jbille Posted October 18, 2006 Share Posted October 18, 2006 I have a form which uploads the following information into mysql:namenumberdescriptionpriceimageHowever I am having trouble uploading the image and then viewing the image on a page with the rest of these results. I've looked but have not been able to find any help on this. Can anyone give me suggestions. Quote Link to comment Share on other sites More sharing options...
simcoweb Posted October 18, 2006 Share Posted October 18, 2006 Please post your code for the image upload. Quote Link to comment Share on other sites More sharing options...
jbille Posted October 18, 2006 Author Share Posted October 18, 2006 [code]$name = $_POST['name']; $number = $_POST['number']; $price = $_POST['price']; $order = $_POST['order']; $category = $_POST['category']; $description = $_POST['description']; $picture = $_POST['picture']; if((empty($name)) || (empty($number)) || (empty($price)) || (empty($category)) || (empty($description))) { echo "Please <a href=\"addproduct.html\">go back</a> and fill in the following information: "; if(empty($name)) echo ("Product Name"); if(empty($number)) echo ("Product Number"); if(empty($price)) echo ("Product Price"); if(empty($category)) echo ("Category must be picked"); if(empty($description)) echo ("Product Description"); exit(1); } $data = addslashes(fread(fopen($picture, "r"), filesize($picture))); $link = mysql_connect(localhost, $user, $pass);if (!$link) { die('Not connected : ' . mysql_error());} mysql_select_db($dbname, $link) or die("Unable to select database"); $query ="INSERT INTO products (category, name, number, price, description, picture) VALUES ('$category', '$name', '$number', '$price', '$description', '$data')"; mysql_query($query) or die('Error, query failed, Call Jimmy (330)268-9271'); mysql_close($link);[/code] Quote Link to comment Share on other sites More sharing options...
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