Search Results

Search found 20931 results on 838 pages for 'mysql insert'.

Page 401/838 | < Previous Page | 397 398 399 400 401 402 403 404 405 406 407 408  | Next Page >

  • Select statement that combines similar rows with certain ids?

    - by vegatron
    hi I have a warehouse_products table which defines how many products in the warehouses so lets say I have 20 records/rows in the table, some rows may contain the same product id but in a different warehouse I need to create select statement that give every product one row, and in this row I must have the quantity in warehouse A and warehouse B .. so in the end I will get for example 10 rows that contain all the data

    Read the article

  • Unknown Column?

    - by Kenny
    ok im trying to get mutual friends between these Two users, user1 and user92 This is the sql that is successful in displaying them SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 THis is how it looks friend 61 72 73 74 75 76 77 78 79 80 81 So now i want to select all users after the number 72, and i try to do it with this sql but its not working? It gives me the error, "unknown coulum name friend in where clause" SELECT IF(user_a = 1 OR user_a = 92, user_b, user_a) friend FROM friendship WHERE friend > 72 and (user_a = 1 OR user_a = 92) OR (user_b = 1 OR user_b = 92) GROUP BY 1 HAVING COUNT(*) > 1 what am i doing wrong? or what is the correct way?? thx

    Read the article

  • Normalise this Table?

    - by Abs
    Hello all, I am creating a social bookmarking app. I am having a re-thought of the DB design in the middle of development. Should I normalise the bookmarks table and remove the tag columns that I have into a separate table. I have 10 tags per bookmark and therefore 10 columns per record (per bookmark). It seems to me that breaking the table into two would just mean I would have to do a join but the way I currently have it, its a straight select - but the table doesn't feel right...? Thanks all

    Read the article

  • 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

    Read the article

  • 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.

    Read the article

  • How to handle customers with multiple addresses in CakePHP

    - by Ryan
    I'm putting together a system to track customer orders. Each order will have three addresses; a Main contact address, a billing address and a shipping address. I do not want to have columns in my orders table for the three addresses, I'd like to reference them from a separate table and have some way to enumerate the entry so I can determine if the addressing is main, shipping or billing. Does it make sense to create a column in the address table for AddressType and enumerate that or create another table - AddressTypes - that defines the address enumeration and link to that table? I have found other questions that touch on this topic and that is where I've taken my model. The problem I'm having is taking that into the cakePHP convention. I've been struggling to internalize the direction OneToMany relationships are formed - the way the documentation states feels backwards to me. Any help would be appreciated, Thanks!

    Read the article

  • How to compare filename of uploaded file and string

    - by user225269
    I use this code to upload image files in xammp server: <?php if ((($_FILES["file"]["type"] == "image/gif") || ($_FILES["file"]["type"] == "image/jpeg") || ($_FILES["file"]["type"] == "image/pjpeg")) && ($_FILES["file"]["size"] < 100000)) { if ($_FILES["file"]["error"] > 0) { echo "Return Code: " . $_FILES["file"]["error"] . "<br />"; } else { echo "Upload: " . $_FILES["file"]["name"] . "<br />"; echo "Type: " . $_FILES["file"]["type"] . "<br />"; echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />"; echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />"; if (file_exists("upload/" . $_FILES["file"]["name"])) { echo $_FILES["file"]["name"] . " already exists. "; } else { move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $_FILES["file"]["name"]); echo "Stored in: " . "upload/" . $_FILES["file"]["name"]; } } } else { echo "Invalid file, File must be less than 100Kb in size with .jpg, .jpeg, or .gif file extension"; } ?> What do I do to compare the file name of the uploaded files with the text inputted by the user? My goal is to be able to compare the user input(ID number) and the file name of the image file which should also be an ID number. So that I will be able to display the image that corresponds with the ID Number provided. What do I need to do?Please give me an idea on how can I achieve this. Thanks

    Read the article

  • how to validate username and password in vb6?

