Oracle 1Z0-051題庫介紹
光靠死背很難應付 Oracle Database: SQL Fundamentals I 的實戰題型,臨場手感需要反覆演練。NewDumps 的測試引擎完整模擬 1Z0-051 正式考試環境,292 道模擬試題練熟之後,上場自然從容。
Oracle 1Z0-051 考試概覽:
| 認證廠商: | Oracle Corporation |
|---|---|
| 考試名稱: | Oracle Database: SQL Fundamentals I |
| 考試代碼: | 1Z0-051 |
| 相關認證: | Oracle Database SQL Certified Associate (1Z0-071 替代考試) |
| 證照有效期限: | 已停考 (不再提供;已由較新的 SQL 認證考試取代) |
| 支援語言: | English |
| 考試形式: | 選擇題 |
| 推薦課程: | Oracle SQL 培訓 |
| 考試報名: | Oracle 認證註冊 |
| 範例考題: | Oracle 1Z0-051 範例考題 |
| 考試方式: | 透過 Oracle 測試合作夥伴進行監考考試 (依地區提供線上或測試中心考試) |
| 必備條件: | 無強制先修條件;建議具備資料庫基本概念 |
| 官方大綱網址: | https://education.oracle.com/ |
Oracle 1Z0-051 考試大綱主題:
| 章節 | 目標 |
|---|---|
| 關聯式資料庫概念 | - 資料庫基礎
|
| 資料操作 | - DML 操作
|
| 單列函數 | - 字元、數字與日期函數
|
| SQL 基礎 | - 資料檢索
|
| 連接與集合運算 | - 資料表關係
|
關於 Oracle 1Z0-051 認證,你可能想問的事
1Z0-051(Oracle Database: SQL Fundamentals I)是 Oracle Corporation 舉辦的認證考試,通過後可取得 Oracle Database SQL Certified Associate (舊制) 認證,認證等級屬於 Associate。本考試與 Oracle Database SQL Certified Associate (1Z0-071 替代考試) 等認證相關,是規劃 Oracle Corporation 認證路徑時的重要一環。準備 Oracle Database: SQL Fundamentals I 時,建議搭配 NewDumps 的 292 道練習題,熟悉題型與出題方向。
報考 1Z0-051 的前置條件為:無強制先修條件;建議具備資料庫基本概念。官方的報考規定可能隨時調整,建議報名前再到官方考試說明頁面確認最新資訊。
以下是官方為 Oracle Database: SQL Fundamentals I 推薦的培訓資源:
完成官方培訓後,再搭配 NewDumps 的 292 道 1Z0-051 練習題反覆演練,能把課程所學轉化為實際的答題能力。
可以。NewDumps 提供 1Z0-051 免費範例試題(Free PDF Demo),下載後即可檢視實際題型與解析品質,滿意再購買完整版。購買後享有 365 天免費更新,期間內題庫內容隨官方考綱同步修訂;更新期滿後若需續更,可享 50% 折扣優惠。
NewDumps 提供「退款保證」:購買後 60 天內參加 1Z0-051 對應考試未通過,可申請全額退款。申請時需於考後 2 天內提交報名證明(准考證)影本與官方成績單(Score Report)PDF,考生姓名須與付款人姓名一致,我們會在 7 天內處理完成;購買後 3 天內應考、未實際參加考試、免費資料與過期訂單不適用。若不想退款,也可選擇免費更換兩個等值考試資料,並保留原購產品的更新服務。交付方面,付款成功後系統會在一分鐘內將產品寄至您的電子郵件信箱,可立即下載使用;若 2 小時內未收到,請聯絡客服協助。產品不限制安裝的電腦數量。
根據官方大綱,1Z0-051 考試共分為 5 個領域,主要包括 SQL 基礎、連接與集合運算、關聯式資料庫概念 等。各領域的詳細子主題與配分,請參考上方的考試大綱區塊,那裡有最完整的說明。
最新的 11g 1Z0-051 免費考試真題:
問題 #1
You want to create an ORD_DETAIL table to store details for an order placed having the following business requirement:
1) The order ID will be unique and cannot have null values.
2) The order date cannot have null values and the default should be the current date.
3) The order amount should not be less than 50.
4) The order status will have values either shipped or not shipped.
5) The order payment mode should be cheque, credit card, or cash on delivery (COD).
Which is the valid DDL statement for creating the ORD_DETAIL table?
