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SQL query for finding records where count > 1
I have a table named PAYMENT . Within this table I have a user ID, an account number, a ZIP code and a date. I would like to find all records for all users that have more than one payment per day with the same account number.
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Primary key/foreign Key naming convention [closed]
...un, have them specify at length why their method is superior for recursive table references.
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XAMPP - MySQL shutdown unexpectedly
...course, destroy all the data which was stored in that file (all the InnoDB tables which were created before setting innodb_file_per_table to enabled). The methods suggested in the log message seem suprtior to me.
– Tgr
Oct 12 '13 at 10:50
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Is there any good dynamic SQL builder library in Java? [closed]
... concept of code generation, where meta classes are generated for database tables and fields. This facilitates a nice, clean DSL but it faces a problem when trying to create queries for databases that are only known at runtime, like in the OP's example above. While jOOQ supports a String based appro...
How do I create an Excel (.XLS and .XLSX) file in C# without installing Microsoft Office?
...asy to use. Plus it has a DataSetHelper that lets you use DataSets and DataTables to easily work with Excel data.
ExcelLibrary seems to still only work for the older Excel format (.xls files), but may be adding support in the future for newer 2007/2010 formats.
You can also use EPPlus, which work...
Sorting rows in a data table
We have two columns in a DataTable , like so:
12 Answers
12
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How can one display images side by side in a GitHub README.md?
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The easiest way I can think of solving this is using the tables included in GitHub's flavored markdown.
To your specific example it would look something like this:
Solarized dark | Solarized Ocean
:-------------------------:|:-------------------------:
.
TABLE:
ID NAME
1 Peter
2 John
3 Greg
4 Peter
SELECT *
FROM TABLE
ORDER BY NAME
=
3 Greg
2 John
1 Peter
4 Peter
SELECT Count(ID), NAME
FROM TABLE
GROUP BY NAME
=
1 Greg
1 John
2 Peter
SELECT NAME
FROM TABLE
GROUP B...
MySQL string replace
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UPDATE your_table
SET your_field = REPLACE(your_field, 'articles/updates/', 'articles/news/')
WHERE your_field LIKE '%articles/updates/%'
Now rows that were like
http://www.example.com/articles/updates/43
will be
http://www.examp...
How to force child div to be 100% of parent div's height without specifying parent's height?
...this with CSS in a browser compatible way is not trivial (but trivial with tables) so find yourself an appropriate pre-packaged solution.
Also, the answer varies on whether you want 100% height or equal height. Usually it's equal height. If it's 100% height the answer is slightly different.
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