dropbop

I cant remember exactly why I did that, I really should start commenting on my code more... or I will end up lost in git myself lol A sub category is any category that has a parent. Categories can be as deep as we want, but we are sticking 1 sub level Parent Cat > Sub Cat Each cat have an ID and a Parent ID. category_id category_name parent_id cat_showing 1 Category One 0 0 2 Category Two 0 0 3 Sub Category One 1 0 4 Sub CategoryTwo 2 0 I suppose what I need to know/learn how to do is, to check the products table for any products that are in the sub categories of the selected (parent) category... So if a visitor was on category?cid=20 and some of its sub cats had no products, I would rather not show them the categories so I dont waste there time looking through pages that have nothing in them.

Hi, I have been working on an eCommerce platform for the last few months. Its almost complete but I need to do one thing and thats to only display sub categories if products exist in them. This is the site craftgo.me (development site), and this is how the categories are produced for a category page. ie. http://www.craftgo.me/category.php?cat_id=96 if($_GET['cat_id']) { $parents_id = $_GET['cat_id']; $cat_num_rows_sql = "SELECT category_id, parent_id FROM mccategories WHERE parent_id = '$parents_id'"; $cat_num_rows_result = mysql_query($cat_num_rows_sql); $cat_row_count = mysql_num_rows($cat_num_rows_result); if($cat_row_count == '0'){ $current_cat_parent_sql = "SELECT parent_id FROM mccategories WHERE category_id = '$cat_id'"; $current_cat_parent_result = mysql_query($current_cat_parent_sql); $cat_parent_row = mysql_fetch_array($current_cat_parent_result); $parent_id = $cat_parent_row['parent_id']; } else { $parent_id = $_GET['cat_id']; } } $catsql = "SELECT category_id, category_name FROM mccategories WHERE cat_showing = '0' AND parent_id = '$parent_id' ORDER BY category_name"; $catresult = mysql_query($catsql); while($catrow = mysql_fetch_array($catresult)) { $category_id = $catrow['category_id']; $cat_name = $catrow['category_name']; echo '<li><a href="category.php?cat_id=' . $category_id . '">' . $cat_name . '</a></li>'; } In the mcproducts table the category id for each product is under 'product_category' Im not even sure where to start to make this work, but if anyone could give me a pointer, that would be super. Many thanks Eoin

That works perfectly! thank you very much... I was initially trying to use JOIN, but from looking at your one, mine was all wrong.. but now I know how to do it properly. I did get an error though, which after googleing it, the error was saying there were 2 columns with 'showing', Column 'showing' in field list is ambiguous So just in case anyone else gets the same error after using a JOIN, it means there is a column in each table with the same name. I haven't tried it yet, but to get around it, you can put the table name before the column name, something like SELECT table1.showing, table2.showing...... please correct me if I'm wrong... I'm still learning myself. Anyway, thanks again scootsha.. problem solved!

I was just pondering on this last night when editing a login page for a script im building.... cheers for the tip. DB

Hi, I have been working on a directory/cms script (first large script to build from scratch). I am stuck on a small part of it where is shows users a list of the listings they have put in the directory. Below is the code that generates the table that shows the list of listings which is pulled from the 'listings' table in my database. <?php ### List of listings user has posted ### if(isset($_SESSION['userlogged'])) { $user_id = $_SESSION['userlogged']; $sqlCommand = "SELECT listing_id, cat_id, listing_title, listing_date, showing FROM listings WHERE showing='0' AND user_id='$user_id' ORDER BY listing_id ASC"; $query = mysql_query($sqlCommand,$con) or die (mysql_error()); $ListingDisplay = ''; while ($row = mysql_fetch_array($query)) { $listing_id = $row["listing_id"]; $cat_id = $row['cat_id']; if ($cat_id == '1') { $cat_name = "Carpenters"; } if ($cat_id == '2') { $cat_name = "Stone Masons"; } if ($cat_id == '3') { $cat_name = "Plumbers"; } if ($cat_id == '4') { $cat_name = "Landscapers"; } $listing_title = $row["listing_title"]; $listing_date = $row["listing_date"]; $ListingDisplay .= ' <tr class="listingaccount"> <td><a href="edit_listing.php?lid=' . $listing_id . '">' . $listing_id . '</a></td> <td>' . $cat_name . '</td> <td>' . $listing_title . '</td> <td>' . $listing_date . '</td> <td><a class="hoverBtn" href="edit_listing.php?lid=' . $listing_id . '"><img src="images/btnedit.png" border="0"></a></td> <td><a class="hoverBtn" href="remove_listing.php?lid=' . $listing_id . '"><img src="images/btndelete.png" border="0"></a></td> </tr>'; } mysql_free_result($query); ?> <table width="680" border="0" align="right" cellpadding="0" cellspacing="0"> <tr> <td colspan="2"><h2>My Listings</h2></td> </tr> <tr> <td> </td> <td> </td> </tr> <tr> <td><a href="account_add_listing.php">Add New Listing</a></td> <td> </td> </tr> <tr> <td> </td> <td> </td> </tr> <tr> <td colspan="2" valign="top"> <table width="100%" border="0" cellspacing="0" cellpadding="0"> <tr> <td>Listing Code</td> <td>Category</td> <td>Title</td> <td>Date Entered</td> <td> </td> </tr> <?php echo $ListingDisplay; ?> </table> </td> </tr> </table> <?php } ?> That all works just fine, but I would like to be able to generate the categories automatically. I am thinking some sort of array, but im not quite sure how to go about it and was hoping maybe someone could give me a starting point I could work from. My categories table is like this: -------------------------- | cat_id | cat_name | -------------------------- Many thanks Eoin

Hi, Im not sure if it related to your issue, but you are missing a ' in this line to the left of $sproductcode $query = "INSERT into orders (customerid, productcode, quantity, productprice, orderdate) values ('$scustomerid', $sproductcode', '$squantity', '$sproductprice', '$sorderdate')"; Hop you get the problem fixed..

I'm actually looking back into it again.. After looking at loads of other ways to resize images, Simple Image seems to be the best.. Thanks jcbones

Ok, here is the finished product... Using the code Seany provided (which I am very grateful for), I have created the dropdown as follows... Below is a jpg of the finished dropdpown <select style="width:150px;" name="category"> <option value="">Select a Category</option> <?php //This selects all rows where Parent_id = 0 and ORDERS by category_name $parent_query = mysql_query("SELECT * FROM categories WHERE parent_id=0 ORDER BY category_name ASC"); while($parent=mysql_fetch_array($parent_query)){ $parent_id = $parent['category_id']; $parent_name = $parent['category_name']; echo "<option style=\"font-weight:bold;\">".$parent_name."</option>"; //This gets all the rows where the PARENT_ID = the parents category_id $child_query = mysql_query("SELECT * FROM categories WHERE parent_id=$parent_id ORDER BY category_name ASC"); while($child=mysql_fetch_array($child_query)){ $child_id = $child['category_id']; $child_name = $child['category_name']; echo "<option style=\"padding-left:15px;\" value=\"".$child_id."\">".$child_name."</option>"; } } ?> </select> [attachment deleted by admin]

Thanks you Seany, I'm very grateful for your help. This looks like it could work perfectly.. now to get creating the dropdown... When I get the dropdown working, I will post the code here in case anyone else comes across a similar issue. Cheers.

Hi, I don know how to explain this in words too well, so i have create 2 images to show what I need. What I need to do is use whats in the first image (table.jpg)(the database) to create whats in the second image (dropdown.jpg)(the drop-down menu) The results of the form will be entered into the database. category_id will be the value thats inserted. Many kind regards Eoin [attachment deleted by admin]

ok, so I have managed to get the basics going.... <?php if(isset($_POST['Submit'])) { $current_image=$_FILES['image']['name']; $extension = substr(strrchr($current_image, '.'), 1); if (($extension!= "jpg") && ($extension != "jpeg")) { die('Unknown extension'); } $time = date("fYhis"); $new_image = "image_large_" . $time . "." . $extension; $new_image_small = "image_small_" . $time . "." . $extension; $destination="uploads/".$new_image; $destinationtwo="uploads/".$new_image_small; $action = copy($_FILES['image']['tmp_name'], $destination); $action = copy($_FILES['image']['tmp_name'], $destinationtwo); if (!$action) { die('File copy failed'); }else{ mysql_query("INSERT INTO customers (small_image, large_image) VALUES ('".$new_image_small."','".$new_image."')"); echo "File copy successful"; } ?> This works so far but is still a bit basic compared to what I need. Here I can upload a file, then save it as 2 files with different name in the uploads folder, and also add the names to the database. What I need to do now, is make one image small (approx 200px) and the other one large (Approx 600px). I have looked at that Simpleimage (Thanks jcbones) but it confuses me, is there a simpler way to change the size of an image that a bit less complicated for me? Thanks Eoin

I have looked around all overr, but cant find what im looking for. Everything I have found is about uploading multile files, what I need to do is upload one file and copy it so there is 3 files. Here is an example of what I need to accomplish. Upload file picture1.jpg From picture1.jpg create 3 files - smallimage_picture1.jpg, largeimage_picture1.jpg & smallimage_picture1_thumb.jpg Each file will need to be copied to a /products directory and also an entry for each file name placed in mysql database table. I don't think this is as simple as it sounds to do in PHP, but i am very willing to learn so if anyone knows of a good tutorial that get get me started that would be super. regards DB

A table was the first thing I tried, but it just showed everything in one long column. I would prefer to use tables to show results tho.

Ah yes!!, well spotted, never even occured to me.... cheers for that. Here is what I have now and with a little help from css, I have it showing as many columns as will fit <?php mysql_connect ($dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error()); mysql_select_db($dbname) or die(mysql_error()); $output = ""; $result = mysql_query("SELECT * FROM mcproducts ORDER BY RAND() LIMIT 10"); while($row = mysql_fetch_array($result)) { $products_local_id = $row['products_local_id']; $productname = $row['product_name']; $thumburl = $row['image_from_url']; $productlink = $row['product_local_url']; $thumbnail = $row['product_image_small']; $currencysymbol = $row['product_currency']; $price = $row['product_price']; $flagicon = $row['product_country_from']; $output = " <div class=\"productimage\"> <a href=".$productlink."".$products_local_id."><img src=".$thumburl."".$thumbnail." width=\"150\" height=\"150\"></a> </div> <div class=\"productdescription\"> <div class=\"pro_name\"> <a href=".$productlink."".$products_local_id.">".$productname."</a> </div> <div class=\"pro_description\"> </div> <div class=\"pro_description\">".$flagicon." ".$currencysymbol." ".$price."</div> </div> "; ?> <html> <head> <link href="style/stylesheet.css" rel="stylesheet" type="text/css" /> </head> <body> <div class="displaybox"> <div class="productimage"> <a href="<?php echo $productlink; ?><?php echo $products_local_id; ?>"><img src="<?php echo $thumburl; ?><?php echo $thumbnail ?>" width="150" height="150"></a> </div> <div class="productdescription"> <div class="pro_name"> <a href="<?php echo $productlink ?><?php echo $products_local_id; ?>"><?php echo $productname; ?></a> </div> <div class="pro_description"> </div> <?php if ($flagicon=="Ireland") { $flagicon = "<img src=\"flags/ireland.jpg\">"; } elseif ($flagicon=="UK") { $flagicon = "<img src=\"flags/uk.jpg\">"; } else echo ""; ?> <div class="pro_description"><?php echo $flagicon; ?><?php echo $currencysymbol; ?> <?php echo $price; ?></div> </div> </div> <?php } ?> </body> </html> Thanks again AyKay47, that was a quick resonse from you... really appreciated indeed DB

