UrbanDweller Posted March 31, 2012 Share Posted March 31, 2012 Hey, I'm in the process of making a site that involves multiple layers of folders. I have a form that uploads an image which i then try to move to the image folder using a define function using the document root + image folder but the image never gets saved to the defined location and the location isnt inserted into the sql query. Its a simple thing I want to do but since Im making a bigger site I want to maximize includes to reduce duplicate code. Include File - defines the document root srvRoot = $_SERVER['DOCUMENT_ROOT']; $serverRoot = str_replace('/', '\/', $serverRoot); define('SRV_ROOT', $srvRoot); Image Upload $image = $_FILES["catImage"]["name"]; $image = addImage($image, SRV_ROOT . "images/shop/img_cat"); function addImage($image, $path){ // FIX FIX FIX $fileExtension = getExtension($image); //this function works fine have tested in previous scripts so not needed $image = substr(md5(rand() * time()), 0, 10).".".$fileExtension; $imgDestination = $path.$image; if($_FILES['catImage']['type'] != "image/gif" && $_FILES['catImage']['type'] != "image/jpg" && $_FILES['catImage']['type'] != "image/jpeg" && $_FILES['catImage']['type'] !="image/png" ){ echo "You may only upload image files"; }else{ //Moves the uploaded file to destination folder via the variable which includes image filename if(move_uploaded_file($_FILES["catImage"]["tmp_name"],$imgDestination)){ return($imgDestination); }else{ header("Location: ../../home.php"); } echo $imgDestination; } } Quote Link to comment Share on other sites More sharing options...
NomadicJosh Posted March 31, 2012 Share Posted March 31, 2012 Couple of questions: What is the point of $serverRoot = str_replace('/', '\/', $serverRoot); and where is the variable $serverRoot used? Where is the function addImage called? I only ask because I need to see how and where you define your second parameter, $path. Quote Link to comment Share on other sites More sharing options...
UrbanDweller Posted March 31, 2012 Author Share Posted March 31, 2012 Oh ignore the $serverRoot str_replace it came with a snippet, just used to switch / with \, its doesnt get used in the script. and $path = SRV_ROOT . "images/shop/cat_img" The idea is to set the image path created from $path + $imageName to a sql table where i can use it to display the image later on from any page on my site due to the path being absolute. Quote Link to comment Share on other sites More sharing options...
NomadicJosh Posted March 31, 2012 Share Posted March 31, 2012 Ok, can you post the whole upload file that shows everything including the database stuff? Quote Link to comment Share on other sites More sharing options...
UrbanDweller Posted April 1, 2012 Author Share Posted April 1, 2012 Here you go, thanks for having a look! config.php // Server root path $srvRoot = $_SERVER['DOCUMENT_ROOT']; //$serverRoot = str_replace('/', '\/', $serverRoot); define('SRV_ROOT', $srvRoot); //define('SRV_ROOT', $srvRoot); // Image directory path define('CATEGORY_IMAGE_PATH', 'lisahumphrey/images/shop/cat_img/'); define('CATEGORY_THUMB_PATH', 'images/shop/cat_thumb/'); database.php $sqlCon = mysql_connect($dbHost, $dbUser, $dbPass) or die ('MYSQL connection Failed. ' . mysql_error()); mysql_select_db($dbName) or die('Cannot select database. ' . mysql_error()); // Database query functions function dbQuery($sql) { return mysql_query($sql) or die('Query failed. ' . mysql_error()); } addCategory.php require_once('../include/database.php'); require_once('../../library/config.php'); function addCategory(){ if(!isset($_POST["catName"], $_POST["catDescription"], $_FILES["catImage"])){ header('Location: category_add.php'); } $name = $_POST["catName"]; $description = $_POST["catDescription"]; $image = $_FILES["catImage"]["name"]; $image = addImage($image, SRV_ROOT); // FIX FIX FIX $parentId = $_POST["parentCat"]; if(isset($_POST["isProduct"])){ $isProduct = 1; }else{ $isProduct = 0; } $sql = "INSERT INTO catagory_tbl (cat_parent_id, cat_isProduct, cat_name, cat_description, cat_thumbnail) VALUES ($parentId, $isProduct, '$name', '$description', '$image')"; $result = dbQuery($sql); header('Location: category_list.php'); } function addImage($image, $path){ // FIX FIX FIX $fileExtension = getExtension($image); $image = substr(md5(rand() * time()), 0, 10).".".$fileExtension; $imgDestination = $path.$image; if($_FILES['catImage']['type'] != "image/gif" && $_FILES['catImage']['type'] != "image/jpg" && $_FILES['catImage']['type'] != "image/jpeg" && $_FILES['catImage']['type'] !="image/png" ){ echo "You may only upload image files"; }else{ //Moves the uploaded file to destination folder via the variable which includes image filename if(move_uploaded_file($_FILES["catImage"]["tmp_name"],$imgDestination)){ //MOVES IMAGE TO IMAGE DIR return($imgDestination); }else{ header("Location: ../../home.php"); } echo $imgDestination; } } function getExtension($fileName){ /* Gets image extension type from image name * Splits image name at the . to get name and extension. */ $fileNameParts = explode(".", $fileName); // Retrieves image extension. $fileExtension = end($fileNameParts); // Converts extension string into all lowercase $fileExtension = strtolower( $fileExtension ); // Returns fileExtension to previous code that called function getExtension return($fileExtension); } Quote Link to comment Share on other sites More sharing options...
