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Commit 4b52e58

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feat: update sql solution to lc problem: No.1294 (doocs#1811)
1 parent 2d73cb2 commit 4b52e58

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3 files changed

+15
-16
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3 files changed

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‎solution/1200-1299/1294.Weather Type in Each Country/README.md‎

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Original file line numberDiff line numberDiff line change
@@ -122,15 +122,15 @@ Morocco 11 月的平均 weather_state 为 (25 + 27 + 31) / 3 = 27.667 所以天
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SELECT
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country_name,
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CASE
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WHEN avg(weather_state) <= 15 THEN 'Cold'
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WHEN avg(weather_state) >= 25 THEN 'Hot'
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WHEN AVG(weather_state) <= 15 THEN 'Cold'
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WHEN AVG(weather_state) >= 25 THEN 'Hot'
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ELSE 'Warm'
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END AS weather_type
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FROM
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Weather AS w
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JOIN Countries AS c ONw.country_id=c.country_id
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WHERE date_format(day, '%Y-%m') = '2019-11'
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GROUP BY w.country_id;
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JOIN Countries USING (country_id)
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WHERE DATE_FORMAT(day, '%Y-%m') = '2019-11'
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GROUP BY 1;
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```
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<!-- tabs:end -->

‎solution/1200-1299/1294.Weather Type in Each Country/README_EN.md‎

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@@ -117,15 +117,15 @@ We know nothing about the average weather_state in Spain in November so we do no
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SELECT
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country_name,
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CASE
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WHEN avg(weather_state) <= 15 THEN 'Cold'
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WHEN avg(weather_state) >= 25 THEN 'Hot'
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WHEN AVG(weather_state) <= 15 THEN 'Cold'
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WHEN AVG(weather_state) >= 25 THEN 'Hot'
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ELSE 'Warm'
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END AS weather_type
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FROM
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Weather AS w
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JOIN Countries AS c ONw.country_id=c.country_id
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WHERE date_format(day, '%Y-%m') = '2019-11'
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GROUP BY w.country_id;
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JOIN Countries USING (country_id)
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WHERE DATE_FORMAT(day, '%Y-%m') = '2019-11'
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GROUP BY 1;
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```
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<!-- tabs:end -->
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Original file line numberDiff line numberDiff line change
@@ -1,13 +1,12 @@
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# Write your MySQL query statement below
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SELECT
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country_name,
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CASE
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WHEN avg(weather_state) <= 15 THEN 'Cold'
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WHEN avg(weather_state) >= 25 THEN 'Hot'
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WHEN AVG(weather_state) <= 15 THEN 'Cold'
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WHEN AVG(weather_state) >= 25 THEN 'Hot'
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ELSE 'Warm'
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END AS weather_type
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FROM
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Weather AS w
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JOIN Countries AS c ONw.country_id=c.country_id
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WHERE date_format(day, '%Y-%m') = '2019-11'
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GROUP BY w.country_id;
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JOIN Countries USING (country_id)
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WHERE DATE_FORMAT(day, '%Y-%m') = '2019-11'
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GROUP BY 1;

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