Question

我是编码新手。试图通过构建个人理财网络管理应用程序来学习。

我有两个下拉菜单-第二个菜单根据第一个下拉菜单中的选择更改其选项。

我没有写完整的这段代码，因此没有完全理解它。

我的问题是这样的：如何将每个下拉列表中的选定选项发送到Mysql数据库？

我试图向底部添加执行此操作的代码，但是无论我如何放置，我都会不断收到第74行的错误“未定义索引”。

再次，请原谅我的无知-尝试通过实际构建东西来学习。这只是我正在尝试的一小段代码-当它起作用时，我会将其放入我的实际文档中。

<?php
$sports_arr = array();
$sports_arr[] = "Basketball";
$sports_arr[] = "Baseball";
$sports_arr[] = "Football";

$position = array();
$position['Basketball'][] = "Power Forward";
$position['Basketball'][] = "Small Forward";
$position['Basketball'][] = "Center";
$position['Soccer'][] = "Center Forward";
$position['Soccer'][] = "Right Wing";
$position['Soccer'][] = "Left Wing";
$position['Football'][] = "Halfback";
$position['Football'][] = "Fullback";
$position['Football'][] = "Wide Reciever";
$position['Football'][] = "Tight End";
$position['Football'][] = "Center";
?>



<!-- Firstdropdown selection -->
<div class="home">
<select id="s1">
    <?php foreach($sports_arr as $sa) { ?>
    <option value="<?php echo $sa; ?>"><?php echo $sa; ?></option>
    <?php } ?>
 </select>



 <!-- Second dropdown selection -->
 <select id="s2">
</select>
</div>

<script type="text/javascript">
var s1= document.getElementById("s1");
var s2 = document.getElementById("s2");
onchange(); //Change options after page load
s1.onchange = onchange; // change options when s1 is changed

function onchange() {
<?php foreach ($sports_arr as $sa) {?>
    if (s1.value == '<?php echo $sa; ?>') {
        option_html = "";
        <?php if (isset($position[$sa])) { ?> // Make sure position is exist
            <?php foreach ($position[$sa] as $value) { ?>
                option_html += "<option><?php echo $value; ?></option>";
            <?php } ?>
        <?php } ?>
        s2.innerHTML = option_html;
    }
<?php } ?>
}
</script>

<input type="submit" name="Submit> value="Submit">


                                             <?php 


/*Connect to Database*/ 
$connect = mysqli_connect("localhost", "root", "", "test_db");
/* place coresponding values into database in same order*/
$sql = "INSERT INTO hello( Sport, Position) 



VALUES


('".$_POST["option.html"]."')"; 

if(mysqli_query($connect, $sql))  
{  
 echo 'Data Inserted';  
}  
 ?>

将2个链接的下拉菜单的结果发布到Mysql

0 个答案: