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  • SQL query root parent child records

    - by Vish
    Hi, We have nested folders with parent-child relationship. We use MySQL MyISAM DB. The data is stored in the DB in the following manner. Every time a child folder is created in the nested structure, the previous parentID is added. I want to get the RootFolderID of a folder which is added in the hierarchy as tabulated below. FoldID ParentID |RootFolderID -----------------|------------------- 1 0 | 0 2 1 | 1 3 2 | 1 4 3 | 1 5 4 | 1 Please let me know how to get the root folderID and populate it in the RootFolderID column after a folder is created each time. Thanks.

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  • Is it ok to hardcode dynamic links in a permanent view?

    - by meder
    Let's say I wanted to showcase 2-3 clickable buttons on my homepage which will be there permanently. These are links to the css, html, and javascript tag listing pages. Is it fine to just hardcode href=/tags/css and href=/tags/html right in my django templates/view? I won't change them for at least a year or so, meaning I don't think I need to add a column to the tags table to distinguish them - is this common or should I try to make it somewhat dynamic? These tags are in a table but so are 1000 other tags.

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  • Deceptive MySQL Query

    - by jerebear
    So I don't consider myself a novice at MySQL but this one has me stumped: I have a message board and I want to pull a list of all the most recent posts grouped by the Thread ID. Here's the table: MB_Posts -ID -Thread_ID -Created_On (timestamp) -Creator_User (user_id) -Subject -Contents -Edited (timestamp) -Reported I've tried many different things to keep it simple but I would like help from the community on this one. Just to kick this out there...this one does not work as expected: SELECT * FROM MB_Posts GROUP BY Thread_ID ORDER BY ID DESC

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  • Oracle SQL Update query takes days to update

    - by B Senthil Kumar
    I am trying to update a record in the target table based on the record coming in from source. For instance, if the incoming record is present in the target table I would update them in the target else I would simply insert. I have over one million records in my source while my target has 46 million records. The target table is partitioned based on calendar key. I implement this whole logic using Informatica. I find that the Informatica code is perfectly fine looking at the Informatica session log but its in the update it takes long time (more than 5 days to update one million records). Any suggestions as to what can be done on the scenario to improve the performance?

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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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  • problem in counting two fields in one query

    - by Mac Taylor
    hey guys i need to count new private messages and old one from a table so first thing come to mind is using mysql_num_rows and easy thing to do // check new pms $user_id = $userinfo['user_id']; $sql = "SELECT author_id FROM bb3privmsgs_to WHERE user_id='$user_id' AND (pm_new='1' OR pm_unread='1')"; $result = $db->sql_query($sql) ; $new_pms = $db->sql_numrows($result); $db->sql_freeresult($result); // check old pms $sql = "SELECT author_id FROM bb3privmsgs_to WHERE user_id='$user_id' AND (pm_new='0' OR pm_unread='0')"; $result = $db->sql_query($sql) ; $old_pms = $db->sql_numrows($result); $db->sql_freeresult($result); but how can i count these two fields just in one statement and shorter lines ?~

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  • Saving related model objects

    - by iHeartDucks
    I have two related models (one to many) in my django app and When I do something like this ObjBlog = Blog() objBlog.name = 'test blog' objEntry1 = Entry() objEntry1.title = 'Entry one' objEntry2 = Entry() objEntry2.title = 'Entry Two' objBlog.entry_set.add(objEntry1) objBlog.entry_set.add(objEntry2) I get an error which says "null value in column and it violates the foreign key not null constraint". None of my model objects have been saved. Do I have to save the "objBlog" before I could set the entries? I was hoping I could call the save method on objBlog to save it all. NOTE: I am not creating a blog engine and this is just an example.

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  • Why does this MySQL Query hang?

    - by zzapper
    SELECT * FROM tbl_order_head AS o INNER JOIN tbl_orders_log AS c ON o.PAYMENT_TRANSACTION_LOG_ID=c.TRANSACTION_ID WHERE o.VISUAL_ID = '77783'; tbl_order_head 67,000 (30 fields) records, tbl_orders_log 17000 (5 fields) records. I don't know if it would eventually return as I am running it on a live server and fear overloading. I am doing similar queries and much more complex queries successfully.

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  • Recursive COUNT Query (SQL Server)

    - by Cosmo
    Hello Guys! I've two MS SQL tables: Category, Question. Each Question is assigned to exactly one Category. One Category may have many subcategories. Category Id : bigint (PK) Name : nvarchar(255) AcceptQuestions : bit IdParent : bigint (FK) Question Id : bigint (PK) Title : nvarchar(255) ... IdCategory : bigint (FK) How do I recursively count all Questions for a given Category (including questions in subcategories). I've tried it already based on several tutorials but still can't figure it out :(

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  • MySQL query for initial filling of order column

    - by Sejanus
    Sorry for vague question title. I've got a table containing huge list of, say, products, belonging to different categories. There's a foreign key column indicating which category that particular product belongs to. I.e. in "bananas" row category might be 3 which indicates "fruits". Now I added additional column "order" which is for display order within that particular category. I need to do initial ordering. Since the list is big, I dont wanna change every row by hand. Is it possible to do with one or two queries? I dont care what initial order is as long as it starts with 1 and goes up. I cant do something like SET order = id because id counts from 1 up regardless of product category and order must start anew from 1 up for every different category.

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  • mySQL query : working with INTERVAL and CURDATE

    - by Tristan
    Hello, i'm building a chart and i want to recieve data for each months Here's my first request which is working : SELECT s.GSP_nom AS nom, timestamp, AVG( v.vote + v.prix ) /2 AS avg FROM votes_serveur AS v INNER JOIN serveur AS s ON v.idServ = s.idServ WHERE s.valide =1 AND v.date > CURDATE() -30 GROUP BY s.GSP_nom ORDER BY avg DESC But, in my case i've to write 12 request to recieve datas for the 12 previous months, is there any trick to avoid writing : // example for the previous month AND v.date > CURDATE() -60 AND v.date < CURDATE () -30 I heard about INTERVAL, i went to the mySQL doc but i didn't manage to implement it. Any ideas / example of using INTERVAL please ? Thank you

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  • Recursive COUNT Query (MS SQL)

    - by Cosmo
    Hello Guys! I've two MS SQL tables: Category, Question. Each Question is assigned to exactly one Category. One Category may have many subcategories. Category Id : bigint (PK) Name : nvarchar(255) AcceptQuestions : bit IdParent : bigint (FK) Question Id : bigint (PK) Title : nvarchar(255) ... IdCategory : bigint (FK) How do I recursively count all Questions for a given Category (including questions in subcategories). I've tried it already based on several tutorials but still can't figure it out :(

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  • Overwrite queryset which builds filter sidebar

    - by cw
    Hi, I'm writing a hockey database/manager. So I have the following models: class Team(models.Model): name = models.CharField(max_length=60) class Game(models.Model): home_team = models.ForeignKey(Team,related_name='home_team') away_team = models.ForeignKey(Team,related_name='away_team') class SeasonStats(models.Model): team = models.ForeignKey(Team) Ok, so my problem is the following. There are a lot of teams, but Stats are just managed for my Club. So if I use "list_display" in the admin backend, I'd like to modify/overwrite the queryset which builds the sidebar for filtering, to just display our home teams as a filter option. Is this somehow possible in Django? I already made a custom form like this class SeasonPlayerStatsAdminForm(forms.ModelForm): team = forms.ModelChoiceField(Team.objects.filter(club__home=True)) So now just the filtering is missing. Any ideas?

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  • help! Linq query

    - by menon
    I am getting error msg on the word Records - Type or namespace could not be found. Please help debugging it, what is missing? if (ProjDDL1.SelectedItem.Value != "--") results = CustomSearch<Records>(results, s => s.Business == ProjDDL1.SelectedItem.Value); Method CustomSearch: private DataTable CustomSearch<TKEY>(DataTable dt, Func<Records, bool> selector) { DataTable results = (dt.AsEnumerable().Where(selector).CopyToDataTable()); return results; }

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  • NHibernate - using custom sql query for a column

    - by stacker
    Is there anyway to use custom sql with NHibernate? I want to use custom sql for a specific column. select id, viewsCount, commentsCount, 0.2 * viewsCount / (select top 1 viewsCount from articles where isActive = 1 order by viewsCount DESC) as priorityViews, 0.8 * commentsCount / (select top 1 commentsCount from articles where isActive = 1 order by commentsCount DESC) as priorityComments, round(0.2 * viewsCount / (select top 1 viewsCount from articles where isActive = 1 order by viewsCount DESC) + 0.8 * commentsCount / (select top 1 commentsCount from articles where isActive = 1 order by commentsCount DESC), 1) as priority from articles

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  • SQL query: Last but one rank for user

    - by Derk
    My table structure looks like this: create table rankings ( id IDENTITY NOT NULL, user_id INT NOT NULL, game_poule_id INT NOT NULL, rank INT NOT NULL, insertDate DATETIME NOT NULL, FOREIGN KEY (user_id) REFERENCES users(id) ON DELETE CASCADE, FOREIGN KEY (game_poule_id) REFERENCES game_poules(id) ON DELETE CASCADE ); All old rankings of users per game are saved in this table. Now I want to have the last but one rank in the table for all users in a gamepoule. Has someone an idea how to achive this? Thanks

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  • Php/Mysql - need help to insert and update multiple rows with a single query

    - by Guanche
    Hello, is there any way how in this situation insert and update DB with single queries? $message = 'Hello to all group members'; $userdata = mysql_query("SELECT memberid, membernick FROM members WHERE groupid='$cid'") or die('Error'); while(list($memberid, $membernick) = mysql_fetch_row($userdata)) { $result1 = mysql_query("INSERT INTO messages VALUES (NULL,'$membernick', '$memberid', '$message')") or die('Error'); $result2 = mysql_query("UPDATE users SET new_messages=new_messages+1, total_messages=total_messages+1 WHERE id='$memberid'") or die('Error'); }

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  • Enable export to XML via HTTP on a large number of models with child relations

    - by Vasil
    I've a large number of models (120+) and I would like to let users of my application export all of the data from them in XML format. I looked at django-piston, but I would like to do this with minimum code. Basically I'd like to have something like this: GET /export/applabel/ModelName/ Would stream all instances of ModelName in applabel together with it's tree of related objects . I'd like to do this without writing code for each model. What would be the best way to do this?

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  • PHP - SQL query to get update time from table status

    - by Tribalcomm
    This is my php code (I already have a connection to the db): $array = mysql_query("SHOW TABLE STATUS FROM mytable;"); while ($array = mysql_fetch_array($result)) { $updatetime = $array['Update_time']; } echo $updatetime; I get: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource. I am running MySQL 5.0.89 and PHP5. I do not want to add a new field to the table... I want to use the table status... Any help? Thanks!

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