使用正则表达式从 Grafana 表达式中检索 Prometheus 指标名称

Question

我尝试了许多不同的regex 模式来获得它，但不是很成功。

这个问题的模式：

<method_name(> metric_name <{filter_condition}> <[time_duration]> <)> <by (some members)>
            ^------------------------------------------------------^
                          method_name(...) can be multiple

如您所见，<...> 可以是可选的，而metric_name 是必须的，我想从这个equation 中检索它。

Case # 1
input: sum(log_search_by_service_total {service_name!~\"\"}) by (service_name, operator)
output: log_search_by_service_total

Case # 2
input: log_request_total
output: log_request_total

Case # 3
input:  sum(delta(log_request_total[5m])) by (args, user_id)
output: log_request_total

Case # 4
input: log_request_total{methodName=~\"getAppDynamicsGraphMetrics|getAppDynamicsMetrics\"}
output: log_request_total

Case # 5
input: sum(delta(log_request_total{className=~\".*ProductDashboardController\",methodName=~\"getDashboardConfig|updateMaintainers|addQuickLink|deleteQuickLink|addDependentMiddleware|addDependentService|updateErrorThreshold\"}[5m])) by (user_id)"
output: log_request_total

Case # 6
input: count_scalar(sum(log_query_request_total) by (user_id))
output: log_query_request_total

这是我在 Java 中尝试过的演示。但似乎我无法获得正确的pattern 来检索我上面提到的模式的确切答案。

如果可能，请分享一些想法。

public static void main(String... args) {
    String[] exprs = {"sum(log_query_task_cache_hit_rate_bucket)by(le)",
            "sum(log_search_by_service_total {service_name!~\"\"}) by (service_name, operator)",
            "log_request_total",
            " sum(delta(log_request_total[5m])) by (args, user_id)",
            "log_request_total{methodName=~\"getAppDynamicsGraphMetrics|getAppDynamicsMetrics\"}",
            "sum(delta(log_request_total{className=~\".*ProductDashboardController\",methodName=~\"getDashboardConfig|updateMaintainers|addQuickLink|deleteQuickLink|addDependentMiddleware|addDependentService|updateErrorThreshold\"}[5m])) by (user_id)",
            "sum(log_request_total{methodName=\"getInstanceNames\"}) by (user_id)",
            "sum(log_request_total{methodName=\"getVpcCardInfo\",user_id!~\"${user}\"}) by (envName)",
            "count_scalar(sum(log_query_request_total) by (user_id))",
            "avg(log_waiting_time_average) by (exported_tenant, exported_landscape)",
            "avg(task_processing_time_average{app=\"athena\"})",
            "avg(log_queue_time_average) by (log_type)",
            "sum(delta(product_dashboard_service_sum[2m]))",
            "ceil(delta(product_dashboard_service_count[5m]))]"
    };
    String[] expected = {
            "log_query_task_cache_hit_rate_bucket",
            "log_search_by_service_total",
            "log_request_total",
            "log_request_total",
            "log_request_total",
            "log_request_total",
            "log_request_total",
            "log_request_total",
            "log_query_request_total",
            "log_waiting_time_average",
            "task_processing_time_average",
            "log_queue_time_average",
            "product_dashboard_service_sum",
            "product_dashboard_service_count"
    };
    Pattern pattern = Pattern.compile(".*?\\(?([\\w|_]+)\\{?\\[?.*");
    testPattern(exprs, expected, pattern);
    pattern = Pattern.compile(".*\\(?([\\w|_]+)\\{?\\[?.*");
    testPattern(exprs, expected, pattern);
    pattern = Pattern.compile(".*?\\(?([\\w|_]+)\\{?\\[?.*");
    testPattern(exprs, expected, pattern);
}

private static void testPattern(String[] exprs, String[] expected, Pattern pattern) {
    System.out.println("\n********** Pattern Match Test *********\n");
    for (int i = 0; i < exprs.length; ++i) {
        String expr = exprs[i];
        Matcher matcher = pattern.matcher(expr);
        if (matcher.find()) {
            System.out.println("\nThe Original Expr: " + expr);
            System.out.println(String.format("Expected:\t %-40s Matched:\t %-40s", expected[i], matcher.group(1)));
        } else {
            System.out.println("expected: " + expected[i] + " not matched");
        }
    }
}

---

更新 2018-08-06

感谢 Bohemian 的帮助，它真的启发了我（因为我一直相信 regex 可以用干净的解决方案变魔术）。

后来，我发现exprs 比我预想的要复杂，如下所示：

Case # 7
input: topk(10,autoindex_online_consume_time_total_sum{app=~"$app", DTO_Name=~"$c_class"})
expected: autoindex_online_consume_time_total_sum
// to get the metric name: autoindex_online_consume_time_total_sum
// still I can make it work with small modifications as ^(?:\w+\()*(?:\d+,)*(\w+)

但以下一种甚至更多不同的复杂组合让我转向了可靠的方法：

Case # 8
input: sum(hue_mail_sent_attachment_bytes_total) by (app)  / sum(hue_mail_sent_mails_with_attachment_total) by (app)
Expected: [hue_mail_sent_attachment_bytes_total, hue_mail_sent_mails_with_attachment_total]

现在要复杂得多...甚至不可预测，因为无法控制来自用户的expr 输入。

所以我通过更可靠和简单的解决方案实现了相同的目标：

1. 首先将 distinct 指标名称存储到数据库中；
2. 当expr 出现时，使用contains(String s) 在内存中检查它；
3. 仍然可能存在问题：如果某些指标名称包含其他指标名称，则会出现过度匹配；

Answer 1

此正则表达式捕获您在第 1 组中的目标

^(?:\w+\()*(\w+)

见live demo。

在java中，获取你的目标：

String metricName = input.replaceAll("^(?:\\w+\\()*(\\w+)", "$1");