NomadicJosh Posted April 1, 2012 Share Posted April 1, 2012 So addCategory.php is what you call into the browser, or is there another file you call in the browser and addCategory.php is your action? What happens when you comment out the beginning and ending lines of each function and then try the upload? Quote Link to comment Share on other sites More sharing options...
UrbanDweller Posted April 1, 2012 Author Share Posted April 1, 2012 catagory.php is called from the form i get the image etc from database.php & config.php are only related to catagory.php Quote Link to comment Share on other sites More sharing options...
NomadicJosh Posted April 1, 2012 Share Posted April 1, 2012 Ok, the form code is what I need to look at. Quote Link to comment Share on other sites More sharing options...
UrbanDweller Posted April 1, 2012 Author Share Posted April 1, 2012 this does talk to database.php <body> <?php require_once("../include/database.php")?> <form id="addCategoryForm" method="POST" action="process_category.php" enctype="multipart/form-data"> <input type="text" id="catName" name="catName" value=""/> <span>Will its display products?</span><input type="checkbox" id="isProduct" name="isProduct" /> <select name="parentCat"> <option name="noParent" value="0">Shop Homepage</option> <?php /* Script below fills select options with avaliable parent categories * from catagory_tbl database using a simple loop. */ $result = mysql_query( "SELECT cat_id, cat_name FROM catagory_tbl WHERE cat_isProduct=0 ORDER BY cat_id" ) or die("Query Failed:" .mysql_error());; // Loop starts if(dbNumRows($result) > 0){ while($row = dbFetchAssoc($result)){ extract($row); echo "<option value='".$cat_id."'>".$cat_name."</option>"; } } // Loop stops ?> </select> <input type="text" id="catDescription" name="catDescription" value=""/> <input type="file" id="catImage" name="catImage" /> <input type="submit" id="catSubmit" name="catSubmit" value="Submit"/> <input type="submit" id="catCancel" name="catCancel" value="Cancel"/> </form> </body> You have now got all my code lol Quote Link to comment Share on other sites More sharing options...
NomadicJosh Posted April 1, 2012 Share Posted April 1, 2012 Let me know if I am reading this wrong. This to me says that when the form is not submitted, go back to category_add.php: if(!isset($_POST["catName"], $_POST["catDescription"], $_FILES["catImage"])){ header('Location: category_add.php'); } However, since you are using a function, it makes not since to me to have that line. If I am reading it correctly, you might want to change it to something like this: function addCategory(){ if(isset($_POST['catSubmit'])) { $name = $_POST["catName"]; $description = $_POST["catDescription"]; $image = $_FILES["catImage"]["name"]; $image = addImage($image, SRV_ROOT); // FIX FIX FIX $parentId = $_POST["parentCat"]; if(isset($_POST["isProduct"])){ $isProduct = 1; }else{ $isProduct = 0; } $sql = "INSERT INTO catagory_tbl (cat_parent_id, cat_isProduct, cat_name, cat_description, cat_thumbnail) VALUES ($parentId, $isProduct, '$name', '$description', '$image')"; $result = dbQuery($sql); header('Location: category_list.php'); } } I am not sure if the product part will work, but you can try this code out. If it doesn't work, then take it out of the function completely and then try it to see if something gets posted to the database. Quote Link to comment Share on other sites More sharing options...
UrbanDweller Posted April 1, 2012 Author Share Posted April 1, 2012 na thats all fine i wrote down the .php files wrong add_category is the form and process_category processes the form Sorry bout that, the e Quote Link to comment Share on other sites More sharing options...
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