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  • Very simple shopping cart, remove button

    - by Kynian
    Im writing sales software that will be walking through a set of pages and on certain pages there are items listed to sell and when you click buy it basically just passes a hidden variable to the next page to be set as a session variable, and then when you get to the end it call gets reported to a database. However my employer wanted me to include a shopping cart, and this shopping cart should display the item name, sku, and price of whatever you're buying, as well as a remove button so the person doing the script doesnt need to go back through the entire thing to remove one item. At the moment I have the cart set to display everything, which was fairly simple. but I cant figure out how to get the remove button to work. Here is the code for the shopping cart: $total = 0; //TEST CODE: $_SESSION['itemname-addon'] = "Test addon"; $_SESSION ['price-addon'] = 10.00; $_SESSION ['sku-addon'] = "1234h"; $_SESSION['itemname-addon1'] = "Test addon1"; $_SESSION ['price-addon1'] = 99.90; $_SESSION ['sku-addon1'] = "1111"; $_SESSION['itemname-addon2'] = "Test addon2"; $_SESSION ['price-addon2'] = 19.10; $_SESSION ['sku-addon2'] = "123"; //end test code $items = Array ( "0"=> Array ( "name" => $_SESSION['itemname-mo'], "price" => $_SESSION ['price-mo'], "sku" => $_SESSION ['sku-mo'] ), "1" => Array ( "name" => $_SESSION['itemname-addon'], "price" => $_SESSION ['price-addon'], "sku" => $_SESSION ['sku-addon'] ), "2" => Array ( "name" => $_SESSION['itemname-addon1'], "price" => $_SESSION ['price-addon1'], "sku" => $_SESSION ['sku-addon1'] ), "3" => Array ( "name" => $_SESSION['itemname-addon2'], "price" => $_SESSION ['price-addon2'], "sku" => $_SESSION ['sku-addon2'] ) ); $a_length = count($items); for($x = 0; $x<$a_length; $x++){ $total +=$items[$x]['price']; } $formattedtotal = number_format($total,2,'.',''); for($i = 0; $i < $a_length; $i++){ $name = $items[$i]['name']; $price = $items[$i]['price']; $sku = $items[$i]['sku']; displaycart($name,$price,$sku); } echo "<br /> <b>Sub Total:</b> $$formattedtotal"; function displaycart($name,$price,$sku){ if($name != null || $price != null || $sku != null){ if ($name == "no sale" || $price == "no sale" || $sku == "no sale"){ echo ""; } else{ $formattedprice = number_format($price,2,'.',''); echo "$name: $$formattedprice ($sku)"; echo "<form action=\"\" method=\"post\">"; echo "<button type=\"submit\" />Remove</button><br />"; echo "</form>"; } } } So at this point Im not sure where to go from here for the remove button. Any suggestions would be appreciated.

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  • Form Not Submitting

    - by John
    Hello, When I try to click on the "submit" button for the form below, nothing happens. Any ideas why not? Thanks in advance, John submit.php: <?php require_once "header.php"; $u = $_SESSION['username']; if (!isLoggedIn()) { // user is not logged in. if (isset($_POST['cmdlogin'])) { // retrieve the username and password sent from login form & check the login. if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox2(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { show_loginform(); } } else { . show_userbox2(); } echo '<div class="submittitle">Submit an item.</div>'; echo '<form action="http://www...com/.../submit2.php" method="post"> <input type="hidden" value="'.$_SESSION['loginid'].'" name="uid"> <div class="submissiontitle"><label for="title">Story Title:</label></div> <div class="submissionfield"><input name="title" type="title" id="title" maxlength="1000"></div> <div class="urltitle"><label for="url">Link:</label></div> <div class="urlfield"><input name="url" type="url" id="url" maxlength="500"></div> <div class="submissionbutton"><input name="submit" type="submit" value="Submit"></div> </form> '; ?> submit2.php: <?php //if($_SERVER['REQUEST_METHOD'] == "POST"){header('Location: http://www...com/.../submit2.php');} require_once "header.php"; if (isLoggedIn() == true) { $remove_array = array('http://www.', 'http://', 'https://', 'https://www.', 'www.'); $cleanurl = str_replace($remove_array, "", $_POST['url']); $cleanurl = strtolower($cleanurl); $cleanurl = preg_replace('/\/$/','',$cleanurl); $title = $_POST['title']; //$url = $_POST['url']; $uid = $_POST['uid']; $title = mysql_real_escape_string($title); $cleanurl = mysql_real_escape_string($cleanurl); $site1 = 'http://' . $cleanurl; $displayurl = parse_url($site1, PHP_URL_HOST); function isURL($url1 = NULL) { if($url1==NULL) return false; $protocol = '(http://|https://)'; $allowed = '[-a-z0-9]{1,63}'; $regex = "^". $protocol . // must include the protocol '(' . $allowed . '\.)'. // 1 or several sub domains with a max of 63 chars '[a-z]' . '{2,6}'; // followed by a TLD if(eregi($regex, $url1)==true) return true; else return false; } if(isURL($site1)==true) mysql_query("INSERT INTO submission VALUES (NULL, '$uid', '$title', '$cleanurl', '$displayurl', NULL)"); else echo "<p class=\"topicu\">Not a valid URL.</p>\n"; } else { // user is not loggedin show_loginform(); } if (!isLoggedIn()) { // user is not logged in. if (isset($_POST['cmdlogin'])) { // retrieve the username and password sent from login form & check the login. if (checkLogin($_POST['username'], $_POST['password'])) { show_userbox(); } else { echo "Incorrect Login information !"; show_loginform(); } } else { // User is not logged in and has not pressed the login button // so we show him the loginform show_loginform(); } } else { // The user is already loggedin, so we show the userbox. show_userbox(); } require_once "footer.php"; ?>

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  • selecting data from table based on date .

    - by mehdi
    i have database table like this +-------+--------------+----------+ | id | ip | date | +-------+--------------+----------+ | 505 |192.168.100.1 |2010-04-03| | 252 |192.168.100.5 |2010-03-03| | 426 |192.168.100.6 |2010-03-03| | 201 |192.168.100.7 |2010-04-03| | 211 |192.168.100.10|2010-04-03| +-------+--------------+----------+ how can i retirive data from this table where month=03 how to write sql to do that . select * from table where month=03 something like that .

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  • Echoing a pseudo column value after a COUNT

    - by rob - not a robber
    Hi Gang... Please don't beat me if this is elementary. I searched and found disjointed stuff relating to pseudo columns. Nothing spot on about what I need. Anyway... I have a table with some rows. Each record has a unique ID, an ID that relates to another entity and finally a comment that relates to that last entity. So, I want to COUNT these rows to basically find what entity has the most comments. Instead of me explaining the query, I'll print it SELECT entity_id, COUNT(*) AS amount FROM comments GROUP BY entity_id ORDER BY amount DESC The query does just what I want, but I want to echo the values from that pseudo column, 'amount' Can it be done, or should I use another method like mysql_num_rows? Thank you!!!

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  • How to join nearly identical several queries into one?

    - by Devyn
    Hi, Assume I have an order_dummy table where order_dummy_id, order_id, user_id, book_id, author_id are stored. You may complain the logic of my table but I somehow need to do it that way. I want to execute following queries. SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 AND book_id = 1 ORDER BY `order_dummy_id` DESC LIMIT 1 SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 AND book_id = 2 ORDER BY `order_dummy_id` DESC LIMIT 1 SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 AND book_id = 3 ORDER BY `order_dummy_id` DESC LIMIT 1 Please keep in mind that several numbers of same book is included in one order. Therefore, I list order_dummy_id by descending and limit 1 so only LATEST ORDER of A BOOK is shown. But my goal is to show other books in that way in one table. I used group by like this ... SELECT * FROM order_dummy WHERE order_id = 1 AND user_id = 1 GROUP BY book_id but it only shows order_dummy_id with ascending result. I have no idea anymore. Looking forward your kindness help!

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  • Product Name Print Several times, How to fix.?

    - by mans
    i had added the following Opencart module for my order report list... http://www.opencart.com/index.php?route=extension/extension/info&extension_id=3597&filter_search=order%20list%20filter%20model&page=4 I have problems with the column "Products". If there are more than one option the products name prints several times. So if I got a product with three options the product name prints three times. Is there any way to fix this problem? i want print product name and model number only once, any idea.? i will attach the results what i got now... this is my sql query... public function getOrders($data = array()) { $sql = "select o.order_id,o.email,o.telephone,CONCAT(o.shipping_address_1, ' ', o.shipping_address_2) AS address,CONCAT(o.firstname, ' ', o.lastname) AS customer,o.payment_zone AS state,o.payment_address_2 AS block, o.payment_address_1 AS address,o.payment_postcode AS postcode,(SELECT os.name FROM " . DB_PREFIX . "order_status os WHERE os.order_status_id = o.order_status_id AND os.language_id = '" . (int)$this->config->get('config_language_id') . "') AS status,o.payment_city AS city,GROUP_CONCAT(pd.name) AS pdtname,GROUP_CONCAT(op.model) AS model,o.date_added,sum(op.quantity) AS quantity,GROUP_CONCAT(opt.value ) AS options, GROUP_CONCAT(opt.order_product_id ) AS ordprdid,GROUP_CONCAT(op.order_product_id ) AS optprdid, GROUP_CONCAT(op.quantity) AS opquantity from `" . DB_PREFIX . "order` o LEFT JOIN " . DB_PREFIX . "order_product op ON (op.order_id = o.order_id) LEFT JOIN " . DB_PREFIX . "product_description pd ON (pd.product_id = op.product_id and pd.language_id = '" . (int)$this->config->get('config_language_id') . "') LEFT JOIN " . DB_PREFIX . "order_option opt ON (opt.order_product_id = op.order_product_id) "; Product Name = GROUP_CONCAT(pd.name) AS pdtname,

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  • Multiple Refinement/Clusters in a single search

    - by Brain Teasers
    Suppose I have options of searching are 1.occupation 2.education 3.religion 4.caste 5.country and many more. when i perform this search, i can get easily calculate refinements. problem is i need to calculate refinements in a way that for individual refinement is calculated in a manner that all search parameter is considered expect the calculated refinement check on the left side ... even i have choose the profession area, still search refinements of this is coming.... same functionality if of all refinements.Please help.I use sphinx for searching seen this type of refinements in http://ww2.shaadi.com/search

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  • Function for putting all database table to an array

    - by jasmine
    I have written a function to print database table to an array like this $db_array= Array( ID=>1, PARENTID =>1, TITLE => LIPSUM, TEXT =>LIPSUM ) My function is: function dbToArray($table) { $allArrays =array(); $query = mysql_query("SELECT * FROM $table"); $dbRow = mysql_fetch_array($query); for ($i=0; $i<count($dbRow) ; $i++) { $allArrays[$i] = $dbRow; } $txt .='<pre>'; $txt .= print_r($allArrays); $txt .= '</pre>'; return $txt; } Anything wrong in my function. Any help is appreciated about my problem. Thanks in advance

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  • PHP getting blank pages after submit a form + signal Segmentation fault (11)

    - by Ole Media
    I few days ago I update my macbook pro to snow leopard, and since then some php files are not showing. This is what happens: I created a php form, when going to 'http://localhost/webform.php' I can see the form just fine. Then, once I submit the form, I just get a blank page. I enable error and warnings reporting under php.ini to make sure I'm not missing something, but still I'm not getting anything, just the blank page. Then I checked under apache log files, and what I notice is that every time I submit the form I see the following line coming up under the apache logs: [Wed Apr 07 21:40:28 2010] [notice] child pid 70223 exit signal Segmentation fault (11) I'm clueless on this one. Any ideas on how to fix it?

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  • hibernate not throwing stale state exception nor it is overwriting data

    - by Reddy
    Our application do the following. 1. Start the transaction. 2. Execute a query using prepared statement 3. Check a condition to see the number of rows updated are equal to the required number. 4. It commits on success of above condition otherwise it will roll back However the problem is that when two threads are simultaneously enter this code. Thread-1 is updating a row in step 2. It checked the condition and committed successfully since the condition is successful. Thread-2 started execution somewhere between steps 1 & 4, and it is failing on at condition checking at step 3 (as it is getting number of updated rows as 0). I expected second thread to throw an exception but it is not. What could be the problem?

