Question

我不知道如何获取数据库中的所有数据并将其显示在通知菜单中，我显示的只是第一条记录。

这就是我现在拥有的一切。

HTML：

<a href="#">
    <div class="user-img">
       <img src="images/1 (2).png" alt="user" class="img-circle" name="imgs" id="imgs">
       <span class="profile-status online pull-right"></span> 
    </div>
    <div class="mail-contnet">
        <h5 id="name"><?php echo $name; ?></h5>
        <span class="mail-desc" id="msgs"><?php echo $msg; ?></span>
        <span class="time" id="dTime"><?php echo $format; ?></span>
   </div>
</a>

PHP：

<?php

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

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$name="";
$mail="";
$msg="";
$date = "";
$format = "";

$sql = "SELECT fullname, email, msg, time,date from msg";
$result = mysqli_query($conn,$sql);
$num = mysqli_num_rows($result);

while ($row = mysqli_fetch_array($result)) {
    if ($num > 0) {
        $name = $row["fullname"];
        $mail = $row["email"];
        $msg = $row["msg"];
        $date = new DateTime($row["date"] . " " . $row["time"]);
        $format = date_format($date, "Y-m-d g:i A");
    }
    else {
        $name = "";
        $mail = "";
        $msg = "";
        $date = "";
    }
}
?>

Answer 1

请检查以下代码

$output = '';
if($num > 0)
{
    while ($row = mysqli_fetch_array($result))
    {

        $name = $row["fullname"];
        $mail = $row["email"];
        $msg = $row["msg"];
        $date = new DateTime($row["date"] . " " . $row["time"]);
        $format = date_format($date, "Y-m-d g:i A");

        $output .= '<a href="#">';
        $output .= '<div class="user-img"><img src="images/1 (2).png" alt="user" class="img-circle" name="imgs" id="imgs"> <span class="profile-status online pull-right"></span> </div>';
        $output .= '<div class="mail-contnet"><h5 id="name">'.$name.'</h5> <span class="mail-desc" id="msgs">'.$msg.'</span> <span class="time" id="dTime">'.$format.'</span></div>';
        $output .= '</a>';
    }
}

echo $output;

Answer 2

你的代码只是不断替换变量的值（$ name，$ mail等）。您应该将其保存在增量变量中。

$data = '';
while ($row = mysqli_fetch_array($result)) {
    if ($num > 0) {        
        $name = $row["fullname"];
        $mail = $row["email"];
        $msg = $row["msg"];
        $date = new DateTime($row["date"] . " " . $row["time"]);
        $format = date_format($date, "Y-m-d g:i A");
    }
    else {
        $name = "";
        $mail = "";
        $msg = "";
        $date = "";
    }
    $data.='<tr>
                <td>'.$name.'</td>
                <td>'.$mail.'</td>
                <td>'.$msg.'</td>
                <td>'.$date.'</td>
    </tr>';
}
echo '<table>';
echo $data;
echo '</table>';

在PHP文件中循环

2 个答案: