php - MySQLi fetch_all 的奇怪错误

Question

这可能会被标记为重复，但我的错误似乎与我在 stackoverflow 上看到的其他错误不同。

下面的代码应该显示图像中的内容:代码:

 if($news_result = $db->query("SELECT * FROM `news` ORDER BY `likes` DESC LIMIT 4")) {                      
    if ($news_result->num_rows) {
        $rows = $news_result->fetch_all(MYSQL_ASSOC);
            $row_count = 0;
            foreach($rows as $row) {
                $row_count++;

                $content_echo = '
                    <div class="row inside" style="width: 99%; margin-left: auto; margin-right: auto;">
                    <a href="http://portuguesegaming.com/post/'.$row['url'].'">
                    <div class="view third-effect"
                            style="
                                background: url(../images/'.$background.'.png);
                                background-size:80%;
                                background-position: center;
                                margin-right: auto;
                                margin-left: auto;">    
                            <h3>'.$row['title'].'</h3><br>
                            <table style="width: 78%;">
                                <tr>
                                    <td>
                                        <h4 style="font-size: 12px;">'.$row['description'].'</h4>
                                    </td>
                                </tr>
                                <tr>
                                    <td colspan="2">
                                        <h4>BY: '.$row['creator'].'</h5>
                                    </td>
                                </tr>
                            </table>
                            <div class="likes" style="margin-top: -20px;"><div class="likes-icon"></div><h4>'.$row['likes'].'</h4></div>
                            <div class="mask">
                                <h6>READ MORE</h6>  
                            </div>
                        </div>
                    </a>
                </div>';

                $separator = '  <br>
                <div class="separator" style="
                        margin-bottom: -25px;
                        margin-top: -25px;
                        background-image: linear-gradient(to right, rgba(254, 254, 254, 0), rgba(254, 254, 254, 0.75), rgba(254, 254, 254, 0));"></div>
                <br>';

                if ($news_result->num_rows > 1 && $row_count > 0 && $row_count < $news_result->num_rows) {
                    $content_echo .= $separator;
                }
                echo $content_echo;
            }
        }
    }

图片:

这适用于我的 wamp 服务器，但是当我将文件上传到 godaddy 的主机时，出现错误:Fatal error: Call to undefined method mysqli_result::fetch_all() in /home/ipsyko/public_html/index.php on line 102我读到了它，他们说这是因为 PHP 版本，goddady 的 PHP 版本是 5.4(native) 并且出现错误，我更改为正常 4.4(not native) 并且没有出现错误，但也没有出现我想要的内容，这很奇怪，然后我更改为 5.5 并发生了与 5.4 中发生的相同的事情不是原生的，我什至检查了“mysqlnd”选项，但没有任何效果。任何人都知道如何解决它？

Answer 1

我认为这段代码可以工作:

<?php
$news_result = $db->query("SELECT * FROM `news` ORDER BY `likes` DESC LIMIT 4");
if(mysqli_num_rows($news_result) >= 1){
  while($row = mysqli_fetch_array($news_result)){

    switch($row['type']){
      case "global":$background = "globalnews";
        break;
      case "console":$background = "console_bg";
        break;
      case "pc":$background = "pc_bg";
        break;
      case "bar":$background = "bar_bg";
        break;
      default:$background = "";
    }

    $content_echo = '<div class="row inside" style="width: 99%; margin-left: auto; margin-right: auto;">
      <a href="http://portuguesegaming.com/post/'.$row['url'].'">
        <div class="view third-effect" style="background: url(../images/'.$background.'.png); background-size:80%; background-position: center; margin-right: auto; margin-left: auto;">    
          <h3>'.$row['title'].'</h3>
          <table style="width: 78%;">
            <tr>
              <td><h4 style="font-size: 12px;">'.$row['description'].'</h4></td>
            </tr>
            <tr>
              <td colspan="2"><h4>BY: '.$row['creator'].'</h4></td>
            </tr>
          </table>
          <div class="likes" style="margin-top: -20px;"><div class="likes-icon"></div><h4>'.$row['likes'].'</h4></div>
          <div class="mask">
            <h6>READ MORE</h6>  
          </div>
        </div>
      </a>
    </div>';

    $separator = '<div class="separator" style="margin-bottom: -25px; margin-top: -25px; background-image: linear-gradient(to right, rgba(254, 254, 254, 0), rgba(254, 254, 254, 0.75), rgba(254, 254, 254, 0));"></div>';
    echo $content_echo.$separator;
  }
}   
?>