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‎README.md

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**[⬆ Back to Top](#sql-coding-challenges-for-beginners)**
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## 37. Daily Leads and Partners
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The `DailySales` table contains the date and the name of products sold and the IDs of the leads and partners they were sold to. Write a SQL query that for each `date_id` and `make_name`, returns the number of distinct `lead_id`'s and distinct `partner_id`'s. Here is an example:
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```
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DailySales table:
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+-----------+-----------+---------+------------+
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| date_id | make_name | lead_id | partner_id |
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+-----------+-----------+---------+------------+
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| 2020年12月8日 | toyota | 0 | 1 |
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| 2020年12月8日 | toyota | 1 | 0 |
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| 2020年12月8日 | toyota | 1 | 2 |
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| 2020年12月7日 | toyota | 0 | 2 |
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| 2020年12月7日 | toyota | 0 | 1 |
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| 2020年12月8日 | honda | 1 | 2 |
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| 2020年12月8日 | honda | 2 | 1 |
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| 2020年12月7日 | honda | 0 | 1 |
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| 2020年12月7日 | honda | 1 | 2 |
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| 2020年12月7日 | honda | 2 | 1 |
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+-----------+-----------+---------+------------+
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```
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```
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Result table:
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+-----------+-----------+--------------+-----------------+
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| date_id | make_name | unique_leads | unique_partners |
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+-----------+-----------+--------------+-----------------+
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| 2020年12月8日 | toyota | 2 | 3 |
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| 2020年12月7日 | toyota | 1 | 2 |
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| 2020年12月8日 | honda | 2 | 2 |
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| 2020年12月7日 | honda | 3 | 2 |
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+-----------+-----------+--------------+-----------------+
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```
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<details><summary>Solution</summary>
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```sql
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SELECT
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date_id,
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make_name,
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COUNT(DISTINCT(lead_id)) AS unique_leads,
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COUNT(DISTINCT(partner_id)) AS unique_partners
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FROM DailySales
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GROUP BY date_id, make_name;
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```
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</details>
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---
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**[⬆ Back to Top](#sql-coding-challenges-for-beginners)**

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