SophiaPretty

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Teaching Since: Jul 2017
Last Sign in: 305 Weeks Ago, 1 Day Ago
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  • MBA,PHD, Juris Doctor
    Strayer,Devery,Harvard University
    Mar-1995 - Mar-2002

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  • Manager Planning
    WalMart
    Mar-2001 - Feb-2009

Category > Computer Science Posted 21 Dec 2017 My Price 10.00

A company needs a table to save its employee's info.

Hello. Can anyone please help me in this relational database MySQL.

 

  1. Use left outer join to write a query to show all person's contact_id, first_name, last_name from mc_contacts table, and title, salary from job_current table. show NULLs if a person doesn't have a job.
  2. A company needs a table to save its employee's info.

•       Write the sql to create one table: id, name, manager_id

•       The boss's manager_id is his own id.

•       Write a query to show id, name,manager name.

  1. Write a query to show all the boys and girls name from boys and girls table.CREATE TABLE `contact_interest` (
    id int not null auto_increment PRIMARY KEY,
      `contact_id` int(11) NOT NULL,
      `interest` varchar(50) NOT NULL,
      CONSTRAINT `c_interest_contact_id_fk` FOREIGN KEY (`contact_id`) REFERENCES `my_contacts` (`contact_id`) ON DELETE NO ACTION ON UPDATE NO ACTION
    ) ENGINE=InnoDB DEFAULT CHARSET=latin1

    /*
    -- Query: SELECT * FROM week8_lab_db.contact_interest
    LIMIT 0, 1000

    -- Date: 2017-03-14 23:17
    */
    INSERT INTO `contact_interest` (`id`,`contact_id`,`interest`) VALUES (1,1,'Movies');
    INSERT INTO `contact_interest` (`id`,`contact_id`,`interest`) VALUES (2,3,'Reading');
    INSERT INTO `contact_interest` (`id`,`contact_id`,`interest`) VALUES (3,6,'Tennis');
    CREATE TABLE `job_current` (
      `contact_id` int(11) NOT NULL,
      `title` varchar(30) NOT NULL,
      `salary` decimal(10,2) DEFAULT NULL,
      PRIMARY KEY (`contact_id`),
      CONSTRAINT `contact_id_fk` FOREIGN KEY (`contact_id`) REFERENCES `my_contacts` (`contact_id`) ON DELETE NO ACTION ON UPDATE NO ACTION
    ) ENGINE=InnoDB DEFAULT CHARSET=latin1

    /*
    -- Query: SELECT * FROM week8_lab_db.job_current
    LIMIT 0, 1000

    -- Date: 2017-03-14 21:33
    */
    INSERT INTO `job_current` (`contact_id`,`title`,`salary`) VALUES (1,'Web Designer',70000.00);
    INSERT INTO `job_current` (`contact_id`,`title`,`salary`) VALUES (2,'Computer Programmer',100000.00);
    INSERT INTO `job_current` (`contact_id`,`title`,`salary`) VALUES (4,'Web Designer',65000.00);
    INSERT INTO `job_current` (`contact_id`,`title`,`salary`) VALUES (5,'Teacher',120000.00);
    CREATE TABLE `my_contacts` (
      `contact_id` int(11) NOT NULL AUTO_INCREMENT,
      `last_name` varchar(30) DEFAULT NULL,
      `first_name` varchar(20) DEFAULT NULL,
      `email` varchar(50) DEFAULT NULL,
      `gender` char(1) DEFAULT NULL,
      `birthday` date DEFAULT NULL,
      `location` varchar(50) DEFAULT NULL,
      `status` varchar(20) DEFAULT NULL,
      `interests` varchar(100) DEFAULT NULL,
      `seeking` varchar(100) DEFAULT NULL,
      `prof_id` int(11) DEFAULT NULL,
      `zip_code` varchar(10) DEFAULT NULL,
      PRIMARY KEY (`contact_id`)
    ) ENGINE=InnoDB AUTO_INCREMENT=7 DEFAULT CHARSET=latin1

    /*
    -- Query: SELECT * FROM week8_lab_db.my_contacts
    LIMIT 0, 1000

    -- Date: 2017-03-14 21:33
    */
    INSERT INTO `my_contacts` (`contact_id`,`last_name`,`first_name`,`email`,`gender`,`birthday`,`location`,`status`,`interests`,`seeking`,`prof_id`,`zip_code`) VALUES (1,'Toth','Anne','Anne_Toth@leapinlimos.com','F','1969-11-18','San Fran, CA',NULL,NULL,NULL,1,'91745');
    INSERT INTO `my_contacts` (`contact_id`,`last_name`,`first_name`,`email`,`gender`,`birthday`,`location`,`status`,`interests`,`seeking`,`prof_id`,`zip_code`) VALUES (2,'Manson','Anne','am86@objectville.net','F','1977-08-09','Seattle, WA',NULL,NULL,NULL,2,'48109');
    INSERT INTO `my_contacts` (`contact_id`,`last_name`,`first_name`,`email`,`gender`,`birthday`,`location`,`status`,`interests`,`seeking`,`prof_id`,`zip_code`) VALUES (3,'Hardy','Anne','anneh@b0tt0msup.com','F','1963-04-18','San Fran, CA',NULL,NULL,NULL,5,'91745');
    INSERT INTO `my_contacts` (`contact_id`,`last_name`,`first_name`,`email`,`gender`,`birthday`,`location`,`status`,`interests`,`seeking`,`prof_id`,`zip_code`) VALUES (4,'Parker','Anne','annep@starbuzzcoffee.com','F','1983-01-10','San Fran, CA',NULL,NULL,NULL,4,'91745');
    INSERT INTO `my_contacts` (`contact_id`,`last_name`,`first_name`,`email`,`gender`,`birthday`,`location`,`status`,`interests`,`seeking`,`prof_id`,`zip_code`) VALUES (5,'Blunt','Anne','anneblunt@breakneckpizza.com','F','1959-10-09','San Fran, CA',NULL,NULL,NULL,6,'48109');
    INSERT INTO `my_contacts` (`contact_id`,`last_name`,`first_name`,`email`,`gender`,`birthday`,`location`,`status`,`interests`,`seeking`,`prof_id`,`zip_code`) VALUES (6,'Jacobs','Anne','anne99@objectville.net','F','1968-02-05','San Jose, CA',NULL,NULL,NULL,3,'91745');
    CREATE TABLE `profession` (
      `id` int(11) NOT NULL AUTO_INCREMENT,
      `profession` varchar(20) DEFAULT NULL,
      PRIMARY KEY (`id`)
    ) ENGINE=InnoDB AUTO_INCREMENT=7 DEFAULT CHARSET=latin1

    /*
    -- Query: SELECT * FROM week8_lab_db.profession
    LIMIT 0, 1000

    -- Date: 2017-03-14 21:33
    */
    INSERT INTO `profession` (`id`,`profession`) VALUES (1,'Artist');
    INSERT INTO `profession` (`id`,`profession`) VALUES (2,'Baker');
    INSERT INTO `profession` (`id`,`profession`) VALUES (3,'Computer Programmer');
    INSERT INTO `profession` (`id`,`profession`) VALUES (4,'Student');
    INSERT INTO `profession` (`id`,`profession`) VALUES (5,'Teacher');
    INSERT INTO `profession` (`id`,`profession`) VALUES (6,'Web Designer');

Answers

(5)
Status NEW Posted 21 Dec 2017 01:12 PM My Price 10.00

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