Tag Archives: Tips

Cannot open files in Microsoft Excel from SharePoint

Recently I uploaded a file in a document library on a SharePoint site. I had created a report and wanted to publish it. After reviewing client requested some changes in it, I tried to open that file from SharePoint directly in Microsoft Excel using “Open in Excel”.

Whatever I tried I always received an error “The file is corrupt and cannot be opened”. Did a little search on the web and found the solution.

  • Open Microsoft Excel
  • Go to File > Options
  • Click on the Trust Center at the bottom of left panel
  • Click on “Trust Center Settings” button on displayed pane
  • Choose “Protected View” from left panel
  • Uncheck all options on displayed pane, and click “OK” till back

sp-excel-trust-center

Now when you open that file again, you will be able to do that, until any further issue, will keep that handy.

Bucketing consecutive numbers in a range

Recently we had a little tricky problem to solve. We were displaying a report in which we had to bucket numbers in a range, such that only consecutive numbers should be in that range, if any break is there, then a new range should start. Our first solution did not worked as required. Most difficult part was identifying numbers in a sequence, and placing them in a bucket. We could not create any simple T-SQL queries which could easily sort these things out. Then we thought of first capturing the bucket of each number so that we can easily work it out, and that was not possible without cursors. Lets have a look how we did that.

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Mod operator puzzle

Recently I saw a puzzle on SQL Server Central, I was intrigued to solve it myself, so here is what I was able to accomplish, and in less than 10 minutes.

DECLARE @i int
SET @i = 0

CREATE TABLE #X (val INT)

WHILE @i < 100
BEGIN
    SELECT @i = @i + 1
    INSERT INTO #X VALUES (@i)
END

SELECT val,
       val % 3,
       val % 5,
       CASE 
            WHEN val % 3 = 0 AND val % 5 = 0 THEN 'BIZZBUZZ'
            WHEN val % 3 = 0 THEN 'BIZZ'
            WHEN val % 5 = 0 THEN 'BUZZ'
       END xval
FROM   #X

DROP TABLE #X

Generate random number for each row

Today I had to add a new date column to one of my table, and populate it with some random entries. I found a script from SQL Team Blog about generating random numbers. I modified it a little and used it to populate these dates based on another date I had in my table.

Here is modified code:

-- Create the variables for the random number generation
DECLARE @Upper int;
DECLARE @Lower int-- This will create a random number between 1 and 365
SET @Lower = 1 -- The lowest random number
SET @Upper = 365 -- The highest random number

-- we'll use this temp table to assign a random number to each ID value
CREATE TABLE #temp (ID int NOT NULL, RandNum float NULL)
INSERT #temp (ID) SELECT ID FROM FOO

-- now, assign a new random value to each key value in #temp
DECLARE @id int
DECLARE Randomizer CURSOR
FOR SELECT RandNum FROM #temp

OPEN Randomizer
FETCH NEXT FROM Randomizer INTO @id

WHILE @@Fetch_Status != -1
BEGIN
UPDATE #temp SET RandNum = rand()
WHERE CURRENT OF Randomizer

FETCH NEXT FROM Randomizer
END

CLOSE Randomizer
DEALLOCATE Randomizer

UPDATE F
SET F.NewDate = dateadd(dd, round((T.RandNum * 100), 0), F.AnyDate)
FROM
FOO F INNER JOIN #temp T ON F.ID = T.ID

DROP TABLE #temp

Sequential Numbering/Counting of Records

Now say your table does not have an identify column, you don’t want to use a temporary table or alter your existing table, but you still would like to have a record number associated with each record. In this case you could use a self join to return a record number for each row. Here is an example that calculates a RecNum column, and displays the LastName for each record in the Northwind.dbo.Employees table. This example uses count(*) to count the number of records that are greater than or equal LastName in this self join.

SELECT COUNT(*) RecNum,
       a.LastName
FROM   Northwind.dbo.Employees a
       JOIN Northwind.dbo.Employees b
            ON  a.LastName >= b.LastName
GROUP BY
       a.LastName
ORDER BY
       a.LastName

The results from this query looks like this:

RecNum LastName
1 Buchanan
2 Callahan
3 Davolio
4 Dodsworth
5 Fuller
6 King
7 Leverling
8 Peacock
9 Suyama

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Converting string to datetime

Here how you can convert string date to a datetime value:

DECLARE @DateString char(14)
SET @DateString = '20060703074815'
SELECT CAST(SUBSTRING(@DateString, 1, 4) + '-' +
SUBSTRING(@DateString, 5, 2) + '-' +
SUBSTRING(@DateString, 7, 2) + 'T' +
SUBSTRING(@DateString, 9, 2) + ':' +
SUBSTRING(@DateString, 11, 2) + ':' +
SUBSTRING(@DateString, 13, 2) AS datetime)