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我是新手,請耐心等待。 我想要一個單選按鈕,當用戶選擇一個時,兩個值將被髮送到數據庫(記錄表)。一種是用戶選擇的服務,另一種是相應的價格。但'服務'和'價格'來自另一個表,這意味着該值取決於'ms_pricelist'數據庫的內容。這裏是我的代碼:一個開關櫃中的兩個值
(這是我的輸入形式讓我們嘗試把重點放在<input type="radio">
按鈕)
<form method="post" action="">
<table style="width: 10%;">
<p style="font-size: 30px; font-family: Verdana;">   Service Contract </p>
<hr style="margin-top: -20px; width: 95%;">
<p style="margin-left: 50px;">Name of Deceased: <input type="text" name="service_deceased" value="<?php echo $query2['service_deceased']; ?>" />
<br><br>
<p style="margin-left: 50px;">Type of Service:
<input type="radio" name="service_type">Sunny Star</button>
<input type="radio" name="service_type">Lemon Tropic</button>
<input type="radio" name="service_type">White Fairy</button><br>
<input type="radio" name="service_type">Chinese Holy</button>
<input type="radio" name="service_type">Red Ginger</button>
<input type="radio" name="service_type">Blue Days</button></p>
      
<p style="margin-left: 50px;">Amount: <input type="text" name="service_amount" value="<?php echo $query2['service_amount']; ?>" />
</p>
<p style="margin-left: 410px;"> NET AMOUNT: <input type="text" name="net_amount" value="<?php echo $query2['net_amount']; ?>" />
<br></p>
<input type="submit" name="submit" value="Update" style="margin-left: 1000px; height: 50px; width: 150px;" />
</form>
這是我的 'ms_pricelist' 數據庫
[ms_priceID | ms_name | ms_price]
--------------------------------------
[ 1 | Sunny Star | 28000]
[ 2 | Lemon Tropic | 48000]
[ 3 | White Fairy | 58000]
[ 4 | Chinese Holy | 98000]
[ 5 | Red Ginger | 168000]
[ 6 | Blue Days | 250000]
'的service_type' 是其中ms_name應該存儲在記錄表中,而ms_price應該存儲在'service_amount'中。
這可能嗎?我一直在尋找近一個月的答案,但仍無法找到答案。有關如何完成它的任何建議?我使用PHP和MYSQL。提前感謝你!
編輯:
下面是形式的PHP代碼。
<?php
include('connect-db.php');
if(isset($_GET['CustomerNumber']))
{
$CustomerNumber=$_GET['CustomerNumber'];
if(isset($_POST['submit']))
{
$crc_no = $_POST['crc_no'];
$date_of_service = $_POST['date_of_service'];
$date_of_interment = $_POST['date_of_interment'];
$authorized_family = $_POST['authorized_family'];
$rel_to_deceased = $_POST['rel_to_deceased'];
$rel_address = $_POST['rel_address'];
$service_phone_no = $_POST['service_phone_no'];
$service_deceased = $_POST['service_deceased'];
$service_type = $_POST['service_type'];
$service_amount = $_POST['service_amount'];
$less_service = $_POST['less_service'];
$net_amount = $_POST['net_amount'];
$query3=mysql_query("UPDATE records SET crc_no='$crc_no', date_of_service='$date_of_service', date_of_interment='$date_of_interment', authorized_family='$authorized_family', rel_to_deceased='$rel_to_deceased', rel_address='$rel_address', service_phone_no='$service_phone_no', service_deceased='$service_deceased', rel_address='$rel_address', service_phone_no='$service_phone_no', service_deceased='$service_deceased', service_type='$service_type', service_amount='$service_amount', less_service='$less_service', net_amount='$net_amount' WHERE CustomerNumber='$CustomerNumber'");
if($query3)
{
header('location: fill.php');
}
}
$query1=mysql_query("SELECT * FROM records where CustomerNumber='$CustomerNumber'");
$query2=mysql_fetch_array($query1);
?>
這個PHP可以正常工作。但單選按鈕(或其他解決方案,你可能會建議)我想是唯一不起作用的東西。
這是** NOT **你應該如何輸入type =「radio」'。 :) – Smuuf
我不明白你的問題,請多指定。 –
如果您需要多張桌子,請向我們展示所有必要的桌子。另外,如果您使用的是PHP,那麼您的代碼在哪裏?沒有看到你嘗試過的東西,將很難幫助你回答你的問題。 –