옵션 태그의 코드가 데이터베이스에 빈 데이터를 보냅니다.

Question

Service Mechanic 고객을 db에 저장하는 사이트를 구축했습니다. 항목을 갱신하는 기능이 있습니다. 선택 필드가 잘 작동하는 하나의 드롭 다운이있었습니다 (여전히 작동합니다). 그러나 사실 이후에 추가 한 두 번째 정보는 정보를 얻기 위해 단지 비어있게됩니다.

여기 내 코드입니다.

include('connect.php'); 

$table='currentJobs'; 
$tableMap=array('Date','Name','Phone','Bike_Year','Bike_Model','Current_Status','SType','Mechanic','Revenue','Notes'); 

$sqlArray=array(); 

foreach ($tableMap AS $key){ 
    $sqlArray[]="`".$key."`='".mysql_real_escape_string(@$_POST[$key],$dbLink)."'"; 
} 

$sql="UPDATE `".$table."` SET ".join(',',$sqlArray)." WHERE `id`='".$_POST['id']."'"; 

if (mysql_query($sql,$dbLink)){ 
    echo 'Data for <i>' .$_POST['Name'].'</i> was successfully updated <br />'.$sql.'.'; 
} 
else { 
    echo 'Sorry, could not process the following sql: <br /><code>'.$sql.'</code>'; 
    echo mysql_errno($dbLink) . ": " . mysql_error($dbLink). "\n"; 
} 

mysql_close($dbLink); 

그것은이를 반환 :

Data for Test User was successfully updated 
UPDATE `currentJobs` SET `Date`='04/14/2013',`Name`='Test User',`Phone`='1234567890',`Bike_Year`='2001',`Bike_Model`='FXD',`Current_Status`='Checked In',`SType`='',`Mechanic`='All',`Revenue`='0',`Notes`='' WHERE `id`='55'. 

내 양식은 다음과 같습니다

<form action="updated.php" method="POST" name="dataForm" id="dataForm"> 
<fieldset> 
    <input type="hidden" name="id" value="<?php echo $row['id']; ?>"/> 
    <label>Arrival<input type="text" name="Date" value="<?php echo $row['Date']; ?>"/></label><br /> 
    <label>Name<input type="text" name="Name" value="<?php echo $row['Name']; ?>"/></label><br /> 
    <label>Phone Number<input type="text" name="Phone" value="<?php echo $row['Phone']; ?>"/></label><br /> 
    <label>Bike Year<input type="text" name="Bike_Year" value="<?php echo $row['Bike_Year']; ?>"/></label><br /> 
    <label>Bike Model<input type="text" name="Bike_Model" value="<?php echo $row['Bike_Model']; ?>"/></label><br /> 
    <label>Current Status<!--Update Status types on two pages--> 
     <select name="Current_Status" form="dataForm" value="<?php echo $row['Current_Status']; ?>"> 
      <option value="Checked In">Checked In</option> 
      <option value="Inline For Service">Inline For Service</option> 
      <option value="In Service">In Service</option> 
      <option value="On Hold - Parts on Order">On Hold - Parts on Order</option> 
      <option value="On Hold - Parts to Paint">On Hold - Parts to Paint</option> 
      <option value="On Hold - Waiting To Hear From Customer">On Hold - Waiting To Hear From Customer</option> 
      <option value="Test Ride">Test Ride</option> 
      <option value="Completed - Awaiting Pick Up">Completed - Awaiting Pick Up</option> 
      <option value="Picked Up">Picked Up</option> 
     </select> 
    </label><br /> 
    <!--update Service Type in two pages.--> 
    <label>Service Type</label> 
     <select name="SType" form="" value="<?php echo $row['SType']; ?>"> 
      <option value="Spec Service Interval">Spec Service Interval</option> 
      <option value="Interim Service">Interim Service</option> 
      <option value="Diagnostics">Diagnostics</option> 
      <option value="Tires">Tires</option> 
      <option value="Engine– Light Work (1-3 hrs)">Engine– Light Work (1-3 hrs)</option> 
      <option value="Engine– Medium Work (3-8 hrs)">Engine– Medium Work (3-8 hrs)</option> 
      <option value="Engine– Heavy Work (8-24 hrs)">Engine– Heavy Work (8-24 hrs)</option> 
      <option value="Drivetrain– Light">Drivetrain– Light</option> 
      <option value="Drivetrain - Heavy">Drivetrain - Heavy</option> 
      <option value="Dyno Tune">Dyno Tune</option> 
     </select> 
    <br /> 
    <label>Mechanic Assigned<input type="text" name="Mechanic" value="<?php echo $row['Mechanic']; ?>"/></label><br /> 
    <label>Final Revenue<input type="text" name="Revenue" value="<?php echo $row['Revenue']; ?>"/></label><br /> 
    <label>Notes<input type="text" name="Notes" value="<?php echo $row['Notes']; ?>"/></label><br /> 
    <input type="submit" value="Update" style="color:#000000;" /> 
</fieldset> 

왜 두 번째는 데이터를 전송하지 선택 것인가?