使用PHP变量打印选择查询的结果

Question

我只是在整个晚上试图显示您在代码中找到的查询中的项目。该程序应该允许用户选择出发和到达，并且在点击按钮之后，页面刷新消除路线的价格。该程序一直有效，直到最后几行，经过一些测试，我的猜测是查询无法读取这两个变量。你有什么建议吗？提前感谢您的耐心和帮助 代码下方，这是我的测试网站的link。

<form action='index.php' method='post'>
Select Departure:
<select name="partenze" class="form-control">
        <option value="">--- Select Departure ---</option>

                    <?php
                        require('prova1.php');
                        $sql1 = "SELECT * FROM partenze"; 
			$sql2 = "SELECT * FROM arrivi"; 
                        $result1 = $mysqli->query($sql1);
 						while($row1 = $result1->fetch_assoc()){
					?>   
					<option value="<?php echo $row1["p_ID"]; ?>"><?php echo $row1["p_localita"]; ?></option>
					<?php } ?>

</select>
<br>

Select Arrival:
<select name="arrivi" class="form-control">
        <option value="">--- Select Arrival ---</option>

                    <?php
                        $result2 = $mysqli->query($sql2);
                        while($row2 = $result2->fetch_assoc()){
					?>   
					<option value="<?php echo $row2["a_ID"]; ?>"><?php echo $row2["a_localita"]; ?></option>
					<?php } ?>

</select>
<input type='submit' name='submit' id='submit' value='Get Selected Values' />
</form>

<?php
	if(isset($_POST['submit'])){
	$selected_val1 = $_POST['partenze'];
	$selected_val2 = $_POST['arrivi'];
	require('prova1.php'); 
    $result3 = $mysqli->query("SELECT * FROM listino WHERE partenza = '".$selected_val1."' AND '".$selected_val2."'");
	while($row3 = $result3->fetch_assoc()){
	echo $row3['prezzo'];
	}
}
?>