需要幫助寫入ID到MySQL數據庫

我想寫一個已存在於另一個MySQL表中的ID到第二個MySQL表，我在下拉菜單中回顯ID和與ID關聯的名稱。需要幫助寫入ID到MySQL數據庫

這裏是源代碼：

<div id="postedit" class="clearfix"> 
      <h2 class="ico_mug">Legg ansatte!</h2> 
      <form name="form1" method="post" action="<?($_SERVER['DOCUMENT_ROOT']."/ansatte.php")?>"> 
      <div><input name="navn" type="text" size="30" tabindex="1" value="Skriv navn!" /></br></br> 
      <div id="form_middle_cont" class="clearfix"> 
      <input name="telefon" type="text" size="30" tabindex="1" value="Skriv inn telefon nummer!" /></br></br> 
      <input name="mobil" type="text" size="30" tabindex="1" value="Skriv inn mobiltelefon nummer!" /></br></br> 
      <input name="epost" type="text" size="30" tabindex="1" value="Skriv inn epost nummer!" /></br></br> 
      <input name="bilde" type="text" size="30" tabindex="1" value="Skriv inn link til bilde!" /></br></br> 
      </div> 
       <h3>Possisjon: </h3> 
     <select name="Possisjon" id="possisjon"> 

     <?PHP // Generate a drop-down list of sections. 

       $result = $connector->query('SELECT position_name FROM position ORDER BY position_id'); 

       // Get an array containing the results. 
       // Loop for each item in that array 
       while ($row = $connector->fetchArray($result)){ 
        echo '<option value="'.$row['position_id'].'">'.$row['position_name'].'</option>'; 
       } 
      ?> 
     </select></br></br> 

     <h3>Stilling:</h3> 
     <input name="stilling" type="text" size="30" tabindex="1" /></br></br> 



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

      </div> 
      </form> 
<?php 
// Create an instance of DbConnector 
$connector = new DbConnector(); 

// Setter alt inn i databasen 

if($_POST) { 

$navn = $_POST['navn']; 
$telefon = $_POST['telefon']; 
$mobil = $_POST['mobil']; 
$epost = $_POST['epost']; 
$image = $_POST['bilde']; 
$stilling = $_POST['stilling']; 
$position = $_POST['possisjon']; 

if($navn == '') 
echo '<div id="fail" class="info_div"><span class="ico_cancel">Angi en tittel!</span></div><br>'; 
if($mobil == '') 
echo '<div id="fail" class="info_div"><span class="ico_cancel">Angi mer en én tag!</span></div><br>'; 
if($epost == '') 
echo '<div id="fail" class="info_div"><span class="ico_cancel">Velg en seksjon!</span></div><br>'; 
if($image == '') 
echo '<div id="fail" class="info_div"><span class="ico_cancel">Oida! Du har ikke skrevet noe innhold i artikkelen, prøv igjen!</span></div><br><br>'; 
if($position == '') 
echo '<div id="fail" class="info_div"><span class="ico_cancel">lalalala</span></div><br><br>'; 

if($navn != '' && $telefon != '' && $mobil != '' && $epost != '' && $image != '' && $stilling != '' && $possisjon !='') { 
mysql_query("INSERT INTO ansatte (navn, telefon, mobil, epost, image, stilling) VALUES ('{$navn}', '{$telefon}', '{$mobil}', '{$epost}', '{$bilde}', '{$stilling}', '{$possisjon}')"); 
echo '<div id="success" class="info_div"><span class="ico_success">Suksess!</span></div>'; 
} 


} 
?>

任何想法嗎？

乾杯，〜PureDarkness〜

來源

2011-05-25 PureDarkness

改變您的查詢

$result = $connector->query('SELECT position_name,position_id 
          FROM position 
          ORDER BY position_id');

現在

可以使用position_id在下面的代碼

來源

2011-05-25 08:52:51 Ibu

需要幫助寫入ID到MySQL數據庫

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