使用php显示来自两个sql表的结果

我有两个表，分别是

post

和

pcgames

if$_获得['game']；is=action

它将从列类别中搜索桌面post和pcgames上的操作

post表格

id|category  |game
------------------
1 | action   |cabal
------------------
2 | strategy |blackdessert
------------------
3 | RPG      |MUlegend
------------------
4 | action   |ragnarok
------------------

id|category  |game
-------------------
1 | action   |solidsnake
-------------------
2 | action   |finalfantasy
-------------------
3 | RPG      |kingdomhearts
-----------------
4 | action   |tekken
-------------------

pcgames表格

id|category  |game
------------------
1 | action   |cabal
------------------
2 | strategy |blackdessert
------------------
3 | RPG      |MUlegend
------------------
4 | action   |ragnarok
------------------

id|category  |game
-------------------
1 | action   |solidsnake
-------------------
2 | action   |finalfantasy
-------------------
3 | RPG      |kingdomhearts
-----------------
4 | action   |tekken
-------------------

结果

no|category  |game
------------------
1 | action   |cabal
------------------
2 | action   |solidsnake
-------------------
3 | action   |finalfantasy
-------------------
4 | action   |tekken
-------------------

PHP

    <?php
    $game = $_GET['game'];
    $ids = mysqli_real_escape_string($conDB,$id);
    $query = "SELECT * FROM `post` WHERE `game` ='" . $game . "'";
    $result = mysqli_query($conDB,$query);
    while($row = mysqli_fetch_array($result)) {
?> 
        <li> <a href="./post_list.html"><?php echo $row['title']; ?></a> </li>
<?php }; ?> 

---

我想你的意思是从具有类别条件的table post和pcgames中获取数据。如果你想这样做。在您的php上尝试以下操作：

<?php
    $game = $_GET['game'];
    $ids = mysqli_real_escape_string($conDB,$id);
    $no = 1;

    //this will be get data from post table
    $query = "SELECT * FROM `post` WHERE `category` ='" . $game . "'";
    $result = mysqli_query($conDB,$query); 

    //this will be get data from pcgames table
    $query1 = "SELECT * FROM `pcgames` WHERE `category` ='" . $game . "'";
    $result1 = mysqli_query($conDB,$query1);       
?>
<table border=1>
    <tr>
        <th>No</th> <th>Category</th> <th>Game</th>
    </tr>
    <?php  
    //fetch data from post table
    while($user_data = mysqli_fetch_array($result)) {         
        echo "<td>".$no++."</td>";
        echo "<td>".$user_data['category']."</td>";
        echo "<td>".$user_data['game']."</td>";     
    }
    //fetch data from pcgames table
    while($user_data1 = mysqli_fetch_array($result1)) {         
        echo "<td>".$no++."</td>";
        echo "<td>".$user_data1['category']."</td>";
        echo "<td>".$user_data1['game']."</td>";     
    }
    ?>
</table>


问题是什么？您是否指示服务器将
.html
文件视为php？你安装了webserver/php吗？我不会等一整晚的。我们不在同一个时区，你和我。除非我们在同一个时区，而且时间接近午夜；所以我在路上。你可以熬夜，我不行。祝你好运。很抱歉，我没有编辑链接，我只是想从pcgames and post表中获取一个数据“action”。你需要告诉问题出在哪里，然后打开
$conDB
或错误报告的值是多少，如果你只是通过添加die或mysqli\u错误进行错误或测试，请使用错误消息