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  • SQL query to retrieve financial year data grouped by the year

    - by mlevit
    Hi, I have a database with lets assume two columns (service_date & invoice_amount). I would like to create an SQL query that would retrieve and group the data for each financial year (July to June). I have two years of data so that is two financial years (i.e. 2 results). I know I can do this manually by creating an SQL query to group by month then run the data through PHP to create the financial year data but I'd rather have an SQL query. All ideas welcome. Thanks

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  • Top x rows and group by (again)

    - by Tibor Szasz
    Hello, I know it's a frequent question but I just can't figure it out and the examples I found didn't helped. What I learned, the best strategy is to try to find the top and bottom values of the top range and then select the rest, but implementing is a bit tricky. Example table: id | title | group_id | votes I'd like to get the top 3 voted rows from the table, for each group. I'm expecting this result: 91 | hello1 | 1 | 10 28 | hello2 | 1 | 9 73 | hello3 | 1 | 8 84 | hello4 | 2 | 456 58 | hello5 | 2 | 11 56 | hello6 | 2 | 0 17 | hello7 | 3 | 50 78 | hello8 | 3 | 9 99 | hello9 | 3 | 1 I've fond complex queries and examples, but they didn't really helped.

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  • Define keys in temporary table creation

    - by imperium2335
    How do I define the keys for a temporary table that is being created from a SELECT statement? I have: CREATE temporary TABLE _temp_unique_parts_trading engine=memory AS (SELECT parts_trading.enquiryref, sellingcurrency, jobs.id AS jobID FROM parts_trading, jobs WHERE jobs.enquiryref = parts_trading.enquiryref GROUP BY parts_trading.enquiryref) But where do I define the keys?

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  • getting sql records

    - by droidus
    when i run this code, it returns the topic fine... $query = mysql_query("SELECT topic FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; } but when I change it to this, it doesn't run at all. why is this happening? $query = mysql_query("SELECT topic, lock FROM question WHERE id = '$id'"); if(mysql_num_rows($query) > 0) { $row = mysql_fetch_array($query) or die(mysql_error()); $topic = $row['topic']; $lockedThread = $row['lock']; echo "here: " . $lockedThread; }

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  • a query is inserted from PHPMYAdmin but not from PHP

    - by iyad al aqel
    i'm writing a php code to insert form values in a forum values $dbServer = mysql_connect("localhost" , "root", "") ; if(!$dbServer) die ("Unable to connect"); mysql_select_db("kfumWonder"); $name= $_POST['name'] ; $password= md5($_POST['password']); $email= $_POST['email'] ; $major= $_POST['major'] ; $dateOfBirth=$_POST['dateOfBirth'] ; $webSite = $_POST['website']; $joinDate= date("Y m d") ; $query = "INSERT INTO user (name, password, email, major, dob, website, join_date) Values ('$name', '$password', '$email', '$major', '$dateOfBirth', '$webSite' , '$joinDate')" ; //echo $query ; $result = mysql_query($query) ; if (! $result ) echo " no results " ; this works perfectly fine when i took the printed query and run it in PHPMyAdmin but when i run this code nothing happens , any ideas !?

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  • Can't use where clause on correlated columns.

    - by Keyo
    I want to add a where clause to make sure video_count is greater than zero. Only categories which are referenced once or more in video_category.video_id should be returned. Because video_count is not a field in any table I cannot do this. Here is the query. SELECT category . * , ( SELECT COUNT( * ) FROM video_category WHERE video_category.category_id = category.category_id ) AS 'video_count' FROM category WHERE category.status = 1 AND video_count > '0' AND publish_date < NOW() ORDER BY updated DESC; Thanks for the help.

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  • User has many computers, computers have many attributes in different tables, best way to JOIN?

