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MySQL: Quick breakdown of the types of joins [duplicate]
...es
END EDIT
In a nutshell, the comma separated example you gave of
SELECT * FROM a, b WHERE b.id = a.beeId AND ...
is selecting every record from tables a and b with the commas separating the tables, this can be used also in columns like
SELECT a.beeName,b.* FROM a, b WHERE b.id = a.beeId...
Check if a temporary table exists and delete if it exists before creating a temporary table
...me in SQL Server 2005, with the extra "foo" column appearing in the second select result:
IF OBJECT_ID('tempdb..#Results') IS NOT NULL DROP TABLE #Results
GO
CREATE TABLE #Results ( Company CHAR(3), StepId TINYINT, FieldId TINYINT )
GO
select company, stepid, fieldid from #Results
GO
ALTER TABLE #R...
Get day of week in SQL Server 2005/2008
...
Use DATENAME or DATEPART:
SELECT DATENAME(dw,GETDATE()) -- Friday
SELECT DATEPART(dw,GETDATE()) -- 6
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Left Join With Where Clause
...ring away rows where the left join doesn't succeed. Move it to the join:
SELECT `settings`.*, `character_settings`.`value`
FROM `settings`
LEFT JOIN
`character_settings`
ON `character_settings`.`setting_id` = `settings`.`id`
AND `character_settings`.`character_id` = '1'
...
How do I calculate tables size in Oracle
...BLE_NAME FORMAT A32
COLUMN OBJECT_NAME FORMAT A32
COLUMN OWNER FORMAT A10
SELECT
owner,
table_name,
TRUNC(sum(bytes)/1024/1024) Meg,
ROUND( ratio_to_report( sum(bytes) ) over () * 100) Percent
FROM
(SELECT segment_name table_name, owner, bytes
FROM dba_segments
WHERE segment_type IN...
How do I select an element with its name attribute in jQuery? [duplicate]
...
$('[name="ElementNameHere"]').doStuff();
jQuery supports CSS3 style selectors, plus some more.
See more
jQuery - Selectors
jQuery - [attribute=""] selector
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How can I remove or replace SVG content?
...
Here is the solution:
d3.select("svg").remove();
This is a remove function provided by D3.js.
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Counting the number of option tags in a select tag in jQuery
How do I count the number of <option> s in a <select> DOM element using jQuery?
8 Answers
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SQL Server query - Selecting COUNT(*) with DISTINCT
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Count all the DISTINCT program names by program type and push number
SELECT COUNT(DISTINCT program_name) AS Count,
program_type AS [Type]
FROM cm_production
WHERE push_number=@push_number
GROUP BY program_type
DISTINCT COUNT(*) will return a row for each unique count. What you want is C...
Oracle “Partition By” Keyword
...d then use that in a calculation against this records salary without a sub select, which is much faster.
Read the linked AskTom article for further details.
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