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أقدم لكم بعض جمل سيكوال لإفادة من يريد تعلم جمل Sql

لمن يريد تعلم جمل سيكوال

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#1 oracle9

oracle9

    عضو

  • الأعضــاء
  • 26 مشاركة

تاريخ المشاركة 12 January 2012 - 01:29 PM

أخوانى أقدم بكم بعض جمل سيكوال لإفادة من يريد تعلم جمل سيكوال
Chapter 1
002 
003 Practice 1
004 1. Initiate an iSQL*Plus session using the user ID and password provided by the instructor.
005   User_id   :  scott
006   Password: tiger
007 
008 2. iSQL*Plus commands access the database.
009 True/False
010 True
011 
012 3. The following select statement executes successfully:
013 select last_name, job_id, salary AS Sal FROM employees;
014 True/False
015 True
016  
017 4. The following select statement executes successfully:
018 select *
019 FROM job_grades;
020 True/False
021 False
022 
023 5. There are four coding errors in this statement. Can you identify them?
024 select employee_id, last_name
025 sal x 12 ANNUAL SALARY
026 FROM employees;
027 False
028 
029 6. Show the structure of the DEPARTMENTS table. select all data from the table.
030 1.desc   departments;
031 2.  select  *
032	 FROM	  departments;
033 
034 
035 7. Show the structure of the EMPLOYEES table. create a query to display the last name, job code, hire date, and employee number for each employee, with employee number appearing first. Provide an alias STARTDATE for the HIRE_DATE column.
036 desc employees;
037 
038 
039 8. Run your query in the file lab1_7.sql.
040 select employee_id, last_name, job_id, hire_date  "Start Date"
041 FROM    employees;
042 
043 9. create a query to display unique job codes from the EMPLOYEES table.
044 select   job_id
045 FROM	  employees;
046 
047 
048 10. Copy the statement from lab1_7.sql into the iSQL*Plus Edit window. Name the column headings Emp #, Employee, Job, and Hire Date, respectively.
049 select  employee_id  "emp#", last_name  "Employee" ,job_id "Job",
050				   hire_date "hire date"
051 FROM	  employees;
052 
053 11. Display the last name concatenated with the job ID, separated by a comma and space, and name the column Employee and Title.
054 select  last_name || ',' || job_id "employee and title"
055 FROM	 employees;
056 
057 12. create a query to display all the data from the EMPLOYEES table. Separate each column by a comma. Name the column THE_OUTPUT
058 select    EMPLOYEE_ID||','
059				    ||FIRST_NAME||','||LAST_NAME||','||EMAIL||','
060				    ||PHONE_NUMBER||','||HIRE_DATE||','||JOB_ID||','
061				    ||SALARY||','||COMMISSION_PCT||  ','
062				    ||MANAGER_ID||','||DEPARTMENT_ID   "THE _OUTPUT"
063 FROM	   employees;
064 
065 
066 
067 
068 
069 Chapter 2
070 
071 Practice 2
072 1. create a query to display the last name and salary of employees earning more than $12,000.
073 select	 last_name ,salary
074 FROM		 employees
075 WHERE	  salary >12000;
076 
077 2. create a query to display the employee last name and department number for employee number
078 176
079 select  last_name,department_id
080 FROM	  employees
081 WHERE   employee_id=176;
082 
083 
084 3. Modify lab2_1.sql to display the last name and salary for all employees whose salary is not in the range of $5,000 and $12,000.
085 select    last_name,salary
086 FROM	    employees
087 WHERE	 salary not between 12000 and 55000;
088 
089 4. Display the employee last name, job ID, and start date of employees hired between February 20, 1998, AND May 1, 1998. Order the query in ascending order by start date.
090 select  last_name,job_id ,hire_date
091 FROM	  employees
092 WHERE   hire_date between '20-feb-98' AND '01-may-98';
093 
094 
095 5. Display the last name and department number of all employees in departments 20 AND 50 in alphabetical order by name.
096 select		 last_name ,department_id
097 FROM		    employees
098 WHERE		 department_id in(20,50)
099 ORDER  BY   last_name;
100 
101 6. Modify lab2_3.sql to list the last name and salary of employees who earn between $5,000 and $12,000, and are in department 20 or 50. Label the columns Employee and Monthly Salary, respectively.
102 select	  last_name "employee",salary "monthly salary"
103 FROM		 employees
104 WHERE	  salary between 12000 and 5000
105 AND		    department_id in(20,50);
106 
107 
108 7. Display the last name and hire date of every employee who was hired in 1994.
109 select	 last_name , hire_date
110 FROM		 employees
111 WHERE	  hire_date like'%94';
112 
113 8. Display the last name and job title of all employees who do not have a manager.
114 select  last_name ,job_id
115 FROM	  employees
116 WHERE   manager_id is null;
117 
118 9. Display the last name, salary, and commission for all employees who earn commissions. Sort data in descending Order of salary and commissions
119 select	    last_name , salary, commission_pct
120 FROM		   employees
121 WHERE	    commission_pct is not null
122 ORDER  BY  salary DESC;
123 
124 
125 10. Display the last names of all employees where the third letter of the name is an a.
126 select	 last_name
127 FROM	    employees
128 WHERE	 last_name like'__a%';
129 
130 
131 
132 
133 
134 11. Display the last name of all employees who have an a and an e in their last name.
135 select   last_name
136 FROM	  employees
137 WHERE   last_name like'%a%'
138 OR		    last_name like '%e%';
139 
140 12. Display the last name, job, and salary for all employees whose job is sales representative or stock clerk and whose salary is not equal to $2,500, $3,500, or $7,000.
141 select	 last_name,job_id,salary
142 FROM	    employees
143 WHERE	  job_id like'ST_CLERK'
144 OR			   job_id like 'SA_REP'
145 AND		    salary not in(2005,3500,7000);
146 
147 
148 13. Modify lab2_6.sql to display the last name, salary, and commission for all employees whose commission amount is 20%.
149 select  last_name "Employee",salary "Monthly Salary",commission_pct
150 FROM	  employees
151 WHERE   commission_pct =.2
152 
153 
154 
155 
156 
157 
158 
159 
160 
161 
162 
163 
164 
165 
166 
167 
168 
169 
170 
171 Chapter 3
172 
173 Practice 3
174 1.Write a query to display the current date. Label the column Date.
175 select  sysdate “data”
176 FROM	  employees;
177 
178 2. FOR each employee, display the employee number, last_name, salary, AND salary increased by 15% and expressed as a whole number. Label the column New Salary.
179 select   EMPLOYEE_ID  , LAST_NAME   , salary,salary*1.15
180				    "NEW _SALARY"
181 FROM	    employees;
182 
183 3. Run your query in the file lab3_2.sql.
184 
185 4. Modify your query lab3_2.sql to add a column that subtracts the old salary from  the new salary. Label the column Increase.
186 select  EMPLOYEE_ID,LAST_NAME, salary,
187		 salary*1.15 "NEW _  SALARY",salary*1.15- salary "Increase"
188 FROM	  employees;
189 
190 5. Write a query that displays the employee’s last names with the first letter capitalized and all other letters lowercase, and the length of the names, for all employees whose name starts with J, A, or M. Give each column an appropriate label. Sort the results by the employees’ last_names.
191 select	   initcap (last_name) "name",length(last_name) "length"
192 FROM		   employees
193 WHERE	    last_name like 'A%'
194 OR				 last_name like 'J%'
195 OR				 last_name like 'M%'
196 ORDER BY  last_name;
197 
198 
199 
200 
201 
202 6. for each employee, display the employee’s last name, and calculate the number of months between today and the date the employee was hired. Label the column MONTHS_WORKED.
203 order your results by the number of months employed. Round the number of months up to the closest whole number.
204 select	    last_name ,
205			  round (months_between(sysdate,
206			  hire_date),0)"MONTHS_WORKE"
207 FROM		   employees
208 ORDER BY   round (months_between( sysdate ,hire_date),0);
209 
210  
211 7. Write a query that produces the following for each employee:
212 <employee last name> earns <salary> monthly but wants <3 times
213 salary>. Label the column Dream Salaries.
214 select  last_name ||'earns' || to_char(salary,'$99,999.00') ||'monthly but
215				   wents'||to_char(3*   salary,'$99,999.00') "dream salaries"
216 FROM	 employees;
217 
218 8. create a query to display the last name and salary for all employees. format the salary to be characters long, left-padded with $. Label the column SALARY.
219 select  last_name, lpad (salary,15,'$') "salary"
220 FROM	 employees;
221 
222 9. Display each employee&#8217;s last name, hire date, and salary review date, which is the first Monday after six months of service. Label the column REVIEW. format the dates to appear in the
223 format similar to &#8220;Monday, the Thirty-First of July, 2000.&#8221;
224 select    last_name  ,hire_date,
225					  to_char(next_day(add_months(hire_date,6) ,'monday'),
226					  'day "the" fmDdspth"of" month yyyy') "review"
227 FROM		 employees;
228 
229 10. Display the last name, hire date, and day of the week on which the employee started. Label the column DAY. order the results by the day of the week starting with Monday.
230 select	    last_name ,hire_date,to_char(hire_date,'day') "day"
231 FROM		   employees
232 ORDER BY   to_char(hire_date,'day')  ;
233 11. create a query that displays the employees&#8217; last names and commission amounts. If an employee does not earn commission, put &#8220;No Commission.&#8221; Label the column COMM.
234 select  last_name ,nvl (to_char(commission_pct),'NO COMMISSION')
235 FROM	  employees;
236 
237  
238 12. create a query that displays the employees&#8217; last names and indicates the amounts of  their annual salaries with asterisks. Each asterisk signifies a thousand dollars. Sort the data in descending order of salary. Label the column EMPLOYEES_AND_THEIR_SALARIES.
239 select  rpad(last_name,trunc(salary/1000,0)+length(last_name),'*')
240				   EMPLOYEES_AND_THEIR_SALARIES
241 FROM	 employees;;
242 
243 
244 13. Using the DECODE function, write a query that displays the grade of all employees based on the value of the column JOB_ID, as per the following data:
245 Job Grade
246 AD_PRES A
247 ST_MAN B
248 IT_PROG C
249 SA_REP D
250 ST_CLERK E
251 select  job_id ,
252 Decode	 (job_id,	 'AD_PRES','A',
253							    'ST_MAN','B',
254							    'IT_PROG','C',
255									 'SA_REP','D',
256							    'ST_CLERK','E',
257															 '0')	   
258 FROM   employees;
259	 
260 
261 
262 
263 
264 
265 Chapter 4
266 
267 Practice 4
268 1. Write a query to display the last name, department number, and department name for all employees.
269 select	 e.last_name, e.department_id ,d.department_name
270 FROM	    employees e, departments d
271 WHERE	 e.department_id = d.department_id;
272 
273 2. create a unique listing of all jobs that are in department 80. Include the location of the department in the output.
274 select   e.job_id , d.location_id
275 FROM	  employees e, departments d
276 WHERE   d.department_id= 80
277 AND		  e.department_id = d.department_id;
278 
279 
280 3. Write a query to display the employee last name, department name, location ID, and city of all employees who earn a commission.
281 select   e.last_name,d.department_name,l.location_id,l.city
282 FROM	   employees e, departments d ,locations l
283 WHERE    e.department_id = d.department_id
284 AND		  d.location_id=l.location_id
285 AND		  commission_pct is not null ;
286 
287 4. Display the employee last name and department name for all employees who have an a (lowercase) in their last names.
288 select    e.last_name,d.department_name
289 FROM	   employees e, departments d
290 WHERE    e.department_id = d.department_id
291 AND		   last_name like '%a%' ;
292 
293  
294   5. Write a query to display the last name, job, department number, and department name for all employees who work in Toronto.
295 select    e.last_name ,e.job_id,e.department_id,d.department_name
