如何在数据库查询 returns 无结果时显示消息？

Question

我有这个代码：

<?php

$platform_id = 7;
$query="SELECT * FROM game WHERE game_id IN (SELECT game_id FROM game_platform WHERE platform_id=".$platform_id.") AND game_release BETWEEN '2015-08-01' AND '2015-08-31' ORDER BY game_release";

$result=mysqli_query($link,$query) or die (mysqli_error());


        echo"<table border=1>
        <tr>
            <th width=40px>Day</th>
            <th width=270px>Title</th>
            <th width=203px>Genre</th>
            <th width=203px>Developer</th>
            <th width=205px>Publisher</th>
            <th width=61px>Retail?</th>
            <th width=30px height=30px></th>
            <th width=104px>Note</th>
        </tr>";

        while($row=mysqli_fetch_assoc($result))
       {
        echo"
        <tr>
            <td> ". (new DateTime($row['game_release']))->format("j") ." </td>
            <td>{$row['game_name']}</td>

            <td>";
            $query="SELECT * FROM genre WHERE genre_id=".$row['game_genre']; 
            $genreresult=mysqli_query($link,$query);                                    $genrerow=mysqli_fetch_assoc($genreresult);
        echo $genrerow['genre_name'];
        echo "</td>
            <td>{$row['game_dev']}</td>
            <td>{$row['game_pub']}</td>
            <td>{$row['game_type']}</td>
            <td><a href=\"{$row['game_site']}\" target=\"_blank\"><img src=\"images/officialwebsite.png\" title=\"Official website\"/></a>
                <a href=\"{$row['game_trailer']}\" target=\"_blank\"><img src=\"images/youtube.png\" title=\"Trailer\"/></a></td>
            <td>{$row['game_note']}</td>
        </tr>";


       }

       echo"</table>";
?>

我想在所述月份没有发布时显示一条消息，因为现在我得到一个空的 table，这对用户来说既不友好也不美观。

我在这个网站上找到了这段代码，但是由于 .isEmpty.

，我得到了一个语法错误

<?php

$platform_id = 7;
$query="SELECT * FROM game WHERE game_id IN (SELECT game_id FROM game_platform WHERE platform_id=".$platform_id.") AND game_release BETWEEN '2015-08-01' AND '2015-08-31' ORDER BY game_release";

$result=mysqli_query($link,$query) or die (mysqli_error());


        echo"<table border=1>
        <tr>
            <th width=40px>Day</th>
            <th width=270px>Title</th>
            <th width=203px>Genre</th>
            <th width=203px>Developer</th>
            <th width=205px>Publisher</th>
            <th width=61px>Retail?</th>
            <th width=30px height=30px></th>
            <th width=104px>Note</th>
        </tr>";

        if($result.isEmpty())
        echo "Unfortunately there are no releases this month!";

        else
        {

        while($row=mysqli_fetch_assoc($result))
        {
        echo"
        <tr>
            <td> ". (new DateTime($row['game_release']))->format("j") ." </td>
            <td>{$row['game_name']}</td>

            <td>";
            $query="SELECT * FROM genre WHERE genre_id=".$row['game_genre']; 
            $genreresult=mysqli_query($link,$query);                                    $genrerow=mysqli_fetch_assoc($genreresult);
        echo $genrerow['genre_name'];
        echo "</td>
            <td>{$row['game_dev']}</td>
            <td>{$row['game_pub']}</td>
            <td>{$row['game_type']}</td>
            <td><a href=\"{$row['game_site']}\" target=\"_blank\"><img src=\"images/officialwebsite.png\" title=\"Official website\"/></a>
                <a href=\"{$row['game_trailer']}\" target=\"_blank\"><img src=\"images/youtube.png\" title=\"Trailer\"/></a></td>
            <td>{$row['game_note']}</td>
        </tr>";


       }
       }

       echo"</table>";
       ?>

我继续阅读，似乎没有其他方法，因为 while 循环没有 'else's?

我该如何解决这个问题？谢谢！

Answer 1

您似乎混合了 javascript 和 php，因为 .isEmpty() 不是 php 函数。

您需要在显示 table 之前检查 row count，例如：

$result=mysqli_query($link,$query) or die (mysqli_error());
if (mysqli_num_rows($result) > 0) {
   // your table generation code
} else {
  echo "Unfortunately there are no releases this month!";
}

如何在数据库查询 returns 无结果时显示消息？

How do I display a message when database query returns no results?

php

while-loop