2013-05-16 291 views
0

我在addproduct.php下面的代碼連接點擊提交按鈕後doaddproduct.php:數據插入到數據庫 - PHP MYSQL

<tr> 
<td>Category Name</td> 
<td>:</td> 
<td> 

<select name="categoryname" id=""> 
<option value="">Select</option> 
<?php 

while($row=mysql_fetch_array($result)) { 
?> 
<option value="<?php echo $row['CategoryID']; ?>"> 

<?php echo $row['CategoryName']; ?></option> 
<?php } ?> 
</select> 

</td> 
</tr> 

所以,在下拉菜單中(addProduct命令。 PHP)從數據庫中提供選項(表:類別,列:categoryname)。例如,categoryID'1'的categoryname是food,categoryid'2'的categoryname是drink,等等。

現在,我想從選擇下拉選項中的數據插入到我的數據庫,通過改變所選擇的類別名稱爲的categoryID(表:的categoryID:產品,列)。例如,我從下拉菜單中選擇「喝酒」,當我點擊提交按鈕後,程序將數據插入到表產品:

|產品ID | CategoryID | ProductName |

| 123231 | 2 |可口可樂|

下面是我的doaddproduct.php我下面的代碼:

<?php 
    include("connect.php"); 

    $productname = $_POST['productname']; 
    $categoryname = $_POST['categoryname']; 
    $stock = $_POST['stock']; 
    $price = $_POST['price']; 

    if($productname == NULL || $productname == ""){ 
     header("location:../addproduct.php?err=You must fill product name"); 
    }else if($categoryname== "none"){ 
     header("location:../addproduct.php?err=You must choose category name"); 
    }else if($stock == NULL || $stock == ""){ 
     header("location:../addproduct.php?err=You must fill stock"); 
    }else if($price == NULL || $price == ""){ 
     header("location:../addproduct.php?err=You must fill price"); 
    }else if($_FILES["file"]["type"] != "image/jpeg" && $_FILES["file"]["type"] != "image/png" && $_FILES["file"]["type"] != "image/jpg"){ 
     header("location:../addproduct.php?err=Extention of your photo must be jpg/jpeg/png"); 
    }else{ 
     $query = $username . "-" .$fullname . "-" . $phone ."-".$email."-".$password."-".$rpassword."-".$address."-".$gender; 
     $qcek = "select * from product where productname = '$productname'"; 
     $result = mysql_query($qcek); 
     if(mysql_num_rows($result) > 0){ 
      header("location:../addproduct.php?err=Product name already used"); 
     }else{ 
      //for upload 
      $ext = substr($_FILES["file"]["name"], strrpos($_FILES["file"]["name"], '.')); 
      move_uploaded_file($_FILES["file"]["tmp_name"],"../photos/" . $username . $ext); 
      $pho = $username . $ext; 

      $password = md5($password); 
      //still confused here 
        $query = "insert into product values('', '', '$productname', '$pho', '$stock', '$price')"; 

      mysql_query($query); 
      header("location:../addproduct.php?err=Success"); 
     } 

    } 

?> 

我應該怎麼辦?

+0

是什麼問題? 你的故事嗎? –

+0

@RajeevRanjan問題的錯誤是我無法改變的選擇的下拉菜單選項(類別名)到CATEGORYID這樣我就可以將其插入到我的新表(這是表產品) – noobprogrammer

+0