ConfigurationHelperService.GetUserConfigurationPath 方法

定义

基于不同项目属性的值获取当前项目的最可能的用户配置文件路径。

public:
 System::String ^ GetUserConfigurationPath(IServiceProvider ^ provider, EnvDTE::Project ^ project, System::Configuration::ConfigurationUserLevel userLevel, bool underHostingProcess, EnvDTE::Configuration ^ buildConfiguration);
public string GetUserConfigurationPath (IServiceProvider provider, EnvDTE.Project project, System.Configuration.ConfigurationUserLevel userLevel, bool underHostingProcess, EnvDTE.Configuration buildConfiguration);
member this.GetUserConfigurationPath : IServiceProvider * EnvDTE.Project * System.Configuration.ConfigurationUserLevel * bool * EnvDTE.Configuration -> string
Public Function GetUserConfigurationPath (provider As IServiceProvider, project As Project, userLevel As ConfigurationUserLevel, underHostingProcess As Boolean, buildConfiguration As Configuration) As String

参数

project
Project

当前的 Project

underHostingProcess
Boolean

true 如果在托管进程中,则为; 否则为 false

buildConfiguration
Configuration

EnvDTE.BuildConfiguration.

返回

String

用户配置路径,或者, null 如果无法从项目中的信息构造有效路径,则为。

适用于