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  • question with its query

    - by user329820
    Hi this is my homework and the question is this: List the average balance of customers by city and short zip code (the first five digits of thezip code). Only include customers residing in Washington State (‘WA’). also the Customer table has 5 columns(Name,Family,CustZip,CustCity,CustAVGBal) I wrote the query like below is this correct? SELECT CustCity,LEFT(CustZip,5) AS NewCustZip,CustAVGBal FROM Customer WHERE CustCity = 'WA' THANKS!!

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  • SQL: Find the max record per group

    - by user319088
    I have one table, which has three fields and data. Name , Top , Total cat , 1 , 10 dog , 2 , 7 cat , 3 , 20 horse , 4 , 4 cat , 5 , 10 dog , 6 , 9 I want to select the record which has highest value of Total for each Name, so my result should be like this: Name , Top , Total cat , 3 , 20 horse , 4 , 4 Dog , 6 , 9 I tried group by name order by total, but it give top most record of group by result. Can anyone guide me, please?

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  • 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

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  • 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!

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  • 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

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  • 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

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  • 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

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  • 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

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  • 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?

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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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  • How to retrieve column total when rows are paginated?

    - by Rick
    Hey guys I have a column "price" in a table and I used a pagination script to generate the data displayed. Now the pagination is working perfectly however I am trying to have a final row in my HTML table to show the total of all the price. So I wrote a script to do just that with a foreach loop and it sort of works where it does give me the total of all the price summed up together however it is the sum of all the rows, even the ones that are on following pages. How can I retrieve just the sum of the rows displayed within the pagination? Thank you! Here is the query.. SELECT purchase_log.id, purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM purchase_log LEFT JOIN cart_contents ON purchase_log.id = cart_contents.purchase_id LEFT JOIN members ON purchase_log.member_id = members.id GROUP BY id ORDER BY id DESC LIMIT 0,30";

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  • Combine SQL statement

    - by ninumedia
    I have 3 tables (follows, postings, users) follows has 2 fields - profile_id , following_id postings has 3 fields - post_id, profile_id, content users has 3 fields - profile_id, first_name, last_name I have a follows.profile_id value of 1 that I want to match against. When I run the SQL statement below I get the 1st step in obtaining the correct data. However, I now want to match the postings.profile_id of this resulting set against the users table so each of the names (first and last name) are displayed as well for all the listed postings. Thank you for your help! :) Ex: SELECT * FROM follows JOIN postings ON follows.following_id = postings.profile_id WHERE follows.profile_id = 1

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  • 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...

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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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  • 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

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  • How to run a set of SQL queries from a file, in PHP?

    - by Harish Kurup
    I have some set of SQL queries which is in a file(i.e query.sql), and i want to run those queries in files using PHP, the code that i have wrote is not working, //database config's... $file_name="query.sql"; $query==file($file_name); $array_length=count($query); for($i=0;$i<$array_length;$i++) { $data .= $query[$i]; } echo $data; mysql_query($data); it echos the SQL Query from the file but throws an error at mysql_query() function...

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  • 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."

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  • 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.

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  • 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

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  • 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

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