0

我创建了一个动态下拉列表，它从数据库中填充数据。直到现在它工作正常。现在我想将下拉列表中的选定数据添加到我的数据库中。我尝试了各种可能的方法，但无法在我的数据库中插入值，以下程序没有给出任何错误消息，但它也没有在我的数据库中插入数据。请检查一下。

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
</body>
</html>

<?php 

session_start();
if(isset($_SESSION['username']))
{


    include 'connect.php';

            $select_query=          'Select * from category';
            $select_query_run =     mysql_query($select_query);

    echo "  
        <form action='insert_product.php' method='POST' ></br>

        Product Name:   <input type='text' name='product_name'  /></br>

        Price       :   <input type= 'text' name= 'price'  /></br>

        Description :   <input type='text' name='description'  />*Seperate by Comma</br>

                        ";



    //  Drop Down Display   




            echo "<select name='category'>";


            while   ($select_query_array=   mysql_fetch_array($select_query_run) )
            {

                    echo "<option value='".$select_query_array['category_id']."' >".
                    htmlspecialchars($select_query_array["name"])."</option>";


                }

         $selectTag= "<input type='submit' value='Insert'  /></select></form>";

         echo $selectTag;
    //Drop Down End!




    if(isset($_POST['product_name']) && isset($_POST['price']) && isset($_POST['description'])  )
    {
        echo $product_name  =       $_POST['product_name'];
        echo $price         =       $_POST['price'];
        echo $description   =       $_POST['description'];
        echo $category      =       $_POST['category'];


//Problem Area 

    $query= "insert into products (name, price, description,  category_id ) 
                VALUES( '$product_name', $price, '$description', $category )";


    if($query_run=      mysql_query($query))
    {

        echo 'Data Inserted';



        }   
        else
        {
            'Error In SQL'.mysql_error();
            }
    }

    else
    {
        echo 'empty Field';
        }

}

else
{
    echo 'You Must Log in To View this Page!';
    }
?>
4

2 回答 2

0

改变：

$select_query=          'Select * from category';
$select_query_run =     mysql_query($select_query);
echo "  
    <form action='insert_product.php' method='POST' ></br>

    Product Name:   <input type='text' name='product_name'  /></br>

    Price       :   <input type= 'text' name= 'price'  /></br>

    Description :   <input type='text' name='description'  />*Seperate by Comma</br>

                    <input type='submit' value='Insert'  />";
echo "<select name='category'>";
while   ($select_query_array=   mysql_fetch_array($select_query_run) )
        {

                echo "<option value='".$select_query_array['category_id']."' >".
                htmlspecialchars($select_query_array["name"])."</option>";


            }

echo "</select></form>";

查询：

$query= "insert into products (name, price, description, category_id, name) VALUES('$product_name', $price, '$description', '".$_POST['category']."' )";
于 2013-07-17T08:07:43.263 回答
0

我认为，您必须将其用作帖子“类别”，而不是 Option Es 的值。

$category = $_POST['category']);
nsert into products (name, price, description, category_id, name) 
                VALUES('$product_name', $price, '$description', '$category"' )";
于 2013-07-17T08:11:50.790 回答