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Prevent multiple instances of a given app in .NET?
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151
Use Mutex. One of the examples above using GetProcessByName has many caveats. Here is a good a...
Which data type for latitude and longitude?
...and PostGIS. I want to store latitude and longitude values in PostgreSQL 9.1.1 database table. I will calculate distance between two points, find nearer points by using this location values.
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linux svn搭建配置及svn命令详解 - 更多技术 - 清泛网 - 专注C/C++及内核技术
...搞定。
启动SVN服务：svnserve -d -r /opt/svn/repos/ --listen-host 127.0.0.1 （注：不指定端口号，默认为3690）
1、安装
[root@www ~]# yum install subversion
[root@www ~]# svn -v 判断是否安装成功
svnserve, version 1.6.11 (r934486) 出...
Database design for audit logging
...IMARY KEY,
Name nvarchar(200) NOT NULL,
CreatedByName nvarchar(100) NOT NULL,
CurrentRevision int NOT NULL,
CreatedDateTime datetime NOT NULL
And the contents:
CREATE TABLE dbo.PageContent(
PageID int NOT NULL,
Revision int NOT NULL,
Title nvarchar(200) NOT NULL,...
REST URI convention - Singular or plural name of resource while creating it
...ll likely error, as this request doesn't make any sense, whereas /resource/123 makes perfect sense.
Using /resource instead of /resources is similar to how you would do this if you were working with, say, a file system and a collection of files and /resource is the "directory" with the individual 1...
In CMake, how can I test if the compiler is Clang?
...or AppleClang
endif()
Also see the AppleClang policy description.
CMake 3.15 has added support for both the clang-cl and the regular clang front end. You can determine the front end variant by inspecting the variable CMAKE_CXX_COMPILER_FRONTEND_VARIANT:
if (CMAKE_CXX_COMPILER_ID STREQUAL "Clang")
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Converting List to List
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edited Apr 15 at 18:23
Solubris
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How can I prevent the backspace key from navigating back?
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linux 下巧妙使用squid代理服务器 - 更多技术 - 清泛网 - 专注C/C++及内核技术
...，加强局域网的安全性等。它的主要作用有以下几点。
1．共享网络
2．加快访问速度，节约通信带宽
3．防止内部主机受到攻击
4．限制用户访问，完善网络管理
原理：
① 客户端A向代理服务器提出访问Internet的请求。
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SQL join: selecting the last records in a one-to-many relationship
...n StackOverflow.
Here's how I usually recommend solving it:
SELECT c.*, p1.*
FROM customer c
JOIN purchase p1 ON (c.id = p1.customer_id)
LEFT OUTER JOIN purchase p2 ON (c.id = p2.customer_id AND
(p1.date < p2.date OR (p1.date = p2.date AND p1.id < p2.id)))
WHERE p2.id IS NULL;
Explana...