Search Results

Search found 101927 results on 4078 pages for 'ms sql server'.

Page 530/4078 | < Previous Page | 526 527 528 529 530 531 532 533 534 535 536 537  | Next Page >

  • saving an image object in sql server database from action method

    - by user3532152
    [HttpPost] public void Test(HttpPostedFileBase file) { UsersContext db = new UsersContext(); byte[] image = new byte[file.ContentLength]; file.InputStream.Read(image, 0, image.Length); CrimeReport i = new CrimeReport { ImageId=1, ImageName="Anjli",ImageContent = image, Active=true }; db.CrimeReports.Add(i); db.SaveChanges(); } I am getting an exception on db.CrimeReports.Add(i);

    Read the article

  • Modifying SQL XML ?olumn

    - by Chinjoo
    I have an XML column in one of my table. For example I have an Employee table with following fields: Name (varhcar) | Address (XML) The Address field is having values like <Address> <Street></Street> <City></City> </Address> I have some n number of rows already in the table. Now I want to insert a new node - Country to all the rows in tha table. With default: <Country>IND</Country>. How can I write the query for this. I want all the existing data to be as it is with adding the country node to all the Address column XML.

    Read the article

  • SQL more elegant combination of boolean checks possible?

    - by Matze
    Call me pedantic but is there a more elegant way to combine all those checks? SELECT * FROM [TABLE1] WHERE [path] = 'RECEIVE' AND [src_ip] NOT LIKE '10.48.20.10' AND [src_ip] NOT LIKE '0.%' AND [src_ip] NOT LIKE '127.%' ORDER BY [date],[time] DESC; To something like this: SELECT * FROM [TABLE1] WHERE [path] = 'RECEIVE' AND [src_ip] NOT LIKE IN ('10.48.20.10','0.%','127.%', .... ) ORDER BY [date],[time] DESC;

    Read the article

  • SQL 2000 Multiple IF Statements

    - by Spidermain50
    I get a error when I try to use multiple IF statements. This is the error... "Msg 156, Level 15, State 1, Procedure fnTNAccidentIndicator, Line 81 Incorrect syntax near the keyword 'END'." This is the structure of my code... USE SS_TNRecords_Accident SET ANSI_NULLS ON GO SET QUOTED_IDENTIFIER ON GO CREATE FUNCTION dbo.fnTNAccidentIndicator ( @inAccidentNumber nvarchar, @inIndicatorMode int ) RETURNS nvarchar AS BEGIN DECLARE @AlcoholInd nvarchar DECLARE @DrugInd nvarchar DECLARE @SpeedInd nvarchar DECLARE @ReturnValue nvarchar SET @AlcoholInd = '1' SET @DrugInd = '2' SET @SpeedInd = '3' SET @ReturnValue = 'N' IF (@inIndicatorMode = @AlcoholInd) BEGIN --select statment IF (@@ROWCOUNT > 0) BEGIN @ReturnValue = 'Y' END END IF (@inIndicatorMode = @DrugInd) BEGIN --select statment IF (@@ROWCOUNT > 0) BEGIN @ReturnValue = 'Y' END END IF (@inIndicatorMode = @SpeedInd) BEGIN --select statment IF (@@ROWCOUNT > 0) BEGIN @ReturnValue = 'Y' END END Return @ReturnValue END GO

    Read the article

  • Getting maximum value from table using LINQ

    - by Tena
    I have a table in my database. I want to get the maximum value of a column named NumOfView. I used this code: var advert=(from ad in storedb.Ads where ad.AdScope == "1" select ad.NumOfView).Max(); It works but when there are two or more same maximum values it doesn't work and this message appears: Sequence contains more than one element What should I do now? Your answers will be very helpfull. Thanks

    Read the article

  • Is this possible in sql server 2005?

    - by chandru_cp
    This is my queries select ClientName,ClientMobNo from Clients select DriverName,DriverMobNo from Drivers It gives me two result tables... But i want to combine both the result tables into a single table... I tried union and union all it doesn't give me what i want.... Note: There is no relationship between the two tables...... There may be 200 clients and 100 drivers...

    Read the article

  • SQL Trigger Need to set x from a value

    - by Eric
    Im stuck on a the type of trigger needed to for this constraint. I will have a price and a commission. The price determines the commission amount, < 100 - 4%, < 200 - 5% etc. My idea. the database contains a separate table that will hold 4 price values , 101, 201, 401, 601, with their own matching comission %, this will be called PC. When i create a property listing i want to calculate the commission they earn depending on the price entered. on insert, i need to check the new.price and compare it to the prices in PC. Once new.price is less than the price tuple, i set the price to that commission value create or replace TRIGGER findCommission BEFORE INSERT OR UPDATE ON HASLISTING FOR each ROW BEGIN IF (:NEW.ASKING_PRICE < 100001) THEN :NEW.COMMISSION = 6.0; END IF; IF (:NEW.ASKING_PRICE < 250001) THEN :NEW.COMMISSION = 5.5; END IF; IF (:NEW.ASKING_PRICE < 1000001) THEN :NEW.COMMISSION = 5.0; END IF; IF (:NEW.ASKING_PRICE > 1000000) THEN :NEW.COMMISSION = 4.0; END IF; END;

    Read the article

  • Sql query number of occurance

    - by phenevo
    Hi, I wanna to have query: Select cars.* from cars where cars.code in ( select carCode from articles where numberofrecords with this car (it is not a column) >1 and Accepted=1 order by date ) How to write it?

    Read the article

  • T-SQL Query to Get SUM(COUNT())

    - by Ted
    Hi, I am planning to get a report for a table with following table structure: ID RequestDate ----------------------------- 1 2010/01/01 2 2010/02/14 3 2010/03/20 4 2010/01/07 5 2009/03/31 I want the results as: I D_Count RequestDate Sum ----------------------------------------- 2 2010/01 2 1 2010/02 3 2 2010/03 5 Pls help.

    Read the article

  • SQL - Query to display average as either "longer than" or "shorter than"

    - by user1840801
    Here are the tables I've created: CREATE TABLE Plane_new (Pnum char(3), Feature varchar2(20), Ptype varchar2(15), primary key (Pnum)); CREATE TABLE Employee_new (eid char(3), ename varchar(10), salary number(7,2), mid char(3), PRIMARY KEY (eid), FOREIGN KEY (mid) REFERENCES Employee_new); CREATE TABLE Pilot_new (eid char(3), Licence char(9), primary key (eid), foreign key (eid) references Employee_new on delete cascade); CREATE TABLE FlightI_new (Fnum char(4), Fdate date, Duration number(2), Pid char(3), Pnum char(3), primary key (Fnum), foreign key (Pid) references Pilot_new (eid), foreign key (Pnum) references Plane_new); And here is the query I must complete: For each flight, display its number, the name of the pilot who implemented the flight and the words ‘Longer than average’ if the flight duration was longer than average or the words ‘Shorter than average’ if the flight duration was shorter than or equal to the average. For the column holding the words ‘Longer than average’ or ‘Shorter than average’ make a header Length. Here is what I've come up with - with no luck! SELECT F.Fnum, E.ename, CASE Length WHEN F.Duration>(SELECT AVG(F.Duration) FROM FlightI_new F) THEN "Longer than average" WHEN F.Duration<=(SELECT AVG(F.Duration) FROM FlightI_new F) THEN 'Shorter than average' END FROM FlightI_new F LEFT OUTER JOIN Employee_new E ON F.Pid=E.eid GROUP BY F.Fnum, E.ename; Where am I going wrong?

    Read the article

  • how to have the right sum of type Time with sql

    - by kawtousse
    hi , i have 2 fields called Timefrom and TimeUntill the duration is calculated in TimeSpent. The colonne timeFrom is like the followine: 10:00:00 the colonne timeUntill is like the following: 12:00:00 the time spent colonne has the value: 02:00:00. My goal is to calculate the sum of timeSpent. PLease help.

    Read the article

  • [SQL] Select 3 lastest order for each customer

    - by Ratiug
    Hi Here is my table CusOrder that collect customer order OrderID Cus_ID Product_ID NumberOrder OrderDate 1 0000000001 9 1 6/5/2553 0:00:00 2 0000000001 10 1 6/5/2553 0:00:00 3 0000000004 9 2 13/4/2553 0:00:00 4 0000000004 9 1 17/3/2553 0:00:00 5 0000000002 9 1 22/1/2553 0:00:00 7 0000000005 9 1 16/12/2552 0:00:00 8 0000000003 9 3 13/12/2552 0:00:00 10 0000000001 9 2 19/11/2552 0:00:00 11 0000000003 9 2 10/11/2552 0:00:00 12 0000000002 9 1 23/11/2552 0:00:00 I need to select 3 lastest order for each customer and I need all customer so it will show each customer and his/her 3 lastest order how can I do it sorry for my bad english

    Read the article

< Previous Page | 526 527 528 529 530 531 532 533 534 535 536 537  | Next Page >