Question

我一直试图找到一种解决方法，以便在访问我的index.php文件时找到mysql错误消息，代码如下：

Warning: mysqli_query() expects parameter 1 to be mysqli, resource given in /Applications/XAMPP/xamppfiles/htdocs/cms/includes/navigation.php on line 24
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, null given in /Applications/XAMPP/xamppfiles/htdocs/cms/includes/navigation.php on line 26

对于我的生活，我无法弄清楚我在这里做错了什么，因为它在我使用静态链接时起作用，但是在测试动态链接时，它只会变得很糟糕。

以下是我用于导航包含文件夹的navigation.php文件的代码：

   <nav class="navbar navbar-inverse navbar-fixed-top" role="navigation">
    <div class="container">

        <!-- Brand and toggle get grouped for better mobile display -->
        <div class="navbar-header">
            <button type="button" class="navbar-toggle" data-toggle="collapse" data-target="#bs-example-navbar-collapse-1">
                <span class="sr-only">Toggle navigation</span>
                <span class="icon-bar"></span>
                <span class="icon-bar"></span>
                <span class="icon-bar"></span>
            </button>
            <a class="navbar-brand" href="#">Start Bootstrap</a>
        </div>

        <!-- Collect the nav links, forms, and other content for toggling -->
        <div class="collapse navbar-collapse" id="bs-example-navbar-collapse-1">
            <ul class="nav navbar-nav">

            <?php

                $query = "SELECT * FROM categories";
                $please_work = mysqli_query($connection, $query);

                while($row = mysqli_fetch_assoc($please_work)){
                    $cat_title = $row['cat_title'];

                    echo "<li><a href='#'>{$cat_title}</a></li>";

                }

                ?>      

            </ul>
        </div>
        <!-- /.navbar-collapse -->
    </div>
    <!-- /.container -->
</nav>

当然我的db.php文件也位于我的包含文件夹中：

<?php

$db['db_host'] = "localhost";
$db['db_user'] = "root";
$db['db_pass'] = "";
$db['db_name'] = "cms";

foreach($db as $key => $value){   
define(strtoupper($key), $value);
}

$connection = mysql_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);

if($connection){

    echo "We are connected";   
}

?>

我不太需要发布我的index.php，因为我知道错误并非来自那里，因为静态内容在使用时正在工作，这意味着include功能正在运行。

有人可以让我免于长期头痛吗？

谢谢，

CB

Answer 1

您正在混合[ProtoContract] class SubMessageRepresentations { [ProtoMember(5, DataFormat = DataFormat.Default)] public SubObject lengthPrefixedObject; [ProtoMember(6, DataFormat = DataFormat.Group)] public SubObject groupObject; } [ProtoContract(ImplicitFields=ImplicitFields.AllFields)] class SubObject { public int x; } using (var stream = new MemoryStream()) { _pbModel.Serialize( stream, new SubMessageRepresentations { lengthPrefixedObject = new SubObject { x = 0x22 }, groupObject = new SubObject { x = 0x44 } }); byte[] buf = stream.GetBuffer(); for (int i = 0; i < stream.Length; i++) Console.Write("{0:X2} ", buf[i]); }和mysql_*个功能。您应该始终使用mysqli_*。

所以，你需要改变这个：

mysqli_*

到此：

$connection = mysql_connect(DB_HOST, DB_USER, DB_PASS, DB_NAME);

mysqli_query和mysqli_fetch_assoc给出错误

1 个答案: