php - 突出显示当天从数据库中获取的表行

Question

使用下面的代码，我从我的数据库中获取了一个表。第一列有一个日历，我想知道是否可以突出显示包含当天的表格行。

我也用这个脚本显示当天:

    <?php
        date_default_timezone_set("Europe/Rome");
         echo " Italy: " . date("h:i:sa");
         echo " day " . date("d/m/Y") . "<br>";
          ?>

--- 表代码 ---

     <?php
     $query = "SELECT * FROM trip"; 
                $result = mysql_query($query);

                echo "<table >"; // start a table tag in the HTML

                while($row = mysql_fetch_array($result)){   //Creates a loop to loop through results
                echo "<tr>
                            <td>" . $row['Day'] . "</td>
                            <td>" . $row['Country'] . "</td>
                            <td>" . $row['Place'] . "</td>
                            <td>" . $row['Flight'] . "</td>
                        </tr>";  //$row['index'] the index here is a field name
                }

                echo "</table>"; //Close the table in HTML
                mysql_close(); //Make sure to close out the database connection

            ?>

---css代码--

table
{
font-family: verdana;

color: white;
background: #003366;
text-align:center;
font-size:16px;
border-collapse: collapse; 

}
tr   {

padding-top:5px;
padding-bottom:5px;
font-size:16px;

}

tr:hover{

color:#003366;
background: white;

}

---添加类(class).TODAY

.今天
{
字体系列:verdana；
背景:红色；
颜色:#003366；
}


Answer 1

只是改变:

echo "<tr>
    <td>" . $row['Day'] . "</td>
    <td>" . $row['Country'] . "</td>
    <td>" . $row['Place'] . "</td>
    <td>" . $row['Flight'] . "</td>
</tr>";  //$row['index'] the index here is a field name
}

与:

if ($row['Day'] ==  date("Y-m-d") ) {
    echo '<tr class="today">';
} else {
    echo "<tr>";
}
echo "<td>" . $row['Day'] . "</td>
    <td>" . $row['Country'] . "</td>
    <td>" . $row['Place'] . "</td>
    <td>" . $row['Flight'] . "</td>
</tr>";  //$row['index'] the index here is a field name
}

并将类 today 添加到您的 css 文件中。

抱歉，我无法调试它，但我认为你有一个想法