2 回答

TA貢獻(xiàn)1818條經(jīng)驗(yàn) 獲得超3個(gè)贊
感謝您的評(píng)論和回答。Severius5,您的回答(現(xiàn)在已刪除?)給了我答案。這段代碼做我需要的:
var rules = NLog.LogManager.Configuration.LoggingRules;
var rule = rules.FirstOrDefault(x => x.LoggerNamePattern == "hostLogger");
NLog.LogManager.GetLogger("hostLogger").Info("Minimum Log Level: {0}", rule.Levels.Min());
它在日志中給了我這個(gè)輸出:
2018-08-03 14:26:11.7439 | 信息 | 最低日志級(jí)別:調(diào)試

TA貢獻(xiàn)1836條經(jīng)驗(yàn) 獲得超3個(gè)贊
也許是這樣的:
var logger = NLog.LogManager.GetLogger("hostLogger");
var minLogLevel = "Disabled";
if (logger.IsTraceEnabled)
minLogLevel = "Trace";
else if (logger.IsDebugEnabled)
minLogLevel = "Debug";
else if (logger.IsInfoEnabled)
minLogLevel = "Info";
else if (logger.IsWarnEnabled)
minLogLevel = "Warn";
else if (logger.IsErrorEnabled)
minLogLevel = "Error";
else if (logger.IsFatalEnabled)
minLogLevel = "Fatal";
logger.Info("Minimum Log Level: {0}", minLogLevel)
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