refactor(topics): 멀티 에이전트용 지식 재편 — _Common(공통 기본기) + Domain_* 구조
에이전트 8종(대화형/프로그래머 C·S/디자이너/설계자/기획자/QA/PD/PM)에게 [공통 기본 능력 + 롤별 Specialty] 2층으로 지식을 주입하기 위한 재분류. 문서 내용·포맷은 무수정, 폴더 이동만 (6,372개 문서 수 보존 확인). - Topic_Programming → Domain_Programming (내부 구조 보존) - Topic_Graphic → Domain_Design - Topic_Business → Domain_Product - Topic_General → Domain_General - _Common 신설: Math(구 Topic_Math_Specialty), Reasoning(구 General/From_Thinking & Reasoning), Reasoning_Creativity(구 General/From_창의성), Communication(Poetic_Blog_Writing + From_writing) - 타 도메인의 From_* 폴더는 유지 (출처 표기일 뿐, 이미 도메인에 맞게 분류된 문서) - 빈 폴더 정리 (memory/procedures) - 에이전트→폴더 매핑은 workspace의 .astra/agent-knowledge-map.json (9개 에이전트) Co-Authored-By: Claude Fable 5 <noreply@anthropic.com>
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id: sql-having
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title: "SQL Having"
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category: "Database"
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status: "draft"
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verification_status: "conceptual"
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canonical_id: ""
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aliases: ["SQL HAVING Clause", "HAVING clause", "SQL 그룹 조건절"]
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duplicate_of: ""
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source_trust_level: "B"
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confidence_score: 0.9
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created_at: 2026-07-04
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updated_at: 2026-07-04
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review_reason: ""
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merge_history: []
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tags: ["sql", "database", "w3schools", "having", "groupby", "aggregate"]
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raw_sources: ["https://www.w3schools.com/sql/sql_having.asp"]
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applied_in: []
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github_commit: ""
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---
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# [[SQL Having]]
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## 🎯 한 줄 통찰 (One-line insight)
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HAVING filters GROUP BY results based on aggregate conditions, applying AFTER grouping — unlike WHERE, which filters individual rows BEFORE grouping. [S1]
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## 🧠 핵심 개념 (Core concepts)
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- **HAVING clause** — filters the results of a GROUP BY query based on aggregate functions. [S1]
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- **WHERE vs. HAVING** — WHERE filters individual rows before grouping; HAVING filters groups after aggregation. [S1]
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- **Syntax** — `SELECT column1, aggregate_function(column2), column3, ... FROM table_name WHERE condition GROUP BY column1, column3 HAVING condition ORDER BY column_name;`. [S1]
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## 🧩 추출된 패턴 (Extracted patterns)
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- **WHERE + GROUP BY + HAVING composition** — a query can filter rows first (WHERE), group them, then filter the resulting groups (HAVING) — e.g. restrict to two named employees, group by last name, then keep only groups with more than 25 orders. [S1]
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## 📖 세부 내용 (Details)
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- Countries with more than 5 customers: `SELECT Country, COUNT(CustomerID) AS [Number of Customers] FROM Customers GROUP BY Country HAVING COUNT(CustomerID) > 5;`. [S1]
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- Same, sorted descending: `... GROUP BY Country HAVING COUNT(CustomerID) > 5 ORDER BY COUNT(CustomerID) DESC;`. [S1]
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- Employees with more than 10 orders, via JOIN + GROUP BY + HAVING: `SELECT Employees.LastName, COUNT(Orders.OrderID) AS NumberOfOrders FROM Orders INNER JOIN Employees ON Orders.EmployeeID = Employees.EmployeeID GROUP BY LastName HAVING COUNT(Orders.OrderID) > 10;`. [S1]
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- Combined WHERE + GROUP BY + HAVING: filter to two named employees first, then keep only those with more than 25 orders. [S1]
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## ⚖️ 모순 및 업데이트 (Contradictions & updates)
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소스에서 모순되는 정보는 발견되지 않음.
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## 🛠️ 적용 사례 (Applied in summary)
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현재 발견된 실제 적용 사례가 없습니다 — "N건 이상인 그룹만 보기" 같은 리포팅 쿼리에서 GROUP BY와 항상 짝을 이뤄 쓰인다. [S1]
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## 💻 코드 패턴 (Code patterns)
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Filter groups by aggregate condition (SQL):
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```sql
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SELECT Country, COUNT(CustomerID) AS [Number of Customers]
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FROM Customers
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GROUP BY Country
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HAVING COUNT(CustomerID) > 5;
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```
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## ✅ 검증 상태 및 신뢰도
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- **상태:** draft
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- **검증 단계:** conceptual
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- **출처 신뢰도:** B (W3Schools — widely used educational reference, not a primary standards body)
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- **신뢰 점수:** 0.90
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- **중복 검사 결과:** 신규 생성 (New discovery)
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## 🔗 지식 그래프 (Knowledge Graph)
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- **상위/루트:** [[SQL Tutorial]]
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- **관련 개념:** [[SQL Group By]], [[SQL Where]], [[SQL Aggregate Functions]]
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- **참조 맥락:** GROUP BY 결과 자체를 조건으로 걸러야 할 때 사용 — WHERE와 역할을 혼동하지 않는 것이 핵심.
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## 📚 출처 (Sources)
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- [S1] W3Schools — SQL HAVING Clause — https://www.w3schools.com/sql/sql_having.asp
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## 📝 변경 이력 (Change history)
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- 2026-07-04: Initial draft synthesized from the W3Schools "SQL HAVING Clause" page (Astra wiki-curation, P-Reinforce v3.1 format).
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