我知道RSACryptoServiceProvider可以使用公钥加密,然后可以使用私钥解密.
是否可以使用私钥加密并使用RSACryptoServiceProvider使用公钥解密?
使用String
类如下面的方法从本地变量构建字符串是否安全?假设从多个线程调用下面的方法.
public static string WriteResult(int value, string name)
{
return string.Format("Result: value={0} name={1}", value, name);
}
public static string WriteResult2(int value, string name)
{
return "Result: value=" + value + " name=" + name;
}
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或者我需要使用StringBuilder
以确保线程安全吗?
我正在尝试使用几个ASP.NET Web API 2.0端点进行基准测试(使用Apache工作台).其中一个是同步和一个异步.
[Route("user/{userId}/feeds")]
[HttpGet]
public IEnumerable<NewsFeedItem> GetNewsFeedItemsForUser(string userId)
{
return _newsFeedService.GetNewsFeedItemsForUser(userId);
}
[Route("user/{userId}/feeds/async")]
[HttpGet]
public async Task<IEnumerable<NewsFeedItem>> GetNewsFeedItemsForUserAsync(string userId)
{
return await Task.Run(() => _newsFeedService.GetNewsFeedItemsForUser(userId));
}
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在观看了Steve Sanderson的演讲之后,我ab -n 100 -c 10 http://localhost....
向每个端点发出了以下命令.
我很惊讶,因为每个端点的基准似乎大致相同.
关闭史蒂夫解释说我期待异步端点更高效,因为它会立即将线程池线程释放回线程池,从而使它们可用于其他请求并提高吞吐量.但数字看起来完全一样.
我在这里误解了什么?
c# asp.net task-parallel-library async-await asp.net-web-api
我在MSDN上看过:
null关键字是表示空引用的文字,不引用任何对象.
但我看到以下代码运行时没有抛出任何异常:
int? i = null;
var s = i.ToString();
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因此,如果变量i
为null,为什么我可以执行它的方法?
这是我的堆栈跟踪:
Exception 'yii\db\Exception' with message 'could not find driver'
in /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/Connection.php:534
#0 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/Connection.php(836): yii\db\Connection->open()
#1 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/Connection.php(823): yii\db\Connection->getMasterPdo()
#2 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/Command.php(208): yii\db\Connection->getSlavePdo()
#3 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/Command.php(816): yii\db\Command->prepare(true)
#4 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/Command.php(350): yii\db\Command->queryInternal('fetchAll', NULL)
#5 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/mysql/Schema.php(198): yii\db\Command->queryAll()
#6 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/mysql/Schema.php(97): yii\db\mysql\Schema->findColumns(Object(yii\db\TableSchema))
#7 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/db/Schema.php(140): yii\db\mysql\Schema->loadTableSchema('migration')
#8 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/console/controllers/MigrateController.php(121): yii\db\Schema->getTableSchema('{{%migration}}', true)
#9 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/console/controllers/BaseMigrateController.php(610): yii\console\controllers\MigrateController->getMigrationHistory(NULL)
#10 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/console/controllers/BaseMigrateController.php(102): yii\console\controllers\BaseMigrateController->getNewMigrations()
#11 [internal function]: yii\console\controllers\BaseMigrateController->actionUp(0)
#12 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/base/InlineAction.php(55): call_user_func_array(Array, Array)
#13 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/base/Controller.php(151): yii\base\InlineAction->runWithParams(Array)
#14 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/console/Controller.php(91): yii\base\Controller->runAction('', Array)
#15 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/base/Module.php(455): yii\console\Controller->runAction('', Array)
#16 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/console/Application.php(161): yii\base\Module->runAction('migrate', Array)
#17 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/console/Application.php(137): yii\console\Application->runAction('migrate', Array)
#18 /opt/lampp/htdocs/advanced/vendor/yiisoft/yii2/base/Application.php(375): yii\console\Application->handleRequest(Object(yii\console\Request))
#19 /opt/lampp/htdocs/advanced/yii(31): yii\base\Application->run()
#20 …
Run Code Online (Sandbox Code Playgroud) 我有一个数据表和一行.我想将行导入数据表,只要数据表中不存在该行.
