Use the following query to answer the question below it: SELECT pat.PAT_ID "Patient" ,enc.DEPARTMENT_ID "Department" FROM PATIENT pat INNER JOIN PAT_ENC enc ON "Patient" = enc.PAT_ID INNER JOIN CLARITY_DEP dep ON "Department" ORDER BY = dep.DEPARTMENT_ID pat. PAT_NAME ,"Department" All tables and columns used in this query are valid. Why won't it run?

Use the following query to answer the question below it: SELECT pat.PAT_ID "Patient" ,enc.DEPARTMENT_ID "Department" FROM PATIENT pat INNER JOIN PAT_ENC enc ON "Patient" = enc.PAT_ID INNER JOIN CLARITY_DEP dep ON "Department" ORDER BY = dep.DEPARTMENT_ID pat. PAT_NAME ,"Department" All tables and columns used in this query are valid. Why won't it run?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Write a query that displays the book title, cost, and year of publication for every book in the system. Sort the results by book title.
JOIN TABLES One row in the PAT_ENC table represents one patient encounter. One row in the ORDER MED table represents one medication order. One patient encounter can have many medication orders but one medication order can only have one patient encounter. In other words, the cardinality of this PAT_ENC to ORDER_MED relationship is one-to- many. You start a query with PAT_ENC. You then add ORDER_MED using an inner join. What is true about the granularity of the result set before and after adding the ORDER_MED table? SELECT ONE OF THE FOLLOWING A. The granularity stays at one row per patient encounter. B. The granularity stays at one row per medication order. C. The granularity changes from one row per patient encounter to one row per medication order on an encounter. D. The granularity changes from one row per medication order to one row per patient encounter.
Write a query that returns the Name of ALL Customers (based on CUS_FNAME and CUS_LNAME in CUSTOMER table, call this CUSTOMER_NAME), Phone number of ALL Customers (based on CUS_AREACODE and CUS_PHONE in CUSTOMER table, call this CUSTOMER_PHONE), for those haven’t flied before (based on the record in CHARTER table).
Write a query to get the maximum and minimum salary from the senators table (in that order).
The given SQL creates a Song table and inserts some songs. The first SELECT statement selects songs released after 1992. The second SELECT statement selects the release year for song with ID 800. Create a third query that combines the two existing queries. The first SELECT should be the outer query, and the second SELECT should be the subquery. The ORDER BY clause should appear after the subquery. Run your solution and verify the new query returns a result table with five rows, all with release years after 1992. CREATE TABLE Song (ID INT,Title VARCHAR(60),Artist VARCHAR(60),ReleaseYear INT,Genre VARCHAR(20),PRIMARY KEY (ID)); INSERT INTO Song VALUES(100, 'Hey Jude', 'Beatles', 1968, 'pop rock'),(200, 'You Belong With Me', 'Taylor Swift', 2008, 'country pop'),(300, 'You\'re Still the One', 'Shania Twain', 1998, 'country pop'),(400, 'Need You Now', 'Lady Antebellum', 2011, 'country pop'),(500, 'You\'ve Lost That Lovin\' Feeling', 'The Righteous Brothers', 1964, 'R&B'),(600, 'That\'s…
please answer the below question: Using data below/atached, write a query to find the last_name that is 4th from the bottom using subquery. (you need to use multiple subqueries to get that) and Using data below/attached, write a query to find the third oldest hire date in the table. use this data table: