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• Sql string adding problem

  - by Woland

  SELECT a.one + ' test ' +b.two from table1 a right join table1 on a.id =b.id The problem is that when one is null then it the whole string is null, is there some kind of trick to bypass this problem msSQL 2005

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• Django. Invalid keyword argument for this function. ManyToMany

  - by sagem_tetra

  I have this error: 'people' is an invalid keyword argument for this function class Passage(models.Model): name= models.CharField(max_length = 255) who = models.ForeignKey(UserProfil) class UserPassage(models.Model): passage = models.ForeignKey(Passage) people = models.ManyToManyField(UserProfil, null=True) class UserProfil(models.Model): user = models.OneToOneField(User) name = models.CharField(max_length=50) I try: def join(request): user = request.user user_profil = UserProfil.objects.get(user=user) passage = Passage.objects.get(id=2) #line with error up = UserPassage.objects.create(people= user_profil, passage=passage) return render_to_response('thanks.html') How to do correctly? Thanks!

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• Detecting Completion of an Array of Threads

  - by Harsha

  I have created an array of threads and started all of them. How to know whether all threads have completed work. I don't want to use thread.wait or thread.join.

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• Using Joins vs Entity associations

  - by shivesh

  I am learning Entity framework and linq-to-entities. It's possible to get cross values from multiple tables using JOINS (join keyword) or using the navigation fields ( associations) in which case the framework knows how to reference the cross data. My question is what to use when?

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• Set the right two characters (if there are any) to capitals?

  - by Hyflex

  Below is my data: Data Here 94/452O Data more 94/4522i Data bla 94/111 Data bla 94/459es Data bla 94/444 items is automatically generated by some previous code but it could come out like: items = ["Data Here 94/452O", "Data more 94/4522i", "Data bla 94/111", "Data bla 94/459es", "Data bla 94/444"] Now currently I'm appending the following: "\n".join(items).replace("4ke", "9") with a few other .replaces however I want it to replace/change the characters on the end of the numbers with a capital letter instead of lowercase... Output: Data Here 94/452O Data more 94/4522I Data bla 94/111 Data bla 94/459ES Data bla 94/444

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• MYSQL - Query to check against other table (hard to explain...)

  - by Sam

  I have a query that gets a list of emails who have subscribed for a newsletter trial which lasts 30 days.. $thirty = time() - 3024000; SELECT c.email FROM tbl_clients AS c JOIN tbl_clientoptions AS o ON o.client = c.id WHERE o.option = 'newsletter' AND c.datecreated $thirty What I want to do is do a check in that same query so it also returns clients OVER 30 days old if they have the tbl_clientoptions.option = 'trialoverride' (ie; a row in the client options table with the value "trialoverride") basic columns are: TBL_CLIENTS id,name,email,datecreated TBL_CLIENTOPTIONS id,client,option

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• has any gae code like 'tribes' in this site .

  - by zjm1126

  in this site : http://cloud27.com/tribes/ you can join other's tribe, and create own tribe if you login. so has some open-source code do this ? thanks

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• PHP MySQL query help

  - by user547794

  Hello, I am trying to use this query to return every instance where the variable $d['userID'] is equal to the User ID in a separate table, and then echo the username tied to that user ID. Here's what I have so far: $uid = $d['userID']; $result = mysql_query("SELECT u.username FROM users u LEFT JOIN comments c ON c.userID = u.id WHERE u.id = $uid;")$row = mysql_fetch_assoc($result); echo $row['username'];

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• Displaying a field as a comma separated list in Reporting Services 2005?

  - by user364720

  See title. Basically, the data in this report is set up such that each value in Field A has multiple corresponding values in Field B, and I need to display Field B as a comma-separated list. According to the internets, this is totally easy via a combination of Join() and LookupSet() in 2008... but I'm on 2005. Anyone know how I can do this?

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• Selecting count(*) while checking for a value in the results

  - by Rob

  SELECT COUNT(*) as Count, IF(sch.HomeTeamID = 34,true,false) AS Hawaii FROM schedule sch JOIN schools s ON s.ID = 83 WHERE (sch.HomeTeamID = 83 OR sch.AwayTeamID = 83) AND sch.SeasonID = 4 I'm trying to use count() to simplify my result but also include a field that represents wether any of the results' specific column contained a certain value. Is this possible? I'd basically like a row response with all the info I need.

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• What is the best way to find the period of a (repeating) list in Mathematica?

  - by Arnoud Buzing

  What is the best way to find the period in a repeating list? For example: a = {4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2} has repeat {4, 5, 1, 2, 3} with the remainder {4, 5, 1, 2} matching, but being incomplete. The algorithm should be fast enough to handle longer cases, like so: b = RandomInteger[10000, {100}]; a = Join[b, b, b, b, Take[b, 27]] The algorithm should return $Failed if there is no repeating pattern like above.

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• Website Design Templates

  - by user347834

  I am looking fr someone to make me two website templates for my site for free. Here is a quick design of what I want:(Took me 2 minutes in Paint) http:/ /i50.tinypic.com/33p9aut.jpg (You have to push backspace on the first link to join up the http:/ and the other /)and http://i50.tinypic.com/2qmogoo.jpg Email me at [email protected] or [email protected] for more information

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• Get top rated item using AVG mysql

  - by user1876234

  I want to find top rated item using AVG function in mysql, right now my query looks like this: SELECT a.title, AVG(d.rating) as rating FROM in8ku_content a JOIN in8ku_content_ratings d ON a.id = d.article_id ORDER BY rating DESC Problem is that it takes AVG of all items and the result is not accurate, what should be changed here to get correct result ? Tables: in8ku_content [id, title] in8ku_content_ratings [id, article_id, rating]

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• I have 2 MySql tables. How to get not all data from them in 1 sql query?

  - by Ole Jak

  I have 2 MySql tables. users with id's and username's ; streams with userId's and streamId's How to get them as / join them into one table conteining only username | streamId as SQL response? With one SQL query.

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• SQL query help - finding rows with no relationships

  - by DD

  Consider a DB with a Client table and a Book table: Client: person_id Book: book_id Client_Books: person_id,book_id How would you find all the Person ids which have no books? (without doing an outer join and looking for nulls)

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• MySQL 'user_id' in where clause is ambiguous problem

  - by HoMe

  How can I correct the problem I keep getting from the code below which states 'user_id' in where clause is ambiguous. Thanks for the help in advance. Here is the mysql table. SELECT user.*, user_info.* FROM user INNER JOIN user_info ON user.user_id = user_info.user_id WHERE user_id='$user_id'

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• Creating an SQL query to list all clients, who spent over 1000$

  - by Lapa

  So, I have three tables with the following rows: Customers (Customer_id and Name), Orders (Customer_id, Product_id, Quantity) and Products (Price). How do I write a query which shows all customers who spent more than 1000$? Do I have to join the tables?

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• Change a link's href value based on time

  - by justSteve

  I'm coding a 'Connect to Meeting' page where i would like the link that allows attendees to join our GoToMeeting event to 'become active' 15 minutes prior to the start time. So the page users visit to see the connection info (meetingID, password) includes the start time of the meeting. I need a button ('Connect To Meeting') to change from inactive to Active when [Now() < (StartTime()-15minutes)].

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• delete taking really long time! mysql

  - by every_answer_gets_a_point

  i am doing this: delete calibration_2009 from calibration_2009 join batchinfo_2009 on calibration_2009.rowid = batchinfo_2009.rowid where batchinfo_2009.reporttime like '%2010%'; both tables have about 500k lines of data i suspect that 250k match the criteria to be deleted so far it has been running for 2 hours!!! is there something wrong?

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• Is there anyway that we can get a label value to a Sql

  - by Pradeep

  SELECT COUNT(*) AS Expr1 FROM Book INNER JOIN Temp_Order ON Book.Book_ID = Temp_Order.Book_ID WHERE (Temp_Order.User_ID = 25) AND (CONVERT (nvarchar, Temp_Order.OrderDate, 111) = CONVERT (nvarchar, GETDATE(), 111)) In here i want to change my User_ID to get from a label.Text this Sql Statement is in a DataView. so in the Wizard it not accepting a text box values or anything. can someone please help me to solve this

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• Return all users from group(s) specified as comma delimited value

  - by todor

  I have the following two table scenario: users id groups 1 1,2,3 2 2,3 3 1,3 4 3 and groups id 1 2 3 How do I return the IDs of all users that belong to group 2 and 1 for example? Should I look into join, a helper group_membership table or function to separate the comma delimited group IDs to get something like this: group_membership user_id group_id 1 1 1 2 1 3 2 2 2 3 ... ...

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• an interesting game, The Wars [closed]

  - by beaver1125

  Recently, I have played an interesting webgame,The Wars. Do you want to have a try ? Join with me now. http://xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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• Problem with PHP & MySQL

  - by Shahd

  I wrote this statements but it is not work :( ... can you tell me why? HTML: <form action="join.php" method="post"> <label name="RoomName">Room1</label> </form> PHP: $roomName = $_POST['RoomName']; $roomID = "SELECT RoomID FROM rooms WHERE RoomName = $roomName";

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• TIME REDUCE(OPTIMISE QUERY)

  - by user2527657

  select a.userid,(select firstName from user where userid=NOTUSED.userid) as z, (select max(login_time) from userLoginTime AS b where userid = a.user_id GROUP BY b.user_id ORDER BY b.user_id) as y From(SELECT DISTINCT a.user_id FROM user AS a LEFT OUTER JOIN (SELECT (userid) FROM userlogintime where serialid=15400012)AS b ON user.user_id = b.user_id where a.Serialid=15400012 AND b.userid IS NULL) NOTUSED, Relation r, user a where r.childuserid = NOTUSED.userid and guarduserid = a.userid

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• How do I bypass pkgadd signature verification?

  - by Brian Knoblauch

  Trying to install CollabNet Subversion Client on Solaris x64, but I'm hung up with: ## Verifying signature for signer <Alexander Thomas(AT)> pkgadd: ERROR: Signature verification failed while verifying certificate <subject=Alexander Thomas(AT), issuer=Alexander Thomas(AT)>:<self signed certificate>. Any way to just bypass the certificate check? None of the options listed in the man page seemed appropriate.

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