Question

我设法从我的网站数据库中获取信息，但我不知道如何在表格中显示结果以及如何设置样式。我尝试将<table>放在整个PHP之外，显然不起作用。我尝试在结果回显之前回显<table>标记，然后在它之后回显</table>标记，但是没有这样做。这是我正在使用的代码：

<?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();
?>

我仍然是PHP的新手，试图了解整个过程是如何工作的。提前谢谢！

Answer 1

试试这个：

<?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 ($result->num_rows > 0) {
        // output data of each row
        echo "<table>";
        echo "<thead>
          <tr>
             <th>Name</th>
             <th>Description</th>
          </tr>
         </thead>";
        echo "<tbody>";
        while($row = $result->fetch_assoc()) {
            echo "<tr>";
            echo "<td>".$row["name"]."</td>";
            echo "<td>".$row["description"]."</td>";
            echo "</tr>";            
        }
        echo "</table>";
    } else {
        echo "0 results";
    }

    $conn->close();
?>

Answer 2

尝试这种方法：

[... some HTML header and code ...]

<table>
    <thead>
        <tr><th>Name</th><th>Description</th><th>Content</th></tr>
    </thead>
    <tbody>
    <?php
        [ ... extract something from the database ...]

        while($row = $result->fetch_assoc()) 
        {
            echo "<tr><td>$row[name]</td><td>$row[description]</td><td>Content</td></tr>\n";
        }
    }
    ?>
    </tbody>
</table>

[... some HTML footer ...]

显然，用您自己的代码替换[...]部分。我们的想法是使用PHP输出HTML代码的动态部分，并且是PHP的 power 和HTML结合使用。

我尽量避免在我的PHP中“回显”很多HTML，但它确实有用（参考其他答案）。

详细信息：设置<table>时，您应设置标题部分<thead> <tfoot>（如果适用）和正文部分<tbody>。参考What is the purpose for HTML's tbody?

Answer 3

<?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 ($result->num_rows > 0) {
    echo '<table>';
    echo '<tr><td>Name</td><td>Description</td><td>Content</td></tr>'
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<tr><td>".$row["name"]."</td><td>".$row["description"]."</td><td>".$row["content"] ."</td></tr>";
    }
    echo '</table>';
} else {
    echo "0 results";
}

$conn->close();

＆GT;

Answer 4

基本表格格式为

例如：

<table>
    <tr>
        <td>row 1 item 1</td>
        <td>row 1 item 2</td>
    </tr>
    <tr>
        <td>row 2 item 1</td>
        <td>row 2 item 2</td>
    </tr>
</table>

所以试试：

if ($result->num_rows > 0) {
    echo "<table>";
    while($row = $result->fetch_assoc()) {
        echo "<tr>";
        echo "<td>".$row["name"]."</td>";
        echo "<td>".$row["description"]."</td>";
        echo "<td>".$row["content"]."</td>";
        echo "</tr>";
    }
    echo "</table>";
} else {
    echo "0 results";
}

Answer 5

<?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); } 
?>

在输出任何内容之前，始终首先执行数据库连接，这样，您可以创建自定义错误消息，而不是显示失败的数据库连接或根本没有内容。

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

在<head></head>内使用样式来设置表格的样式，称为CSS。

<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>

从数据库输出内容。

<?php

$conn->close();

?>

最后关闭数据库连接。一起来：

<?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();

?>

如何在PHP

5 个答案: