從數組更改下拉框到複選框

Question

我想從下拉菜單中選擇一個複選框，我想更改它，因爲我希望首先選擇數組中的列表，然後可以通過複選框將數據存儲到數據庫之前選中下拉腳本如下

<?php 
session_start(); 
define('DEFAULT_SOURCE','Site_A'); 
define('DEFAULT_VALUE',100); 
define('DEFAULT_STC','BGS'); 
include('class/stockconvert_class.php'); 
$st = new st_exchange_conv(DEFAULT_SOURCE); 
if(isset($_GET['reset'])) { 
    unset($_SESSION['selected']); 
    header("Location: ".basename($_SERVER['PHP_SELF'])); 
    exit(); 
} 
?> 
<form action="do.php" method="post"> 
<label for="amount">Amount:</label> 
<input type="input" name="amount" id="amount" value="1"> 
<select name="from"> 
<?php 
$stocks = $st->stocks(); 
asort($stocks); 
foreach($stocks as $key=>$stock) 
{ 
    if((isset($_SESSION['selected']) && strcmp($_SESSION['selected'],$key) == 0) || (!isset($_SESSION['selected']) && strcmp(DEFAULT_STC,$key) == 0)) 
    { 
    ?> 
    <option value="<?php echo $key; ?>" selected="selected"><?php echo $stock; ?></option> 
    <?php 
    } 
    else 
    { 
    ?> 
    <option value="<?php echo $key; ?>"><?php echo $stock; ?></option> 
    <?php 
    } 
} 
?> 
</select> 
<input type="submit" name="submit" value="Convert"> 
</form> 

和我改成複選框如下

<?php 
session_start(); 
define('DEFAULT_SOURCE','Site_A'); 
define('DEFAULT_VALUE',100); 
define('DEFAULT_STC','BGS'); 
include('class/stockconvert_class.php'); 
$st = new st_exchange_conv(DEFAULT_SOURCE); 
if(isset($_GET['reset'])) { 
    unset($_SESSION['selected']); 
    header("Location: ".basename($_SERVER['PHP_SELF'])); 
    exit(); 
} 
?> 
<form action="do.php" method="post"> 
<label for="amount">Amount:</label> 
<input type="input" name="amount" id="amount" value="1"><input type="submit" name="submit" value="Convert"> 
<?php 
$stocks = $st->stocks(); 
asort($stocks); 
foreach($stocks as $key=>$stock) 
{ 
    if((isset($_SESSION['selected']) && strcmp($_SESSION['selected'],$key) == 0) || (!isset($_SESSION['selected']) && strcmp(DEFAULT_STC,$key) == 0)) 
    { 
    ?> 
    <br><input type="checkbox" id="scb1" name="from[]" value="<?php echo $key; ?>" checked="checked"><?php echo $stock; ?> 
    <?php 
    } 
    else 
    { 
    ?> 
    <br><input type="checkbox" id="scb1" name="from[]" value="<?php echo $key; ?>"><?php echo $stock; ?> 
    <?php 
    } 
} 
?> 
</form> 

但不工作，我是否需要顯示其他相關的代碼？

感謝，如果有人幫助，感激

更新： 確定後第一顯然是不太明顯的，所以我會添加錯誤

的問題的錯誤是 致命錯誤：調用未定義的方法st_exchange_conv ::用C轉換（）：\ XAMPP \ htdocs中\上線路測試\ do.php 21

線21是$st->convert($from,$key,$date);

session_start(); 
if(isset($_POST['submit'])) 
{ 
    include('class/stockconvert_class.php'); 
    $st = new st_exchange_conv(DEFAULT_SOURCE); 
    $from = mysql_real_escape_string(stripslashes($_POST['from'])); 
    $value = floatval($_POST['amount']); 
    $date = date('Y-m-d H:i:s'); 
    $_SESSION['selected'] = $from; 
    $stocks = $st->stocks(); 
    asort($stocks); 

    foreach($stocks as $key=>$stock) 
    { 
     $st->convert($from,$key,$date); 
     $stc_price = $st->price($value); 
     $stock = mysql_real_escape_string(stripslashes($stock)); 
     $count = "SELECT * FROM oc_stock WHERE stock = '$key'"; 
     $result = mysql_query($count) or die(mysql_error()); 
     $sql = ''; 
     if(mysql_num_rows($result) == 1) 
     { 
      $sql = "UPDATE oc_stock SET stock_title = '$stock', stc_val = '$stc_price', date_updated = '$date' WHERE stock = '$key'"; 
     } 
     else 
     {  
     $sql = "INSERT INTO oc_stock(stock_id,stock_title,stock,decimal_place,stc_val,date_updated) VALUES ('','$stock','$key','2',$stc_price,'$date')"; 
     } 
     $result = mysql_query($sql) or die(mysql_error().'<br />'.$sql); 
    } 
    header("Location: index.php"); 
    exit(); 
} 

爲什麼我要將其從下拉菜單更改爲複選框？ 因爲通過複選框列表我將能夠選擇哪些我檢查它是數據庫的入口，那麼它對我來說似乎並不簡單，我尋求一些幫助<謝謝這麼多對於你的隊友。

Answer 1

您還沒有刪除開頭<select>標記。

但您刪除了<submit>按鈕。

您將name從「from」更改爲「from [from]」。

編輯：你的補充後：

使用你只能選擇一個from值的下拉列表。現在您將其更改爲複選框，從而可以選擇多個條目。這會導致在do.php的腳本中收到一個數組from[]。您的功能無法以任何方式處理數組或多個選擇。

您必須重新設計do.php，將表單更改回下拉列表或使用比率按鈕。