    - by srikanth
    i have created a database in mysql5.0. i want to display the data from it. it has table named login. it has 2 columns username and password. in form i have 2 text fields username and password i just want to validate input with database values and display message box. connection from vb to database is established successfully. but its not validating input. its giving error as 'object required'. please any body help i'm new to vb. i'm using vb6 and mysql5.0 thank you

    Read the article

  • SQL Querying for Threaded Messages

    - by Harper
    My site has a messaging feature where one user may message another. The messages support threading - a parent message may have any number of children but only one level deep. The messages table looks like this: Messages - Id (PK, Auto-increment int) - UserId (FK, Users.Id) - FromUserId (FK, Users.Id) - ParentMessageId (FK to Messages.Id) - MessageText (varchar 200) I'd like to show messages on a page with each 'parent' message followed by a collapsed view of the children messages. Can I use the GROUP BY clause or similar construct to retrieve parent messages and children messages all in one query? Right now I am retrieving parent messages only, then looping through them and performing another query for each to get all related children messages. I'd like to get messages like this: Parent1 Child1 Child2 Child3 Parent2 Child1 Parent3 Child1 Child2

    Read the article

  • Resetting AUTO_INCREMENT on myISAM without rebuilding the table

    - by Artem
    Please help I am in major trouble with our production database. I had accidentally inserted a key with a very large value into an autoincrement column, and now I can't seem to change this value without a huge rebuild time. "ALTER TABLE tracks_copy AUTO_INCREMENT = 661482981" Is super-slow. How can I fix this in production? I can't get this to work either (has no effect): myisamchk tracks.MYI --set-auto-increment=661482982 Any ideas? Basically, no matter what I do I get an overflow: SHOW CREATE TABLE tracks CREATE TABLE tracks ( ... ) ENGINE=MYISAM AUTO_INCREMENT=2147483648 DEFAULT CHARSET=latin1

    Read the article

  • How can I get columns name from select query in php?

    - by Farshad Mehrvarzan
    I want to execute a SELECT query but I don't how many columns to select. Like: select name, family from persons; How can I know which columns to select? "I am currently designing a site for the execute query by users. So when the user executes this query, I won't know which columns selected. But when I want to show the results and draw a table for the user I should know which columns selected."

    Read the article

  • Sql Query to get total rows and total rows matching specific condition

    - by mrNepal
    OK, Here is what my table looks like ------------------------------------------------ id type ----------------------------------------------- 1 a 2 b 3 a 4 c 5 c 7 a 8 a ------------------------------------------------ Now, I need a query that can give me this output... ----------------------------------------------------------------- count(*) | count(type=a) | count(type=b) | count(type=c) ----------------------------------------------------------------- 8 4 1 3 ------------------------------------------------------------------ I only know to get the total set using count(*), but how to do the remaining

    Read the article

  • Zend_Table_Db and Zend_Paginator num rows

    - by Uffo
    I have the following query: $this->select() ->where("`name` LIKE ?",'%'.mysql_escape_string($name).'%') Now I have the Zend_Paginator code: $paginator = new Zend_Paginator( // $d is an instance of Zend_Db_Select new Zend_Paginator_Adapter_DbSelect($d) ); $paginator->getAdapter()->setRowCount(200); $paginator->setItemCountPerPage(15) ->setPageRange(10) ->setCurrentPageNumber($pag); $this->view->data = $paginator; As you see I'm passing the data to the view using $this->view->data = $paginator Before I didn't had $paginator->getAdapter()->setRowCount(200);I could determinate If I have any data or not, what I mean with data, if the query has some results, so If the query has some results I show the to the user, if not, I need to show them a message(No results!) But in this moment I don't know how can I determinate this, since count($paginator) doesn't work anymore because of $paginator->getAdapter()->setRowCount(200);and I'm using this because it taks about 7 sec for Zend_Paginator to count the page numbers. So how can I find If my query has any results?

    Read the article

  • Keeping some choices in the Table for the Field of Type Dropdown

    - by Mercy
    Hi, i am having a Table named Attributes which has id form_id label size sequence_no Type 1 1 Name 200 1 Text 2 1 Age 150 2 Number 3 1 Address 300 3 Textarea 4 1 Gender 200 4 Dropdown I am having the doubt how can i keep the Choices of the Field of type "Dropdown" in the Table Eg. For Gender the choices will Male , Female.. Please give me the suggestions...

    Read the article

  • Concatinating Multiple Rows in SQL

    - by Dave C
    Hello, I have a table structure that looks like this: ID String ----------- 1 A 1 Test 1 String 2 Dear 2 Person I need the final output to look like this: ID FullString -------------------- 1 A, Test, String 2 Dear, Person I am really lost on how to approach this... I looked on a couple examples online but they seemed to be VERY complex... this seems like it should be a real easy problem to solve in sql. Thank you for all assistance!

    Read the article

  • select from multiple tables but ordering by a datetime field

    - by Chris Mccabe
    I have 3 tables that are unrelated (related that each contains data for a different social network). Each has a datetime field dated- I'm already grouping by hour as you can see below (this one below for linked_in) SELECT count(*), date_format(dated, '%Y:%m:%d %H') as hour FROM upd8r_linked_in_accts WHERE CAST(dated AS DATE) = '".$start_date."' GROUP BY hour I would like to know how to do a total across all 3 networks- the tables for the three are CREATE TABLE IF NOT EXISTS `upd8r_facebook_accts` ( `id` int(11) NOT NULL AUTO_INCREMENT, `owner_id` varchar(50) NOT NULL, `user_id` int(11) NOT NULL, `fb_id` bigint(30) NOT NULL, `dated` datetime NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=80 ; CREATE TABLE IF NOT EXISTS `upd8r_linked_in_accts` ( `id` int(11) NOT NULL AUTO_INCREMENT, `owner_id` varchar(50) NOT NULL, `user_id` int(11) NOT NULL, `linked_in` varchar(200) NOT NULL, `oauth_secret` varchar(100) NOT NULL, `first_count` int(11) NOT NULL, `second_count` int(11) NOT NULL, `dated` datetime NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=200 ; CREATE TABLE IF NOT EXISTS `upd8r_twitter_accts` ( `id` int(11) NOT NULL AUTO_INCREMENT, `owner_id` varchar(50) NOT NULL, `user_id` int(11) NOT NULL, `twitter` varchar(200) NOT NULL, `twitter_secret` varchar(100) NOT NULL, `dated` datetime NOT NULL, PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ; something like this ? (SELECT count(*), date_format(dated, '%Y:%m:%d %H') as hour FROM upd8r_linked_in_accts WHERE CAST(dated AS DATE) = '".$start_date."') UNION ALL (SELECT count(*), date_format(dated, '%Y:%m:%d %H') as hour FROM upd8r_facebook_accts WHERE CAST(dated AS DATE) = '".$start_date."') UNION ALL (SELECT count(*), date_format(dated, '%Y:%m:%d %H') as hour FROM upd8r_twitter_accts WHERE CAST(dated AS DATE) = '".$start_date."') UNION ALL GROUP BY hour

    Read the article

  • SQL LEFT JOIN help

    - by Stolz
    My scenario: There are 3 tables for storing tv show information; season, episode and episode_translation. My data: There are 3 seasons, with 3 episodes each one, but there is only translation for one episode. My objetive: I want to get a list of all the seasons and episodes for a show. If there is a translation available in a specified language, show it, otherwise show null. My attempt to get serie 1 information in language 1: SELECT season_number AS season,number AS episode,name FROM season NATURAL JOIN episode NATURAL LEFT JOIN episode_trans WHERE id_serie=1 AND id_lang=1 ORDER BY season_number,number result: +--------+---------+--------------------------------+ | season | episode | name | +--------+---------+--------------------------------+ | 3 | 3 | Episode translated into lang 1 | +--------+---------+--------------------------------+ expected result +-----------------+--------------------------------+ | season | episode| name | +-----------------+--------------------------------+ | 1 | 1 | NULL | | 1 | 2 | NULL | | 1 | 3 | NULL | | 2 | 1 | NULL | | 2 | 2 | NULL | | 2 | 3 | NULL | | 3 | 1 | NULL | | 3 | 2 | NULL | | 3 | 3 | Episode translated into lang 1 | +--------+--------+--------------------------------+ Full DB dump http://pastebin.com/Y8yXNHrH

    Read the article

  • Error in computed Field of select Query

    - by Shehzad Bilal
    This Query is giving me an error of #1054 - Unknown column 'totalamount' in 'where clause' SELECT (amount1 + amount2) as totalamount FROM `Donation` WHERE totalamount > 1000 I know i can resolve this error by using group by clause and replace my where condition with having clause. But is there any other solution beside using having clause. If group by is the only solution then I want to know why I have to use group by clause even I havent use any aggregate function thanks.

