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For coding answers &  Sql 🏃🏃🏃🏃
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SELECT
   DATE_FORMAT(date_field, '%Y-%m') AS month_year,
   'Jobs' AS source,
   COUNT(*) AS total_number
FROM
   jobs
GROUP BY
   month_year
UNION ALL
SELECT
   DATE_FORMAT(date_field, '%Y-%m') AS month_year,
   'Freelancers' AS source,
   COUNT(*) AS total_number
FROM
   freelancers
GROUP BY
   month_year
ORDER BY
   month_year ASC,
   source ASC;

Freelancer code @coding_000❤️
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Q - E-commerce website admin


SELECT c.title AS category_title, p.title, SUM(p.stock_number) AS total_stock
FROM categories c
JOIN products p ON c.id = p.category_id
GROUP BY c.title, p.title
HAVING SUM(p.stock_number) > 10
ORDER BY category_title ASC, p.title ASC, total_stock DESC;
🔥3
SELECT
   DATE_FORMAT(date_field, '%Y-%m') AS month_year,
   'Jobs' AS source,
   COUNT(*) AS total_number
FROM
   jobs
GROUP BY
   month_year
UNION ALL
SELECT
   DATE_FORMAT(date_field, '%Y-%m') AS month_year,
   'Freelancers' AS source,
   COUNT(*) AS total_number
FROM
   freelancers
GROUP BY
   month_year
ORDER BY
   month_year ASC,
   source ASC;

share @coding_000
👍3🔥1🥰1
Freelancer  code


SELECT
    DATE_FORMAT(DATE, '%Y-%m') AS month_year,
    source,
    SUM(total_number) AS total_number
FROM (
    SELECT
        DATE(created_at) AS DATE,
        'jobs' AS source,
        COUNT(*) AS total_number
    FROM JobPosts
    WHERE YEAR(created_at) = 2022
    GROUP BY DATE(created_at)
   
    UNION ALL
   
    SELECT
        DATE(created_at) AS DATE,
        'Freelancers' AS source,
        COUNT(*) AS total_number
    FROM Freelancers
    WHERE YEAR(created_at) = 2022
    GROUP BY DATE(created_at)
) AS t
GROUP BY
    DATE_FORMAT(DATE, '%Y-%m'),
    source
ORDER BY
    DATE_FORMAT(DATE, '%Y-%m'),
    source;




Freelancer code 👆👆👆👆

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Check above 👆👆 If ur unable to find the questions answer
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SELECT
    DATE_FORMAT(dt, '%Y-%m') AS month_year,
    'Jobs' AS source,
    COUNT(*) AS total_number
FROM jobs
WHERE YEAR(dt) = 2022
GROUP BY month_year

UNION ALL

-- For freelancer registrations
SELECT
    DATE_FORMAT(dt, '%Y-%m') AS month_year,
    'Freelancers' AS source,
    COUNT(*) AS total_number
FROM freelancers
WHERE YEAR(dt) = 2022
GROUP BY month_year

ORDER BY month_year, source;


leading freelancer code
🔥2