2017-03-08 17 views
-2

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;"> &nbsp 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> &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 
    <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> 
    &nbsp&nbsp&nbsp&nbsp&nbsp&nbsp 

    <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。

+0

這是** NOT **你應該如何輸入type =「radio」'。 :) – Smuuf

+0

我不明白你的問題,請多指定。 –

+0

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。 –

回答

0

好的。您的描述很模糊,您的代碼非常有限。 從我的理解你需要做的是,你需要檢查哪個單選按鈕被選中,然後你查詢ms_pricelist數據庫的行,其中ms_name =你選中的單選按鈕。然後使用該行的ms_price值。