此代码是我的搜索表单的演示部分。实际的搜索表单包含3个下拉列表，从下拉列表中选择的值将作为关键字进行搜索。此下拉列表包含芒果，苹果，葡萄等水果。搜索代码运行正常。但是问题在于下拉列表中显示的第一个选项是“选择”（不包含任何值），低于该值会开始实际的水果列表，但是在为第一个页面加载时仍会选择第一个水果的值时间
。我想做的是，当页面首次加载时，即下拉列表中的值是Select（选择），则什么都不会显示，之后，当用户从下拉列表中选择值并点击Submit（提交）按钮时，结果才出现应该得到展示

<div class="col-md-3 col-sm-5">
    <div class="media">
        <div class="media-body">
            <?php
                $servername = "localhost";
                $username = "root";
                $password = "";
                $dbname = "db";

                // Create connection
                $con = mysqli_connect($servername, $username, $password, $dbname);
                // Check connection
                if (!$con) {
                    die("Connection failed: " . mysqli_connect_error());
                }

                $sql = "SELECT fruits FROM fruits";
                $result = $con->query($sql);
                echo "<label for='fruits'>Treatment Type: </label>";
                echo "<select name='fruits' id='fruits' class='form-control'><option value=''>--Select--</option>";
                while($row = $result->fetch_assoc()) {
                echo "<option value='" . $row['fruits'] . "'>" . $row['fruits'] . "</option>";
                }
                echo "</select>";
            ?>
        </div>
    </div>
</div>

搜索部分的代码

<?php
    $con=mysqli_connect("","","","");// Check connection
    if (mysqli_connect_errno()) 
        {
            echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
        $fruits = mysqli_real_escape_string($con, $_POST['fruits']);

        $sql1 = "SELECT * FROM treatment WHERE fruits LIKE '%$fruits%'";
        $result = mysqli_query($con, $sql1);
        echo "<table class='table table-striped table-bordered responsive'>
            <thead>
                <tr>
                    <th>Name</th>
                    <th>Type</th>
                    <th>Fruits</th>
                </tr>
            </thead>";

            if (mysqli_num_rows($result) > 0) 
                {
                    while($row = mysqli_fetch_assoc($result)) 
                        {
                            echo "<tbody data-link='row' class='rowlink'>";
                                echo "<tr>";
                                    echo "<td><a href='#'>" . $row['name'] . "</a></td>";
                                    echo "<td>" . $row['type'] . "</td>";       
                                    echo "<td>" . $row['fruits'] . "</td>";

                                echo "</tr>";
                            echo "</tbody>";    
                        }

                }
            else 
                {
                    echo "0 results";
                }
            echo "</table>";
            mysqli_close($con);
?>

如果有人可以指导我将不胜感激

PS（编辑部分）

<div class="col-md-3 col-sm-5">
    <div class="media">
        <div class="media-body">
            <?php
                $servername = "localhost";
                $username = "root";
                $password = "";
                $dbname = "db";

                // Create connection
                $con = mysqli_connect($servername, $username, $password, $dbname);
                // Check connection
                if (!$con) 
                {
                    die("Connection failed: " mysqli_connect_error());
                }

                $sql = "SELECT fruits FROM fruits";
                $result = $con->query($sql); ?>
                echo "<label for="fruits">Treatment Type: </label>";
                echo "<select name="fruits" id="fruits" class="form-control">
                <option value="" <?php if(!isset($_POST['fruits'])) { ?>selected<?php } ?>>--Select--</option>";
                <?php 
                while($row = $result->fetch_assoc()) { ?>

                echo "<option value="<?php echo $row['fruits']; ?>" <?php if(isset($_POST['fruits']) && $_POST['fruits'] == $row['fruits']) { ?>selected<?php } ?>><?php echo $row['fruits']; ?></option>";
                <?php } ?>
            </select>
        </div>
    </div>
</div>