Something like: DECLARE @artSales TABLE (artid int, dt date, price money); INSERT @artSales VALUES (1, ‘20170102’, 10), (1, ‘20170108’, 10), (1, ‘20170112’, 8.50), (1, ‘20170115’, 8.50), (2, ‘20170102’, 20), (2, ‘20170109’, 20), (2, ‘20170112’, 35), (2, ‘20170116’, 40), (3, ‘20170101’, 500), (3, ‘20170111’, 500), (3, ‘20170130’, 500); SELECT artid, dt, oldPrice = price, PriceChange = … Read more
You can achieve it like following. SELECT ‘8=’ + CAST([8_length] AS VARCHAR(10)) AS [8_length] FROM [Your_Table] Note: If the column type is already VARCHAR, in that case you don’t need to use CAST solved I want to fetch data from table and append column name code to the value as key value pair. How to … Read more
Use CHAR_INDEX twice: SELECT *, SUBSTRING(path, pos1 + 1, pos2 – pos1 – 1) FROM tests CROSS APPLY (SELECT NULLIF(CHARINDEX(‘\’, path), 0)) AS ca1(pos1) CROSS APPLY (SELECT NULLIF(CHARINDEX(‘\’, path, pos1 + 1), 0)) AS ca2(pos2) — NULLIF is used to convert 0 value (character not found) to NULL Test on db<>fiddle 1 solved How to … Read more
you can try pivot. in my example here i have concatenated part and location together because i have no idea how to do this with two columns. the point that this is not good style remains, but maybe this helps you out a little. i named your table ttt in my example: select ‘serialno’ as … Read more
List of critical model per store and main resources in stock if any or cero select a.sloc, a.model, a.stock critical, isnull(b.stock,0) ‘Main Resources’ from stock a left join stock b on a.model=b.model and b.sloc=”Main” where a.stock<5 For the last part of your request, I’m not sure what is needed for a new opened store. EDIT … Read more
For each element in your header or footer, whether it be an image or a text box or anything else, right click on each item and select the ‘properties’ option at the bottom of the menu. Then select the ‘visibility’ tab and change from ‘show’ to ‘show or hide based on an expression’. Click the … Read more
In the query designer drag a dimension on top of your report. Chose an Operator and a Filter and check the checkbox by Parameters and BOOOM you have your parameter. If you go now into your report designer you find the parameter you just checked in the folder Parameters on the left navigation pane (on … Read more
UPDATE t1 SET t1.IdTwo = (SELECT top 1 t2.IdTwo FROM TableTwo t2 WHERE t2.IdOne = t1.IdOne AND t2.Active = 1) FROM TableOne t1 1 solved Update value of row based on result from another table SQL Server
this is Ugly, but will eventually work: COLUMN = ‘jksjdksls#$#$@@kskjfjf,”123,456,789″ lsnslkdswfnslsjfls’ left( right(COLUMN,len(COLUMN)-instr(COLUMN,””””)), instr( right(COLUMN,len(COLUMN)-instr(COLUMN,””””)), “”””) -1 ) –> 123,456,789 This is what is done: We take this string ‘jksjdksls#$#$@@kskjfjf,”123,456,789″ lsnslkdswfnslsjfls’ find the first occurence of ” with instr(COLUMN,””””) –> returns 24 take the right end of the string with. Therefore we need to take … Read more
You use: SELECT Tablename, CONCAT(fieldName1,’, ‘,fieldName2,’, ‘,fieldName3) AS name FROM Tablename; 1 solved Simple SQL concatenation
I would read the “not certified on any” to be a check for the existence of a row. If a matching row exists, then don’t return the employee. Only return the emplouee if a matching row doesn’t exist. How would you find a matching row, to find out if an employee is “certified on any”? … Read more
You seem want : select t.* from table t where not exists (select 1 from table t1 where t1.group = t.group and t1.name=”xxxx”) 1 solved Multi not like expression sql
You will first want to ORDER BY such that all rows that have no have a middle name at located at the bottom of each id. So for example ORDER BY first ORDER BY middle then perform SELECT DISTINCT(*) FROM <table_name> WHERE (first IS NOT NULL) AND (last IS NOT NULL). This will grab the … Read more
I have created a table named S65828793 with your provided data. First I have numbered the rows in ascending sequence of departure time. Then from that I have identified the flights that have been another flight within two consecutive days from opposite direction and marked those as returning flight. Then at last I have excluded … Read more
try this: SELECT …, CASE WHEN LFLT.ID = 3 THEN 0 ELSE LFLT.ID END AS OrderIndex FROM … ORDER BY OrderIndex 1 solved Custom order in SQL Procedure