Термины и тесты 🙂↔️По рекламе писать в чат канала
⚡️админ @bushidoVi ⚡️
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Post #627
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Словарь тестировщика | Тестирование trace_id:"*" AND span_status:"error"
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Словарь тестировщика | Тестирование trace_id:"*" AND span_status:"error"
trace_id:"*" AND span_status:"error"
Словарь тестировщика | Тестирование {service="auth"} |= "timeout" | json | line_format "{{.user_id}}"
{service="auth"} |= "timeout" | json | line_format "{{.user_id}}"Словарь тестировщика | Тестирование sum by (service) (rate({level="error"}[5m]))
sum by (service) (rate({level="error"}[5m]))Словарь тестировщика | Тестирование SELECT user_id, COUNT(*) AS errors FROM logs WHERE level = 'ERROR' AND ts >= NOW() - INTERVAL '3 DAY' GROUP BY user_id HAVING COUNT(*) > ( SELECT AVG(cnt) FROM ( SELECT user_id, COUNT(*) AS cnt FROM logs WHERE level = 'ERROR' GROUP BY…
SELECT user_id, COUNT(*) AS errors
FROM logs
WHERE level = 'ERROR'
AND ts >= NOW() - INTERVAL '3 DAY'
GROUP BY user_id
HAVING COUNT(*) > (
SELECT AVG(cnt)
FROM (
SELECT user_id, COUNT(*) AS cnt
FROM logs
WHERE level = 'ERROR'
GROUP BY user_id
) t
);
Словарь тестировщика | Тестирование SELECT id, ts, LEAD(ts) OVER (ORDER BY ts) AS next_ts FROM events;
SELECT id, ts,
LEAD(ts) OVER (ORDER BY ts) AS next_ts
FROM events;
Словарь тестировщика | Тестирование SELECT email FROM users GROUP BY email HAVING COUNT(*) > 1;
SELECT email
FROM users
GROUP BY email
HAVING COUNT(*) > 1;
Словарь тестировщика | Тестирование SELECT AVG(score) FROM reviews WHERE product_id = 15;
SELECT AVG(score) FROM reviews WHERE product_id = 15;