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Retaining file permissions with Git
...te" is not exactly relevant (as you said in your question: different users/groups). But the notion of "executable" doesn't depend on users and groups and can be reused from system to (remote) system.
– VonC
Jul 9 '10 at 15:11
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Making button go full-width?
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For button groups you'll need to add btn-block to the btn-group wrapper as well.
– Jesse
Sep 23 '15 at 19:56
1
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How to match all occurrences of a regex
...ression to capture all matches using Regex#match and iterate over captured groups. Here you write a partial match function and want it applied mutiple times on a given string, this is not the responsibility of Regexp. I suggest you check the implementation of scan for a better understanding: ruby-do...
SQL Server query - Selecting COUNT(*) with DISTINCT
...rogram_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 COUNT(DISTINCT <expression>): evaluates expression for each row in a group and returns the number of unique, non-null v...
TSQL Pivot without aggregate function
... Min(Case DBColumnName When 'Date' Then Data End) Date
From table
Group By CustomerId
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Bootstrap 3 Collapse show state with Chevron icon
...min.js" type="text/javascript" ></script>
<div class="panel-group" id="accordion">
<div class="panel panel-default">
<div class="panel-heading">
<h4 class="panel-title">
<a class="accordion-toggle" data-toggle="collapse" data-parent="#ac...
Why does the order in which libraries are linked sometimes cause errors in GCC?
...ortant detail to this excellent answer: Using "-( archives -)" or "--start-group archives --end-group" is the only sure-fire way of resolving circular dependencies, since each time the linker visits an archive, it pulls in (and registers the unresolved symbols of) only the object files that resolve ...
How to remove all MySQL tables from the command-line without DROP database permissions? [duplicate]
...ts to execute it:
SET FOREIGN_KEY_CHECKS = 0;
SET @tables = NULL;
SELECT GROUP_CONCAT('`', table_schema, '`.`', table_name, '`') INTO @tables
FROM information_schema.tables
WHERE table_schema = 'database_name'; -- specify DB name here.
SET @tables = CONCAT('DROP TABLE ', @tables);
PREPARE st...
JOIN two SELECT statement results
...te], 0) AS [# Late]
FROM
(SELECT ks, COUNT(*) AS '# Tasks' FROM Table GROUP BY ks) t1
LEFT JOIN
(SELECT ks, COUNT(*) AS '# Late' FROM Table WHERE Age > Palt GROUP BY ks) t2
ON (t1.ks = t2.ks);
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SQL query for finding records where count > 1
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Use the HAVING clause and GROUP By the fields that make the row unique
The below will find
all users that have more than one payment per day with the same account number
SELECT
user_id ,
COUNT(*) count
FROM
PAYMENT
GROUP BY
account,
use...