search will return the lower-case version
Anonymous Quiz
2%
SUBSTR()
2%
TRIM()
3%
UPPER()
93%
LOWER()
Takes the string we hand it in parentheses and returns a part of the string.
Anonymous Quiz
88%
SUBSTR()
12%
TRIM()
Update all row to trim spaces in the ‘state’ field
Anonymous Quiz
85%
UPDATE contributors SET state = TRIM(state)
15%
UPDATE contributors SET state = state.trim()
How many distinct ZIP Codes, for instance, are there in the table?
Anonymous Quiz
19%
SELECT COUNT(zip) FROM contributors
81%
SELECT COUNT(DISTINCT zip) FROM contributors
What is the maximum amount that any of our contributors has given?
Anonymous Quiz
94%
SELECT MAX(amount) FROM contributors
6%
SELECT MAXIMUM(amount) FROM contributors
What is the total donors’ amount of contributions from Georgia?
Anonymous Quiz
28%
SELECT COUNT(amount) FROM contributors WHERE state = ‘GA’
72%
SELECT SUM(amount) FROM contributors WHERE state = ‘GA’
What is the average amount contributed?
Anonymous Quiz
93%
SELECT AVG(amount) FROM contributors
7%
SELECT AVERAGE(amount) FROM contributors
What if we were interested in learning who had given the largest amount? We then use …
Anonymous Quiz
21%
super queries
79%
sub queries
SELECT id FROM contributors WHERE amount = (SELECT MAX(amount) FROM contributors
Anonymous Quiz
16%
Using Two Queries
84%
Using sub queries
Select the total amount from each state
Anonymous Quiz
13%
SELECT state, SUM(amount) FROM contributors GROUP BY amount
63%
SELECT state, SUM(amount) FROM contributors GROUP BY state
10%
SELECT amount, SUM(state) FROM contributors GROUP BY amount
14%
SELECT amount, SUM(state) FROM contributors GROUP BY state
For aggregates, the equivalent of a WHERE clause is the HAVING clause. Immediately following the GROUP BY
Anonymous Quiz
93%
T
7%
F
كدا الحمدلله نص المنهج اتبعت اول 5 محاضرات ان شاء الله نكمل الباقي النهارده
.......... function is used to count the number of rows, but it counts all rows in a table including the rows with NULL values, whereas ............. counts only the non-NULL values in the specified column.
Anonymous Quiz
70%
count(*) - count()
30%
count() - count(*)
Which of the following should be used to find the mean of the salary?
Anonymous Quiz
17%
Mean(salary)
56%
Avg(salary)
27%
Sum(salary) / Count(salary)