我检查了DocumentDB上的MSDN for .Net(here)并找到了3个有效的构造函数。但是他们都没有使用连接字符串,这听起来很奇怪。
是否真的没有办法用连接字符串而不是端点+ authKey组合来实例化客户端,或者我错过了什么?
例如,大多数其他Microsoft服务都使用此概念,即https://docs.microsoft.com/en-us/azure/storage/storage-configure-connection-string#parsing-a-connection-string。 在我们的例子中,如果所有Azure相关的东西都以相同的方式初始化,那将是超级的。只是更清洁,而不是停止显示。
P.S。 请停止告诉我有关Uri和authKey参数的现有构造函数,问题(略有)不同。我可以按照自己提供的链接,无需帮助。感谢。
答案 0 :(得分:8)
我创建了一个用于解析连接字符串的类,类似于CloudStorageAccount.Parse的工作方式。我试图尽可能地遵循他们的模式,以防他们决定开源它,这可能有希望在没有太大变化的情况下做出贡献。
public static class DocumentDbAccount
{
public static DocumentClient Parse(string connectionString)
{
DocumentClient ret;
if (String.IsNullOrWhiteSpace(connectionString))
{
throw new ArgumentException("Connection string cannot be empty.");
}
if(ParseImpl(connectionString, out ret, err => { throw new FormatException(err); }))
{
return ret;
}
throw new ArgumentException($"Connection string was not able to be parsed into a document client.");
}
public static bool TryParse(string connectionString, out DocumentClient documentClient)
{
if (String.IsNullOrWhiteSpace(connectionString))
{
documentClient = null;
return false;
}
try
{
return ParseImpl(connectionString, out documentClient, err => { });
}
catch (Exception)
{
documentClient = null;
return false;
}
}
private const string AccountEndpointKey = "AccountEndpoint";
private const string AccountKeyKey = "AccountKey";
private static readonly HashSet<string> RequireSettings = new HashSet<string>(new [] { AccountEndpointKey, AccountKeyKey }, StringComparer.OrdinalIgnoreCase);
internal static bool ParseImpl(string connectionString, out DocumentClient documentClient, Action<string> error)
{
IDictionary<string, string> settings = ParseStringIntoSettings(connectionString, error);
if (settings == null)
{
documentClient = null;
return false;
}
if (!RequireSettings.IsSubsetOf(settings.Keys))
{
documentClient = null;
return false;
}
documentClient = new DocumentClient(new Uri(settings[AccountEndpointKey]), settings[AccountKeyKey]);
return true;
}
/// <summary>
/// Tokenizes input and stores name value pairs.
/// </summary>
/// <param name="connectionString">The string to parse.</param>
/// <param name="error">Error reporting delegate.</param>
/// <returns>Tokenized collection.</returns>
private static IDictionary<string, string> ParseStringIntoSettings(string connectionString, Action<string> error)
{
IDictionary<string, string> settings = new Dictionary<string, string>();
string[] splitted = connectionString.Split(new char[] { ';' }, StringSplitOptions.RemoveEmptyEntries);
foreach (string nameValue in splitted)
{
string[] splittedNameValue = nameValue.Split(new char[] { '=' }, 2);
if (splittedNameValue.Length != 2)
{
error("Settings must be of the form \"name=value\".");
return null;
}
if (settings.ContainsKey(splittedNameValue[0]))
{
error(string.Format(CultureInfo.InvariantCulture, "Duplicate setting '{0}' found.", splittedNameValue[0]));
return null;
}
settings.Add(splittedNameValue[0], splittedNameValue[1]);
}
return settings;
}
}
答案 1 :(得分:6)
DocumentDB SDK没有使用连接字符串的构造函数重载。它们支持使用端点+主密钥和端点+权限/资源令牌进行初始化。
如果您想查看单个连接字符串参数,请在此处提出/ upvote:https://feedback.azure.com/forums/263030-documentdb
答案 2 :(得分:6)
实际上,您可以通过环形交叉路口来做到这一点。
internal class CosmosDBConnectionString
{
public CosmosDBConnectionString(string connectionString)
{
// Use this generic builder to parse the connection string
DbConnectionStringBuilder builder = new DbConnectionStringBuilder
{
ConnectionString = connectionString
};
if (builder.TryGetValue("AccountKey", out object key))
{
AuthKey = key.ToString();
}
if (builder.TryGetValue("AccountEndpoint", out object uri))
{
ServiceEndpoint = new Uri(uri.ToString());
}
}
public Uri ServiceEndpoint { get; set; }
public string AuthKey { get; set; }
}
然后
var cosmosDBConnectionString = new CosmosDBConnectionString(connectionString)
var client = new DocumentClient(
cosmosDBConnectionString.ServiceEndpoint,
cosmosDBConnectionString.AuthKey)
这取自Azure WebJobs扩展SDK,这是Azure Functions V2能够仅使用连接字符串的方式。省去了尝试自己解析字符串的麻烦。
答案 3 :(得分:2)
对我有用:
#include <stdio.h>
#include <string.h>
void login(){
char username[100];
char pass[100];
int i = 0;
while( i < 3){
printf("Enter the username : ");
scanf("%s", &username);
printf("Password : ");
scanf("%s", &pass);
printf("%s %s \n", username, pass);
if ((strcmp(username, "a")==0) && (strcmp(pass, "a")==0)){
printf("You are sucessfull login into your account");
BilaDahLogin();
break;
}
else{
printf("The username or password incorrect\nTry again\n");
i += 1;
}
}
}