Query a list of city names from station for cities that have an even id number. print the results in any order, but exclude duplicates from the answer.

WHERE co . CONTINENT = 'Asia' ; 15. Average Population of Each Continent | Easy | Given the CITY and COUNTRY tables, query the names of all the continents (COUNTRY.Continent) and their respective average city populations (CITY.Population) rounded down to the nearest integer. Note: CITY.CountryCode and COUNTRY.Code are matching key columns. Input Format The CITY and COUNTRY tables are described as follows: SELECT CITY , STATE FROM STATION ; 17. Weather Observation Station 2 | Easy | Query the following two values from the STATION table: • The sum of all values in LAT_N rounded to a scale of 2 decimal places. • The sum of all values in LONG_W rounded to a scale of 2 decimal places. Input Format The STATION table is described as follows: where LAT_N is the northern latitude and LONG_W is the western longitude. Output Format Your results must be in the form: where LAT_N is the northern latitude and LONG_W is the western longitude. For example, if there are three records in the table with CITY values 'New York', 'New York', 'Bengalaru', there are 2 different city names: 'New York' and 'Bengalaru'. The query returns 1, because total number of records - number of unique city names = 3 - 2 = 1 Solution FROM STATION ; 20. Weather Observation Station 5 | Easy | Query the two cities in STATION with the shortest and longest CITY names, as well as their respective lengths (i.e.: number of characters in the name). If there is more than one smallest or largest city, choose the one that c...

LinkedIn and 3rd parties use essential and non-essential cookies to provide, secure, analyze and improve our Services, and to show you relevant ads (including professional and job ads) on and off LinkedIn. Learn more in our Select Accept to consent or Reject to decline non-essential cookies for this use. You can update your choices at any time in your Hey Everyone, I have solved SQL questions from HackerRank. ➡ Problem 1: Query a list of CITY names from STATION for cities that have an even ID number. Print the results in any order, but exclude duplicates from the answer. 💡 Approach: · Used Distinct keyword to retrieve only unique city names · id % 2 = 0 with this query will get only even numbers link: • 📊 Exciting Power BI Project! 🚀 HI everyone I have created one more report of Power BI Maven Northwind Challenge, I had the amazing opportunity to create a top-level KPI dashboard. This dashboard provides a comprehensive overview of our company's performance in key areas. Let me share some of the highlights: 1️⃣ Sales Trends: The dashboard presents an in-depth analysis of our sales trends, allowing us to identify patterns, spot growth opportunities, and make data-driven decisions. We track sales revenue over time, compare performance across regions, and analyze sales by product category. 2️⃣ Product Performance: We dive deep into our product performance metrics to understand which products are driving our success and which ones might need some extra attention. The dashboard ...

STATION Field Type ID NUMBER CITY VARCHAR2(21) STATE VARCHAR2(2) LAT_N NUMBER LONG_W NUMBER where LAT_Nis the northern latitude and LONG_Wis the western longitude. Solution – Weather Observation Station 3 in SQL SELECT DISTINCT CITY FROM STATION WHERE MOD(ID, 2) = 0; Disclaimer: The above Problem ( Hacker Rank but the Solution is Provided by CodingBroz. This tutorial is only for Educational and Learning Purpose.

Code answers related to "Query a list of CITY names from STATION for cities that have an even ID number. Print the results in any order, but exclude duplicates from the answer. The STATION table is described as follows: Station.jpg" • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Query a list of CITY names from STATION for cities that have an even ID number. Print the results in any order, but exclude duplicates from the answer. The STATION table is described as follows: CREATE TABLE STATION ( Id int, CITY varchar(50), STATE varchar(50), LAT_N int, LONG_W int ) I have this SQL which throws an error when I run: SELECT CITY, STATE FROM STATION WHERE ID % 2 = 0 GROUP BY ID, CITY, STATE What DBMS are you using? If you are getting an error it may be because you are using a syntax not supported by your DBMS. The only likely operation for this would be the modulo operator. Does your DBMS support % for modulo? Are you supposed to select the state? Aren't the expected duplicates supposed to exist only because we neglect the state? You probably want to remove the GROUP BY clause and SELECT DISTINCT city instead. Thanks for contributing an answer to Stack Overflow! • Please be sure to answer the question. Provide details and share your research! But avoid … • Asking for help, clarification, or responding to other answers. • Making statements based on opinion; back them up with references or personal experience. To learn more, see our

/*Query all columns for all American cities in CITY with populations larger than 100000. The CountryCode for America is USA. */ select * from CITY where POPULATION > 100000 AND COUNTRYCODE = 'USA '; /*Query the names of all American cities in CITY with populations larger than 120000. The CountryCode for America is USA. */ select NAME from CITY where POPULATION > 120000 AND COUNTRYCODE = 'USA '; /*Query all columns (attributes) for every row in the CITY table. */ select * from CITY; /*Query all columns for a city in CITY with the ID 1661. */ select * from CITY where ID = 1661; /*Query all attributes of every Japanese city in the CITY table. The COUNTRYCODE for Japan is JPN. */ select * from CITY where COUNTRYCODE = 'JPN '; /*Query the names of all the Japanese cities in the CITY table. The COUNTRYCODE for Japan is JPN. */ select NAME from CITY where COUNTRYCODE = 'JPN '; /*Query a list of CITY and STATE from the STATION table. */ select CITY, STATE from STATION; /*Query a list of CITY names from STATION with even ID numbers only. You may print the results in any order, but must exclude duplicates from your answer. */ select distinct CITY from STATION where ID% 2 = 0; /*Let N be the number of CITY entries in STATION, and let N'be the number of distinct CITY names in STATION; query the value of N-N' from STATION. In other words, find the difference between the total number of CITY entries in the table and the number of distinct CITY entries in the table. */ select count(CITY)...