    - by krismeld
    I have a table for users: USERS: ID | NAME | ---------------- 1 | JOHN | 2 | STEVE | a table for computers: COMPUTERS: ID | USER_ID | ------------------ 13 | 1 | 14 | 1 | a table for processors: PROCESSORS: ID | NAME | --------------------------- 27 | PROCESSOR TYPE 1 | 28 | PROCESSOR TYPE 2 | and a table for harddrives: HARDDRIVES: ID | NAME | ---------------------------| 35 | HARDDRIVE TYPE 25 | 36 | HARDDRIVE TYPE 90 | Each computer can have many attributes from the different attributes tables (processors, harddrives etc), so I have intersection tables like this, to link the attributes to the computers: COMPUTER_PROCESSORS: C_ID | P_ID | --------------| 13 | 27 | 13 | 28 | 14 | 27 | COMPUTER_HARDDRIVES: C_ID | H_ID | --------------| 13 | 35 | So user JOHN, with id 1 owns computer 13 and 14. Computer 13 has processor 27 and 28, and computer 13 has harddrive 35. Computer 14 has processor 27 and no harddrive. Given a user's id, I would like to retrieve a list of that user's computers with each computers attributes. I have figured out a query that gives me a somewhat of a result: SELECT computers.id, processors.id AS p_id, processors.name AS p_name, harddrives.id AS h_id, harddrives.name AS h_name, FROM computers JOIN computer_processors ON (computer_processors.c_id = computers.id) JOIN processors ON (processors.id = computer_processors.p_id) JOIN computer_harddrives ON (computer_harddrives.c_id = computers.id) JOIN harddrives ON (harddrives.id = computer_harddrives.h_id) WHERE computers.user_id = 1 Result: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | HARDDRIVE TYPE 25 | But this has several problems... Computer 14 doesnt show up, because it has no harddrive. Can I somehow make an OUTER JOIN to make sure that all computers show up, even if there a some attributes they don't have? Computer 13 shows up twice, with the same harddrive listet for both. When more attributes are added to a computer (like 3 blocks of ram), the number of rows returned for that computer gets pretty big, and it makes it had to sort the result out in application code. Can I somehow make a query, that groups the two returned rows together? Or a query that returns NULL in the h_name column in the second row, so that all values returned are unique? EDIT: What I would like to return is something like this: ID | P_ID | P_NAME | H_ID | H_NAME | ----------------------------------------------------------- 13 | 27 | PROCESSOR TYPE 1 | 35 | HARDDRIVE TYPE 25 | 13 | 28 | PROCESSOR TYPE 2 | 35 | NULL | 14 | 27 | PROCESSOR TYPE 1 | NULL | NULL | Or whatever result that make it easy to turn it into an array like this [13] => [P_NAME] => [0] => PROCESSOR TYPE 1 [1] => PROCESSOR TYPE 2 [H_NAME] => [0] => HARDDRIVE TYPE 25 [14] => [P_NAME] => [0] => PROCESSOR TYPE 1

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  • SQL Query Update is not working

    - by Brett Powell
    Hey guys, I am using pawn script for something, and everything works great except for one of my queries. For some reason, it will not work, and I am hoping it is simple enough someone can spot my mistake as I have been banging my head on it for days. http://ampaste.net/m6a887d30 The two highlighted lines are the queries that are not working. The other one works fine, but the values for 'class1kills' and 'class2kills' remain at 0. Here is a screenshot from phpmyadmin incase I did something silly. http://brutalservers.net/sql.png

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  • Problem with PHP/Java bridge.

    - by Jack
    I am using Tomcat 6. I am running a php script using the JavaBridge. I get the following error when I run my code. Fatal error: Call to undefined function mysqli_connect() in C:\Program Files\apache-tomcat-6.0.26\webapps\JavaBridge\xxxx\xxxxx.php on line 534 Please help.

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  • Correct Sql Script for Formula

    - by Madan Madan
    Can anyone help me write SQL script for the following formula? If DEP = 1 If DROP 1 PLV = 334.86 * exp(0.3541 * ACTIVE_DAYS) + 0.25 * DROP + 20 * DEP Else If DROP < 0 PLV = DROP + 70 * ACTIVE_DAYS Else PLV = 0.25 * DROP + 70 * ACTIVE_DAYS The SQL script which I have is the following SELECT IF(dep=1, if(dep=1, (334.86 * exp(0.3541 * act_days)) + (0.25 * 'drop') + (20 * dep), if('drop'<0, 'drop' + (70 * act_days), (0.25 * 'drop') + (70 * act_days))),'0') as PLV But the above query is not right as something is missing where the formula says Else PLV = 0.26 * DROP Thanks,

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  • How phpmyadmin, php and everything else works?

    - by Tom
    Ok, I guess I got crazy, but really. How phpmyadmin works? Does it have his own phpmyadmin or what? And how php works? Why writing echo 'hello'; it returns hello in the browser? I am really interested on how these things really works, maybe you know any books or smth to figure it out? Thank you.

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