业的妹m找个实习工作
  • ¥15 用友U8：向一个无法连接的网络尝试了一个套接字操作，如何解决？
  • ¥30 我的代码按理说完成了模型的搭建、训练、验证测试等工作(标签-网络|关键词-变化检测)
  • ¥50 mac mini外接显示器 画质字体模糊
  • ¥15 TLS1.2协议通信解密
  • ¥40 图书信息管理系统程序编写
  • ¥20 Qcustomplot缩小曲线形状问题