    Read the article

  • Understanding Nested If.. Else statements

    - by user1174762
    For some reason my PHP login script keeps returning "invalid email/password combination", yet i know I am entering the correct email and password. Does anyone see what I might be doing wrong? <?php $email= $_POST['email']; $password= $_POST['password']; if (!empty($email) && !empty($password)) { $connect= mysqli_connect("localhost", "root", "", "si") or die('error connecting with the database'); $query= "SELECT user_id, email, password FROM users WHERE email='$email' AND password='$password'"; $result= mysqli_query($connect, $query) or die('error with query'); if (mysqli_num_rows($result) == 1) { $row= mysqli_fetch_array($result); setcookie('user_id', $row['user_id']); echo "you are now logged in"; } else { echo "invalid username/password combination"; } } else { echo" you must fill out both username and password"; } ?>

    Read the article

  • SQL Server - Get Inserted Record Identity Value when Using a View's Instead Of Trigger

    - by CuppM
    For several tables that have identity fields, we are implementing a Row Level Security scheme using Views and Instead Of triggers on those views. Here is a simplified example structure: -- Table CREATE TABLE tblItem ( ItemId int identity(1,1) primary key, Name varchar(20) ) go -- View CREATE VIEW vwItem AS SELECT * FROM tblItem -- RLS Filtering Condition go -- Instead Of Insert Trigger CREATE TRIGGER IO_vwItem_Insert ON vwItem INSTEAD OF INSERT AS BEGIN -- RLS Security Checks on inserted Table -- Insert Records Into Table INSERT INTO tblItem (Name) SELECT Name FROM inserted; END go If I want to insert a record and get its identity, before implementing the RLS Instead Of trigger, I used: DECLARE @ItemId int; INSERT INTO tblItem (Name) VALUES ('MyName'); SELECT @ItemId = SCOPE_IDENTITY(); With the trigger, SCOPE_IDENTITY() no longer works - it returns NULL. I've seen suggestions for using the OUTPUT clause to get the identity back, but I can't seem to get it to work the way I need it to. If I put the OUTPUT clause on the view insert, nothing is ever entered into it. -- Nothing is added to @ItemIds DECLARE @ItemIds TABLE (ItemId int); INSERT INTO vwItem (Name) OUTPUT INSERTED.ItemId INTO @ItemIds VALUES ('MyName'); If I put the OUTPUT clause in the trigger on the INSERT statement, the trigger returns the table (I can view it from SQL Management Studio). I can't seem to capture it in the calling code; either by using an OUTPUT clause on that call or using a SELECT * FROM (). -- Modified Instead Of Insert Trigger w/ Output CREATE TRIGGER IO_vwItem_Insert ON vwItem INSTEAD OF INSERT AS BEGIN -- RLS Security Checks on inserted Table -- Insert Records Into Table INSERT INTO tblItem (Name) OUTPUT INSERTED.ItemId SELECT Name FROM inserted; END go -- Calling Code INSERT INTO vwItem (Name) VALUES ('MyName'); The only thing I can think of is to use the IDENT_CURRENT() function. Since that doesn't operate in the current scope, there's an issue of concurrent users inserting at the same time and messing it up. If the entire operation is wrapped in a transaction, would that prevent the concurrency issue? BEGIN TRANSACTION DECLARE @ItemId int; INSERT INTO tblItem (Name) VALUES ('MyName'); SELECT @ItemId = IDENT_CURRENT('tblItem'); COMMIT TRANSACTION Does anyone have any suggestions on how to do this better? I know people out there who will read this and say "Triggers are EVIL, don't use them!" While I appreciate your convictions, please don't offer that "suggestion".

    Read the article

  • Best way to construct this query?

    - by Andrew
    I have two tables set up similar to this (simplified for the quest): actions- id - user_id - action - time users - id - name I want to output the latest action for each user. I have no idea how to go about it. I'm not great with SQL, but from what I've looked up, it should look something like the following. not sure though. SELECT `users`.`name`, * FROM users, actions JOIN < not sure what to put here > ORDER BY `actions`.`time` DESC < only one per user_id > Any help would be appreciated.

    Read the article

  • Allowed Values list in drupal CCK Fields

    - by GaxZE
    Hello, I'm basically looking to simply print out each of the allowed values in a CCK field.. i know the allowed values are stored inside a text field within the table: 'content_node_field'. the values are then stored within 'global_settings' I'm looking to somehow print out each individual allowed value using a PHP loop. however with all values being stored within one text field.. im finding it hard to print out each value individually.

    Read the article

< Previous Page | 397 398 399 400 401 402 403 404 405 406 407 408  | Next Page >