A. CREATE TABLE ord_details
(ord_id NUMBER(2),
ord_date DATE NOT NULL DEFAULT SYSDATE,
ord_amount NUMBER(5, 2) CONSTRAINT ord_amount_min
CHECK (ord_amount >= 50),
ord_status VARCHAR2(15) CONSTRAINT ord_status_chk
CHECK (ord_status IN ('Shipped', 'Not Shipped')),
ord_pay_mode VARCHAR2(15) CONSTRAINT ord_pay_chk
CHECK (ord_pay_mode IN ('Cheque', 'Credit Card',
'Cash On Delivery')));
B. CREATE TABLE ord_details
(ord_id NUMBER(2) CONSTRAINT ord_id_pk PRIMARY KEY,
ord_date DATE DEFAULT SYSDATE NOT NULL,
ord_amount NUMBER(5, 2) CONSTRAINT ord_amount_min
CHECK (ord_amount >= 50),
ord_status VARCHAR2(15) CONSTRAINT ord_status_chk
CHECK (ord_status IN ('Shipped', 'Not Shipped')),
ord_pay_mode VARCHAR2(15) CONSTRAINT ord_pay_chk
CHECK (ord_pay_mode IN ('Cheque', 'Credit Card',
'Cash On Delivery')));
C. CREATE TABLE ord_details
(ord_id NUMBER(2) CONSTRAINT ord_id_nn NOT NULL,
ord_date DATE DEFAULT SYSDATE NOT NULL,
ord_amount NUMBER(5, 2) CONSTRAINT ord_amount_min
CHECK (ord_amount > 50),
ord_status VARCHAR2(15) CONSTRAINT ord_status_chk
CHECK (ord_status IN ('Shipped', 'Not Shipped')),
ord_pay_mode VARCHAR2(15) CONSTRAINT ord_pay_chk
CHECK (ord_pay_mode IN ('Cheque', 'Credit Card',
'Cash On Delivery')));
D. CREATE TABLE ord_details
(ord_id NUMBER(2) CONSTRAINT ord_id_uk UNIQUE NOT NULL,
ord_date DATE DEFAULT SYSDATE NOT NULL,
ord_amount NUMBER(5, 2) CONSTRAINT ord_amount_min
CHECK (ord_amount > 50),
ord_status VARCHAR2(15) CONSTRAINT ord_status_chk
CHECK (ord_status IN ('Shipped', 'Not Shipped')),
ord_pay_mode VARCHAR2(15) CONSTRAINT ord_pay_chk
CHECK (ord_pay_mode IN ('Cheque', 'Credit Card',
'Cash On Delivery')));
問題 #2
You need to calculate the number of days from 1st January 2007 till date . Dates are stored in the default format of dd-mon-rr. Which two SQL statements would give the required output? (Choose two.)
A. SELECT TO_CHAR(SYSDATE. 'DD-MON-YYYY') - '01-JAN-2007' FROM DUAL:
B. SELECT TO_DATE(SYSDAT *DD/MONTH/YYYY') - '01/JANUARY/2007' FROM DUAL:
C. SELECT SYSDATE - '01-JAN-2007' FROM DUAL:
D. SELECT SYSDATE - TOJDATE('01-JANUARY-2007') FROM DUAL:
E. SELECT SYSDATE - TOJDATE(X)1/JANUARY/2007") FROM DUAL:
問題 #3
The PART_CODE column in the SPARES table contains the following list of values:
Which statement is true regarding the outcome of the above query?
A. It produces an error.
B. It displays only the values A%_WQ123 and A%BWQ123 .
C. It displays only the values A%BWQ123 and AB_WQ123.
D. It displays all values.
E. It displays only the values A%_WQ123 and AB_WQ123 .
問題 #4
View the Exhibit and examine the data in the PROJ_TASK_DETAILS table.
The PROJ_TASK_DETAILS table stores information about tasks involved in a project and the relation between them.
The BASED_ON column indicates dependencies between tasks. Some tasks do not depend on the completion of any other tasks.
You need to generate a report showing all task IDs, the corresponding task ID they are dependent on, and the name of the employee in charge of the task it depends on.
Which query would give the required result?
A. SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p JOIN proj_task_details d ON (p.task_id = d.task_id);
B. SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p FULL OUTER JOIN proj_task_details d ON (p.based_on =
d.task_id);
C. SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p LEFT OUTER JOIN proj_task_details d ON (p.based_on =
d.task_id);
D. SELECT p.task_id, p.based_on, d.task_in_charge FROM proj_task_details p JOIN proj_task_details d ON (p.based_on = d.task_id);
問題 #5
Which two statements are true about constraints? (Choose two.)
A. A UNIQUE index gets created for columns with PRIMARY KEY and UNIQUE constraints.
B. The PRIMARY KEY and FOREIGN KEY constraints create a UNIQUE index.
C. The UNIQUE constraint does not permit a null value for the column.
D. The NOT NULL constraint ensures that null values are not permitted for the column.
問題與答案:
| 問題 #1 答案: B | 問題 #2 答案: D,E | 問題 #3 答案: B | 問題 #4 答案: C | 問題 #5 答案: A,D |
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