数据库值检索错误显示答案

【问题标题】：database value retrieval error showing数据库值检索错误显示
【发布时间】：2015-12-24 06:01:03
【问题描述】：

警告：mysql_fetch_array() 期望参数 1 是资源... 当我尝试从数据库中获取以下第一个特色配置文件的值时，会显示上述错误..... 如果我添加：- 或死（mysql_error（））；然后它显示没有选择数据库.. 请帮忙代码：-

<div class="grid_1">
      <div class="container">
        <h1>Featured Profiles</h1>
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            <i class="fa fa-heart pink-heart"></i>
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        </div>
<?php
    $sql="SELECT (file,user_name,user_id,user_occupation) FROM users1";
    $result_set=mysql_query($sql) or die(mysql_error());
        while($row=mysql_fetch_object($result_set))
    {
        ?>

        <ul id="flexiselDemo3">
          <li><div class="col_1"><a href="view_profile.html">
            <img src="images/<?php echo $row->file ?>" alt="" class="hover-animation image-zoom-in img-responsive"/>
             <div class="layer m_1 hidden-link hover-animation delay1 fade-in">
                <div class="center-middle">About Him</div>
             </div>
             <h3><span class="m_3">Profile ID : <?php echo $row->user_id ?></span><br>28, Christian, Australia<br>Corporate</h3></a></div>
          </li>
          <li><div class="col_1"><a href="view_profile.html">
            <img src="images/2.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
             <div class="layer m_1 hidden-link hover-animation delay1 fade-in">
                <div class="center-middle">About Her</div>
             </div>
             <h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
          </li>
          <li><div class="col_1"><a href="view_profile.html">
            <img src="images/3.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
             <div class="layer m_1 hidden-link hover-animation delay1 fade-in">
                <div class="center-middle">About Him</div>
             </div>
             <h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
          </li>
          <li><div class="col_1"><a href="view_profile.html">
            <img src="images/4.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
             <div class="layer m_1 hidden-link hover-animation delay1 fade-in">
                <div class="center-middle">About Her</div>
             </div>
             <h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
          </li>
          <li><div class="col_1"><a href="view_profile.html">
            <img src="images/5.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
             <div class="layer m_1 hidden-link hover-animation delay1 fade-in">
                <div class="center-middle">About Him</div>
             </div>
             <h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
          </li>
          <li><div class="col_1"><a href="view_profile.html">
            <img src="images/6.jpg" alt="" class="hover-animation image-zoom-in img-responsive"/>
             <div class="layer m_1 hidden-link hover-animation delay1 fade-in">
                <div class="center-middle">About Her</div>
             </div>
             <h3><span class="m_3">Profile ID : MI-387412</span><br>28, Christian, Australia<br>Corporate</h3></a></div>
          </li>
        </ul>
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         $(window).load(function() {
            $("#flexiselDemo3").flexisel({
                visibleItems: 6,
                animationSpeed: 1000,
                autoPlay:false,
                autoPlaySpeed: 3000,            
                pauseOnHover: true,
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        });
       </script>
       <script type="text/javascript" src="js/jquery.flexisel.js"></script>

        <?php
            }
        ?>

    </div>
</div>

【问题讨论】：

从你的查询中结束( )!!!你的数据库连接在哪里？？？
我已经在上面的代码中做了这部分：-
把你的查询括起来，它将是SELECT file,user_name,user_id,user_occupation FROM users1
它显示错误（警告：mysql_query() 期望参数 1 是字符串，对象给定）我这样做了：-
您在代码中混合了mysql and mysqli。只使用 mysqli 而不是同时使用

标签： php mysql database

【解决方案1】：

我认为问题出在你的 $result 函数中，它应该有 2 个参数，而你只给了一个参数，所以你的代码应该是这样的

<?php
$connection=mysqli_connect('localhost','yourmysqlusername','yourmysqlpassword','yourmysqldatabase');
     $sql="SELECT (file,user_name,user_id,user_occupation) FROM users1";
    //you have to add $connection as 2nd parameter in $result function
        $result_set=mysql_query($connection,$sql) or die(mysql_error());
            while($row=mysql_fetch_object($result_set))
?>

【讨论】：

按照您的步骤后，它显示错误：- 警告：mysql_query() 期望参数 1 是字符串，给定对象

【解决方案2】：

确保您首先使用mysql_select 连接到数据库，然后使用mysql_select_db 选择数据库。在此之后，您需要使用mysql_query 触发查询。

另外，您可以mysqli_connect 合并前两个步骤。见this

【讨论】：