I was wondering how does one go about showing results from SELECT query in columns in a html table. I have a list of products in a table, and would like to show them on the page in 4 columns. I have done many searches on google to try and find the sulution, but the majority of what im finding instead is about displaying a table from phpmyadmin as a table in html. If its a large operation to do this, I would be very happy if someone could poiint me in the direction of a tutorial maybe. Here is the code I have so far to display the products, but for some reason, it only show 1 row instead of all the rows from my table. <?php $dbhost = "localhost"; $dbuser = "user"; $dbpass = "pass"; $dbname = "dbname"; mysql_connect ($dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error()); mysql_select_db($dbname) or die(mysql_error()); $result = mysql_query("SELECT * FROM mcproducts"); while($row = mysql_fetch_array($result)) { $products_local_id = $row['products_local_id']; $productname = $row['product_name']; $thumburl = $row['image_from_url']; $productlink = $row['product_local_url']; $thumbnail = $row['product_image_small']; $currencysymbol = $row['product_currency']; $price = $row['product_price']; $flagicon = $row['product_country_from']; } ?> <html> <head> <link href="style/stylesheet.css" rel="stylesheet" type="text/css" /> </head> <body> <div class="displaybox"> <div class="productimage"> <a href="<?php echo $productlink; ?><?php echo $products_local_id; ?>"><img src="<?php echo $thumburl; ?><?php echo $thumbnail ?>" width="150" height="150"></a> </div> <div class="productdescription"> <div class="pro_name"> <a href="<?php echo $productlink ?><?php echo $products_local_id; ?>"><?php echo $productname; ?></a> </div> <div class="pro_description"> </div> <?php if ($flagicon=="Ireland") { $flagicon = "<img src=\"flags/ireland.jpg\">"; } elseif ($flagicon=="UK") { $flagicon = "<img src=\"flags/uk.jpg\">"; } else echo ""; ?> <div class="pro_description"><?php echo $flagicon; ?><?php echo $currencysymbol ?> <?php echo $price ?></div> </div> </div> </body> </html> Many thanks, DB

The plan is to have a number of websites where people can list there items for sale. Each on its own domain (.co.uk, .ie, .eu etc) each has its own database. So what I am trying to do is show some random products from each website on a main website (the .com). I have looked into trying to open multiple connections so I could query each database, which I achieved. But even with 2 or more connections open, the results were mixed up.. ie: would show the product name from one database and the currency symbol from another. And after trying to figure it out for the last 5 hours, I thought there must be another way, to populate the database tables on the .com and make life a bit easier later.. I dont mind putting in a load of hard work if it will help later on. This is what I was struggling with all day: <?php error_reporting(E_ALL); ini_set('display_errors', '1'); $host1="localhost"; $base1="datbase1"; $user1="database1_user"; $password1="database1pass"; $host2="localhost"; $base2="datbase2"; $user2="datbase2_user"; $password2="database2pass"; $conection1 = @mysql_connect($host1, $user1, $password1) or die("Error reaching destination<br>".mysql_error()." nr eroare: ".mysql_errno()); print "Succesfuly connected 1! <br>"; $Db1 = @mysql_select_db($base1, $conection1) or die("Error reaching destination database:<br>" . mysql_error(). "<br>" . mysql_errno()); print "Database1 reached!<br>"; $conection2 = @mysql_connect($host2, $user2, $password2) or die("Error reaching source<br>".mysql_error()." nr eroare: ".mysql_errno()); print "Succesfuly connected 2!<br>"; $Db2 = @mysql_select_db($base2, $conection2) or die("Error reaching source database:<br>" . mysql_error(). "<br>" . mysql_errno()); print "Database2 reached!!<br>"; $output = ''; $query = "SELECT * FROM mcproducts ORDER BY product_name DESC"; $result = mysql_query($query, $conection1) or die('Query failed: ' . mysql_error().'||'.mysql_error()); $result = mysql_query($query, $conection2) or die('Query failed: ' . mysql_error().'||'.mysql_error()); while($row = mysql_fetch_array($result)) { $productname = $row['product_name']; $thumburl = $row['image_from_url']; $thumbnail = $row['product_image_small']; $currencysymbol = $row['product_currency']; $price = $row['product_price']; echo $output = "<table width=\"150\"> <tr><td>".$productname."</td></tr> <tr><td width=150 height=150 style=overflow:none;><img src=http://www.craftgo.ie/products/".$thumbnail." width=150 height=150></td></tr> <tr><td>".$currencysymbol."".$price."</td></tr> </table>"; } mysql_close($conection1); mysql_close($conection2); ?> That will show me a list of items from database 2 and the currency from database 1 and image urls from database 1 that should be from database 2. I have been trying to figure it out, but keep failing. So instead of doing it that way, i thought if I could have the data inserted into the .com database at time of processing a form (ie: add new user, add new product product, etc). This means I could create everything on one website and duplicate it for each other website. I hope that helps a bit more to explain what im trying to do. Thanks DB

I am trying to put together a product listing page for a site. The products on the page will be pulled from a main database (3 tables). The main database will be populated from a number of other databases (exactly the same table layout on each), all on the same server, but on their own domains under cpanel hosting. So, what I am trying to figure out is, what is the best way to populate the main database, 1) have the data inserted using the add product form, 2) have a cron run a php script to copy the new data from the sub databases and add to the main database? or some other way? I was thinking about doing it on submittion of the add product form so the data would be sent to a sub database and the main database. But then I was thinking that would give 2 different product_id's, I need them to be the same. The other way, having a php script copy from sub databases to the main one could be good, but than I thought about the product_id's again, there would be clashes. I have been going round in circles with this so I thought I would ask the gurus what they thought would be a good way to do it, so here I am... asking.... What would you recommend the best way to populate a database with multiple other databases without any conflict with products_id's.... but wait, thats not all... I want to find a way to do it so that each product on the page, when clicked will bring the website visitor to the actual product page on which ever domain name that product is listed. And since each products url is made up using the product_id, i need to find another way to do it, without using the product ID. Would something like having a column like usa_produt_id, uk_product_id etc work, and some sort of if statement to choose which one to take the id from? Ok, im probably confusing you now, im confusing myself anyway lol Any suggestions on what might be a good way to roll with this would be hugely welcoming. I will keep plugging away myself and see if I come up with any way of doing it. I am in to new depts here with MySQL and php, so any advice or help would be great. Thanks a mil DB

Im having no luck with this at all.. I cant seem to pull the file in no matter what I try. Almost every out put is either the filename or else blank. Thanks anyway for the help thorpe, but I don't think i'm fully understanding how its done. I have been over at stackoverflow, php.net and a few other places, but there are so many different answers and explanations that I am totally confused now. There must be a simple way to plonk a php file into a define and call it on a page. If it helps, here is my php & mysql versions MySQL: 5.5.8 PHP Version: 5.3.5 Eoin

Hi thorpe, Thanks for the reply. Thats actually what i have been trying, but don't think im doing it right, this is what I have (one of many attempts).. $mainpagetabs = file_get_contents('/tabs/frontpagetabs.php'); define('TEXT_GREETING_GUEST', $mainpagetabs); I have tried a few different ways using single and double quote too. I think I still have a bit to learn Eoin

Is there a way to include a file using a define. I am using zencart and it has text defined when a person is not logged in, but I would like to place something more than just text. The define from the language file is: if (STORE_STATUS == '0') { define('TEXT_GREETING_GUEST', 'Welcome new customer'); } else { define('TEXT_GREETING_GUEST', 'Welcome, please enjoy our online showcase.'); } define('TEXT_GREETING_PERSONAL', 'Hello <span class="greetUser">%s</span>! Would you like to see our <a href="%s">newest additions</a>?'); I have a php file with a set of jquery tabs and the info pulled from the database for each tab. This is for people not logged in or approved to shop which would go here: define('TEXT_GREETING_GUEST', 'notloggedin.php'); I then have another file I would like to include when a customer has created an account and logged in which would go here: define('TEXT_GREETING_PERSONAL', 'loggedin.php'); Here is the code from the file to include when not logged in (have not created the 'logged in' file yet) <?php require_once ('includes/configure.php'); $tabtitles = ""; $tabcontents = ""; $sqloo = mysql_query("SELECT * FROM front_page_tabs ORDER BY tab_id ASC"); $tabCount = mysql_num_rows($sqloo); // count the output amount if ($tabCount > 0) { while($row = mysql_fetch_array($sqloo)){ $id = $row["tab_id"]; $tabtitle = $row["tab_title"]; $tabcontent = $row["tab_content"]; $tabtitles .= ' <li><a href="#tab' . $id . '">' . $tabtitle . '</a></li>'; $tabcontents .= ' <div id="tab' . $id . '" class="tab_content">' . $tabcontent . '</div> '; } } else { $dynamicList = "Please add a Tab Title"; } #mysql_close(); ?> <ul class="tabs"> <?php echo $tabtitles; ?> </ul> <div class="tab_container"> <?php echo $tabcontents; ?> </div> So what I need to know is can a file be included in a php define. I have tried several ways but nothing is working for me. Cheers, Eoin