我怎样才能做到这一点?
当我创建一个SelecList时,我希望能够手动添加SelecListItem并执行此操作,我使用以下代码:
List<SelectListItem> Provinces = new List<SelectListItem>();
Provinces.Add(new SelectListItem() { Text = "Northern Cape", Value = "NC" });
Provinces.Add(new SelectListItem() { Text = "Free State", Value = "FS" });
Provinces.Add(new SelectListItem() { Text = "Western Cape", Value = "WC" });
SelectList lstProvinces = new SelectList(Provinces);
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而不是这个:
var lstProvinces = new SelectList(new[] { "Northern Cape", "Free State", "Western Cape" });
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在我创建SelectList之后,我通过ViewBag将它传递给DropDownListFor:
Html.DropDownListFor(m => m.StartPointProvince, (SelectList)ViewBag.Provinces)
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但是,当我使用第一种方法创建SelectList时,它不起作用 - 它将3个值添加到下拉列表中,但所有值显示为:
*输出截图
但是,当我使用第二种方法时,它工作正常.我希望使用第一种方法,因为我希望能够指定每个项目的Text AND值.
假设一个具有多个并发生成器的系统,每个生成器都努力使用以下可通过其名称唯一标识的公共实体来持久保存一些对象图:
CREATE TABLE CommonEntityGroup(
Id INT NOT NULL IDENTITY(1, 1) PRIMARY KEY,
Name NVARCHAR(100) NOT NULL
);
GO
CREATE UNIQUE INDEX IX_CommonEntityGroup_Name
ON CommonEntityGroup(Name)
GO
CREATE TABLE CommonEntity(
Id INT NOT NULL IDENTITY(1, 1) PRIMARY KEY,
Name NVARCHAR(100) NOT NULL,
CommonEntityGroupId INT NOT NULL,
CONSTRAINT FK_CommonEntity_CommonEntityGroup FOREIGN KEY(CommonEntityGroupId)
REFERENCES CommonEntityGroup(Id)
);
GO
CREATE UNIQUE INDEX IX_CommonEntity_CommonEntityGroupId_Name
ON CommonEntity(CommonEntityGroupId, Name)
GO
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例如,生产者A保存了一些CommonEntityMeeting
s,而生产者B保存了CommonEntitySet
s.他们中的任何一个都必须坚持CommonEntity
与他们的特定物品相关.
基本上,关键点是:
c# sql-server entity-framework optimistic-concurrency sql-insert
我试图找到一种方法来确定git存储库中的文件是否由git-lfs正确管理.
到目前为止,我发现比较结果git lfs ls-files
并且git ls-files ':(attr:filter=lfs)'
似乎给了我想要的东西.
Add-Type -AssemblyName 'System.Linq';
[string[]] $actualLfsFilePaths = git lfs ls-files | `
ForEach-Object `
{
#'12345678 * my.dll' - not all lfs versions support -n flag, so it is better to be conservative
$_.Split(' ', 3)[2]
};
[string[]] $shouldBeUnderLfsFilePaths = git ls-files ':(attr:filter=lfs)';
$filePathsNotUnderLfs = [System.Linq.Enumerable]::ToArray(
[System.Linq.Enumerable]::Except($shouldBeUnderLfsFilePaths, $actualLfsFilePaths)
);
$count = $filePathsNotUnderLfs.Count;
Write-Host "There are $($count) lfs files that are not under lfs";
foreach ($filePathNotUnderLfs in $filePathsNotUnderLfs)
{
Write-Host "`t'$filePathNotUnderLfs'";
} …
Run Code Online (Sandbox Code Playgroud) 我目前在开发服务器上设置了 Magento(远程,SSH 访问)。我想了解 magento 根目录中的“法师”文件是什么?它是如何工作的?
c# ×6
asp.net ×2
php ×2
.net ×1
async-await ×1
cryptography ×1
datarow ×1
datatable ×1
git ×1
git-lfs ×1
git-ls-files ×1
magento ×1
mysql ×1
nullable ×1
pdo ×1
powershell ×1
rsa ×1
selectlist ×1
sql-insert ×1
sql-server ×1
tfs ×1
yii2 ×1