Assume you have four datasets: "employees", "departments", "projects", and "assignments". The scheme of these databases are as follows: The "employees" dataset contains employee information including a unique "employee_ID", "employee_name", and "department_ID". The "departments" dataset contains department information including a unique "department_ID" and "department_name". The "projects" dataset contains project information including a unique "project_ID", "project_name", and "department_ID". The "assignments" dataset contains information about which employees are assigned to which projects, including the "employee_ID" and "project_ID" associated with each assignment. Write a SQL query to retrieve the name and department of all employees who are assigned to a project that is not in the same department as their own department.

时间: 2023-07-06 14:35:51 浏览: 89
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Here's a SQL query to retrieve the name and department of all employees who are assigned to a project that is not in the same department as their own department: ``` SELECT e.employee_name, d.department_name FROM employees e INNER JOIN assignments a ON e.employee_ID = a.employee_ID INNER JOIN projects p ON a.project_ID = p.project_ID INNER JOIN departments d ON e.department_ID = d.department_ID WHERE p.department_ID <> e.department_ID; ``` In this query, we are joining the "employees", "assignments", "projects", and "departments" datasets to retrieve the necessary information. We are selecting the "employee_name" and "department_name" columns from the "employees" and "departments" tables, respectively. Then we are using INNER JOINs to join the tables based on their corresponding IDs. Finally, we are adding a WHERE clause to filter the results and only show employees who are assigned to a project that is not in the same department as their own department. This is done by comparing the "department_ID" column in the "projects" table with the "department_ID" column in the "employees" table.
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Serializability of a class is enabled by the class implementing the java.io.Serializable interface. Classes that do not implement this interface will not have any of their state serialized or deserialized. All subtypes of a serializable class are themselves serializable. The serialization interface has no methods or fields and serves only to identify the semantics of being serializable. To allow subtypes of non-serializable classes to be serialized, the subtype may assume responsibility for saving and restoring the state of the supertype's public, protected, and (if accessible) package fields. The subtype may assume this responsibility only if the class it extends has an accessible no-arg constructor to initialize the class's state. It is an error to declare a class Serializable if this is not the case. The error will be detected at runtime. During deserialization, the fields of non-serializable classes will be initialized using the public or protected no-arg constructor of the class. A no-arg constructor must be accessible to the subclass that is serializable. The fields of serializable subclasses will be restored from the stream. When traversing a graph, an object may be encountered that does not support the Serializable interface. In this case the NotSerializableException will be thrown and will identify the class of the non-serializable object. Classes that require special handling during the serialization and deserialization process must implement special methods with these exact signatures: private void writeObject(java.io.ObjectOutputStream out) throws IOException private void readObject(java.io.ObjectInputStream in) throws IOException, ClassNotFoundException; private void readObjectNoData() throws ObjectStreamException;

用c++解决Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair of currencies. Each point has its own exchange rates, exchange rate of A to B is the quantity of B you get for 1A. Also each exchange point has some commission, the sum you have to pay for your exchange operation. Commission is always collected in source currency. For example, if you want to exchange 100 US Dollars into Russian Rubles at the exchange point, where the exchange rate is 29.75, and the commission is 0.39 you will get (100 - 0.39) * 29.75 = 2963.3975RUR. You surely know that there are N different currencies you can deal with in our city. Let us assign unique integer number from 1 to N to each currency. Then each exchange point can be described with 6 numbers: integer A and B - numbers of currencies it exchanges, and real RAB, CAB, RBA and CBA - exchange rates and commissions when exchanging A to B and B to A respectively. Nick has some money in currency S and wonders if he can somehow, after some exchange operations, increase his capital. Of course, he wants to have his money in currency S in the end. Help him to answer this difficult question. Nick must always have non-negative sum of money while making his operations. Input The first line contains four numbers: N - the number of currencies, M - the number of exchange points, S - the number of currency Nick has and V - the quantity of currency units he has. The following M lines contain 6 numbers each - the description of the corresponding exchange point - in specified above order. Numbers are separated by one or more spaces. 1 ≤ S ≤ N ≤ 100, 1 ≤ M ≤ 100, V is real number, 0 ≤ V ≤ 103. For each point exchange rates and commissions are real, given with at most two digits after the decimal point, 10-2 ≤ rate ≤ 102, 0 ≤ commission ≤ 102. Let us call some sequence of the exchange operations simple if no exchange point is used more than once in this sequence. You may assume that ratio of the numeric values of the sums at the end and at the beginning of any simple sequence of the exchange operations will be less than 104. Output If Nick can increase his wealth, output YES, in other case output NO.

操作系统代码实现:Number Project Name Content Summary State Type 一、Process Scheduling Algorithm Simulation 1、 Simulate the operation of the round-robin algorithm for process scheduling. 2、 Create at least 15 processes and output their scheduling situation under the scheduling algorithm mentioned above and output it to theterminal to check the execution of the algorithm. 3、 The output should include the arrival time of the processes, the end time, and the average execution time. Essential. General 二、Readers-Writer Problem Implmentation 1、 A data set is shared among several concurrent processes: Readers – only read the data set; they do not perform any updates. Writers – can both read and write. 2、 Problem – allow multiple readers (at most 8) to read at the same time. Only one single writer can access the shared data at the same time. Essential. General 三、Program for Least Recently used Algorithm 1、 Create a page access sequence (page number range 0-18) using a random function. The sequence length is 54 and assume that the number of main memory frames allocated to the thread is 6, that is, M = 6. 2、 Implement the LRU algorithm for page replacement on the above access sequence. 3、 Output the page replacement sequence and the page fault rate. Essential. General Requirements 1、 For each experiment project, submit a design report and code. The code should be implemented in C++. The requirements are as follows: a) The content of the design report should include the design ideas and implementation. b) The results of the design report should include testing and running results (screenshots of screen outputs). c) The conclusion of the design report should summarize the problems encountered, the solutions and experiences during the implementation process.

Use c # to complete the following code,Creating a Car class A class is a self-contained piece of software that is able to function on it own. We will model a car in software. Car Class Properties Year : int «private set» Manufacturer : string «private set» Model : string «private set» IsDrivable : bool «private set» Price : double «private set» Methods«constructor» Car(year : int, manufacturer : string, model : string, price : double, isDrivable: bool = true) PrintInfo() : void Description of field members: 1. year: this private member represents the year value of this object 2. manufacturer: this private member represents the manufacturer of this car 3. model: this private member represents the model of this car 4. isDrivable: this private member indicates if this car be driven 5. price: this private member represents the selling price of this car Constructor: 1. public Car(int year, string manufacturer, string model, double price, bool isDrivable = true): this is the constructor of this class. It takes five arguments and assigns them to the appropriate fields. You will assume that all the arguments will be sane (i.e. the values will not break your program). Apology: UML and some newer languages specify the type after the identifier. C# on the other hand requires that you specify the type and then the identifier. To restrict access fields are normally decorated with the private modifier. Programming II Car: fields, constructor, ToString() n.k.p Page 2 of 2 Description of action member: 1. public void PrintInfo(): this method does not take any argument but it will print all information about this object. You get to decide how the output will look like. It is expected that all the values be displayed. In your main method write the code to do the following: 1. Create at least four cars and print them. Remember to call the constructor with 4 or 5 parameters.

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