Thank you PFMaBiSmAd, that's worked perfectly

I have been trying to figure this out for a while and I cant find anything that specificly shows how this is done.. I have a php script that exports certain fields from a mysql database table to a csv file. One of the fields in the table is 'products_image'. From the code below the result is: products_artist, products_title, products_label, products_image Townes Van Zandt, Rain On A Conga Drum, Exile, image.jpg What I would like to show is a full url of the image. Its not showing like that in the database so I would need to be able to add it to the script somehow so it outputs as below: products_artist, products_title, products_label, products_image Townes Van Zandt, Rain On A Conga Drum, Exile, http://www.domainname.com/images/image.jpg <?php $conn = mysql_connect( 'localhost', 'user', 'pass' ) or die( mysql_error( ) ); mysql_select_db( 'database_name', $conn ) or die( mysql_error( $conn ) ); $query = sprintf( 'SELECT products_artist, products_title, products_label, products_image FROM products' ); $result = mysql_query( $query, $conn ) or die( mysql_error( $conn ) ); header( 'Content-Type: text/csv' ); header( 'Content-Disposition: attachment;filename=export.csv' ); $row = mysql_fetch_assoc( $result ); if ( $row ) { echocsv( array_keys( $row ) ); } while ( $row ) { echocsv( $row ); $row = mysql_fetch_assoc( $result ); } function echocsv( $fields ) { $separator = ''; foreach ( $fields as $field ) { if ( preg_match( '/\\r|\\n|,|"/', $field ) ) { $field = '"' . str_replace( '"', '""', $field ) . '"'; } echo $separator . $field; $separator = ','; } echo "\r\n"; } ?> If anyone can point me in the right direction that would be great. Many Thanks DB

You might need to set the date and time variables.. something like this for example $date = date ("l, F jS, Y"); $time = date ("h:i A"); You could then add them to your email message like this: $body = <<<EOD <br /><hr><br /> <strong>Salutation:</strong> $salutation <br /> <strong>First Name:</strong> $first_name <br /> <strong>Last Name: </strong>$last_name <br /> <strong>Email:</strong> $email <br /> <strong>Zip Code:</strong> $zip_code <br /> <strong>Newsletter:</strong> $newsletter <br /> <strong>Registration Date:</strong> $date at $time <br /> EOD; This should send the date and time the form was processed. Which should match the date from $registration_date which gets entered into the database Hope that helps in some way.

I have actually linked my mysql database to MSaccess for now where I can design the queries visually, and maybe even be able to create a cron in windows to send the emails.... This will also give me a sort of 'hard copy' to work from when I do get the time to properly code this feature into the project. Although, this will be just temporary to get me going for now so I can move on a bit further with my project. I will come back to it at a later stage, but at the moment I need to move on and this feature i think would take too much time. Thanks for the reply anyway, much appreciated DB

Hi, I have a database (mysql). In the database is a table with categories (categories_id, categories_name etc). I also have another table (products table) which gets filled using a html form. That table holds category info too. What I am trying to do (without any luck) is when the form is processed, it will populate the table with the new data, then I need it to compare the 2 tables to see if the category fields from both tables have a match. If there is a match in the tables then send an email to people who has asked to be notified when new product has been added that matches a particular criteria. For example, if someone adds a product to the category 'Landscapers', with a location 'Dorset'... I would like to be able to email all the registered tradespeople in Dorset about the new product. I know this should be something quite simple, but I cant seem to get my head around it. Any pointers would be fantastic. Many thanks DB