s.esmaeiliandClaude Opus 5.5
92831ed33e
DATA-MIGRATION-MAP.md: - §10.7 is rewritten from the full dump loaded into MySQL. The legacy debt formula is confirmed (1,590/1,591 balances). The earlier claims that new_city is a cities5 id, that birthdays are clean and that types 3/4/5 are unknown were wrong and are corrected in a table. Favourites come from major_favorites, not fav. The role-number clash (legacy 86 «حسابدار» vs target 86 «کارشناس فروش») is recorded. - §10.8 adds the tool's usage, safety rules, test result, and what it deliberately does not migrate. - §11 and §5.5 are brought up to date. The rejected count is 7, not 14. migration-precheck-3.sql: two bugs that would have failed on the server are fixed: the reserved word `both` and a collation clash between cities.title and customers.new_city. Co-Authored-By: Claude Opus 5.5 <noreply@anthropic.com>
66 lines
3.8 KiB
SQL
66 lines
3.8 KiB
SQL
-- ============================================================================
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-- پیش‌بررسی سوم — فقط دو سؤال باقی‌مانده
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-- ============================================================================
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--
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-- ⚠️ فقط SELECT COUNT. صفر INSERT/UPDATE/DELETE. هیچ ستون متنی در خروجی.
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--
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-- اجرا:
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-- mysql -u USER -p -h HOST -P 3305 oapi < migration-precheck-3.sql > out3.txt
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--
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-- مرجع: docs/DATA-MIGRATION-MAP.md §۵.۹
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-- ============================================================================
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-- ---------------------------------------------------------------------------
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-- سؤال: new_city باید به cityId نگاشت شود یا ستون جدا بگیرد؟
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--
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-- تصمیم مالک: «یا در فیلد شهر بنشیند یا در یک فیلد آدرس متنی مپ شود».
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-- کدام‌یک، به این بستگی دارد که آیا این ۱۷۶ ردیف city_id هم دارند یا نه:
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--
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-- فقط new_city (city_id تهی) → ستون متنی لازم است، چون شهر در جدول cities نیست
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-- هر دو را دارند → new_city تکراری است، می‌توان رهایش کرد
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-- ---------------------------------------------------------------------------
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SELECT '=== new_city vs city_id ===' AS section;
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SELECT
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SUM(new_city IS NOT NULL AND new_city <> '' AND city_id IS NULL) AS only_new_city,
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SUM(new_city IS NOT NULL AND new_city <> '' AND city_id IS NOT NULL) AS both_set,
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SUM((new_city IS NULL OR new_city = '') AND city_id IS NOT NULL) AS only_city_id,
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SUM((new_city IS NULL OR new_city = '') AND city_id IS NULL) AS neither
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FROM customers WHERE deleted_at IS NULL;
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-- از آن ۱۷۶ ردیف، چندتا متنی‌شان با عنوان یک شهر موجود در `cities` یکی است؟
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-- اگر بالا باشد، اکثرشان قابل تبدیل به city_id هستند و ستون متنی لازم نیست.
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SELECT 'new_city matches an existing cities.title' AS metric, COUNT(*) AS n
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FROM customers c
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-- The two columns have different collations (unicode_ci vs general_ci), so
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-- both sides are pinned to one; without it MySQL refuses the comparison.
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JOIN cities ci ON TRIM(ci.title) COLLATE utf8mb4_general_ci = TRIM(c.new_city) COLLATE utf8mb4_general_ci
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WHERE c.deleted_at IS NULL AND c.new_city IS NOT NULL AND c.new_city <> ''
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AND c.city_id IS NULL;
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-- چند مقدار متمایز؟ (عدد است، نه متن — اگر کم باشد نگاشت دستی ممکن است)
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SELECT 'distinct_new_city_values' AS metric, COUNT(DISTINCT TRIM(new_city)) AS n
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FROM customers WHERE deleted_at IS NULL AND new_city IS NOT NULL AND new_city <> '';
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-- ---------------------------------------------------------------------------
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-- سؤال: birth_date — آیا تاریخ‌ها معتبرند؟
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--
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-- ستون timestamp است. ولی اگر تاریخ جلالی به‌اشتباه به‌عنوان میلادی
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-- ذخیره شده باشد، سال‌ها ۱۳xx می‌شوند که در timestamp جا نمی‌شوند،
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-- یا به مقادیر عجیب تبدیل شده‌اند.
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-- ---------------------------------------------------------------------------
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SELECT '=== birth_date sanity ===' AS section;
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SELECT
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COUNT(birth_date) AS total_set,
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SUM(YEAR(birth_date) < 1900) AS year_before_1900,
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SUM(YEAR(birth_date) BETWEEN 1900 AND 1950) AS year_1900_1950,
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SUM(YEAR(birth_date) BETWEEN 1951 AND 2010) AS year_1951_2010,
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SUM(YEAR(birth_date) > 2010) AS year_after_2010,
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CAST(MIN(YEAR(birth_date)) AS CHAR) AS min_year,
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CAST(MAX(YEAR(birth_date)) AS CHAR) AS max_year
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FROM customers WHERE deleted_at IS NULL;
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SELECT '=== END ===' AS section;
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