我正在尝试从以下数据库数据中获取数据:
并使用此CSS&HTML代码显示数据:

    <div class="event">
        <img src="http://dummyimage.com/80x70/f00/fff.png" alt="picture" />
        <p>Room 2</p>
        <p class="patient-name">Jon Harris</p>
        <p class="event-text">This is a pixel. A flying pixel!</p>
        <p class="event-timestamp">feb 2 2011 - 23:01</p>
    </div>

.event {
    display:block;
    background: #ececec;
    width:380px;
    padding:10px;
    margin:10px;
    overflow:hidden;
    text-align: left;
}
.event img {
    display:block;
    float:left;
    margin-right:10px;
}

.event p {
    font-weight: bold;
}

.event img + p {
    display:inline;
}

.patient-name {
    display:inline;
    color: #999999;
    font-size: 9px;
    line-height:inherit;
    padding-left: 5px;
}
.event-text{
    color: #999999;
    font-size: 12px;
    padding-left: 5px;
}
.event-timestamp{
    color: #000;
    padding-left: 5px;
    font-size: 9px;
}

这是我的PHP代码:
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
  <title>DASHBOARD - Arduino 3</title>
  <link rel="stylesheet" type="text/css" href="css/style.css">
</head>
<body>
    <?php
    $con = mysql_connect("localhost","*****","***");
        if(!con)
        {
            die('Could not connect: ' . mysql_error());
        }

        mysql_select_db("arduino_db",$con);

        $result = mysql_query("SELECT * FROM events");
        //Start container
        echo " <div id='background_container'> ";

        while($row = mysql_fetch_array($result))
        {
            echo "<div class='event'>";
            echo "<img src='img/ev_img/red.jpg' alt='picture' />";
            echo "<p>" . $row['inneboende'] . "</p>";
            echo "<p class='patient-name'>" . "$row['overvakare']" . "</p>";
            echo "<p class='event-text'>" . "$row['handelse']" . "</p>";
            echo "<p class='event-timestamp'>" . "$row['tid']" . "</p>";
            echo "</div>";
        }

        //end container
        echo "</div>"
        mysql_close($con);

    ?>
</body>
</html>

我得到的只是一张白纸,我不明白为什么。

最佳答案

此处缺少分号(我已添加):

//end container
echo "</div>";

您应该删除这些行上$row数组变量周围的引号,如下所示:
echo "<p class='patient-name'>" . $row['overvakare'] . "</p>";
echo "<p class='event-text'>" . $row['handelse'] . "</p>";
echo "<p class='event-timestamp'>" . $row['tid'] . "</p>";

有了这些变化,它就工作了(我测试过)。
我想这是你发布了你的实际连接细节的好处:)
正如评论中所提到的,现在您必须更改密码。
另外,在开发过程中,您应该确保启用了错误报告,see this answer以各种方式执行此操作。

07-24 17:29
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