296 FROM	   employees e,departments d,locations l
297 WHERE    l.city like 'Toronto'
298 AND		  e.department_id=d.department_id;
299 Chapter 5
300 
301 Practice 5
302 Determine the validity of the following three statements. Circle either True or False.
303 1. Group functions work across many rows to produce one result per group.
304 True/False
305 True
306 
307 2. Group functions include nulls in calculations.
308 True/False
309 False
310 
311 3. The WHERE clause restricts rows prior to inclusion in a group calculation.
312 True/False
313 False
314 
315 4. Display the highest, lowest, sum, and average salary of all employees. Label the columns Maximum, Minimum, Sum, and Average, respectively. Round your results to the nearest whole number.
316 select	 max(salary)"Maximum",min(salary)"Minimum",
317					 sum (salary)"Sum", avg(salary)"Average"
318 FROM	    employees;
319 
320 5. Modify the query in lab5_4.sql to display the minimum, maximum, sum, and average salary for each job type.
321 select	    job_id,max(salary)"Maximum",min(salary)"Minimum",
322						 sum (salary)"Sum", avg(salary)"Average"
323 FROM		    employees
324 GROUP BY   job_id;
325 
326 6. Write a query to display the number of people with the same job.
327 select	    job_id,count(employee_id)
328 FROM		   employees
329 GROUP BY   job_id;;
330 
331 
332 7. Determine the number of managers without listing them. Label the column Number of Managers. Hint: Use the MANAGER_ID column to determine the number of managers.
333 select   count (DISTINCT(manager_id))  " Number of Manager"
334 FROM	   employees
335 
336 
337 8. Write a query that displays the difference between the highest and
338 lowest salaries. Label the column DIFFERENCE.
339 select    max(salary)-min(salary) difference
340 FROM	   employees;;
341 
342 
343 9. Display the manager number and the salary of the lowest paid employee for that manager. Exclude anyone whose manager is not known. Exclude any groups WHERE the minimum salary is $6,000 or less. Sort the output in descending Order of salary.
344 select	    manager_id ,min(salary)
345 FROM		   employees
346 HAVING	   min(salary)>=6000
347 AND			   manager_id is not null
348 GROUP BY   manager_id
349 ORDER BY   min(salary) DESC;
350 
351 
352 10. Write a query to display each department&#8217;s name, location, number of employees, and the average salary for all employees in that department. Label the columns Name, Location, Number of People, and Salary, respectively. Round the average salary to two decimal places.
353 select	    d.department_name "Name",d.location_id "Location",
354					    count(e.employee_id) "Nember of People",avg(e.salary)
355						 "Salary"
356 FROM		   departments d,employees e
357 WHERE	    d.department_id= e.department_id
358 GROUP BY  d.department_name,location_id ;
359 
360 
361 11. create a query that will display the total number of employees and, of that total, the number of employees hired in 1995, 1996, 1997, and 1998. create appropriate column headings.
362 select   count(employee_id) "Total",
363				    sum(decode(to_char(hire_date,'yyyy'),1995,1,0))"1995",
364				    sum(decode(to_char(hire_date,'yyyy'),1996,1,0))"1996",
365				    sum(decode(to_char(hire_date,'yyyy'),1997,1,0))"1997",
366				    sum(decode(to_char(hire_date,'yyyy'),1998,1,0))"1998"
367 FROM	  employees;
368 
369 12. create a matrix query to display the job, the salary for that job based on department number, and the total salary for that job, for departments 20, 50, 80, and 90, giving each column an appropriate heading.
370 
371 select		 job_id "Job",
372					    sum(decode (department_id,20,salary,0))"dept 20",
373					    sum(decode (department_id,50,salary,0))"dept 50",
374					    sum(decode (department_id,80,salary,0))"dept 80",
375					    sum(decode (department_id,90,salary,0))"dept 90",
376					    sum(salary)
377 FROM		    employees
378 GROUP BY  job_id;


#2 oracle9

oracle9

    عضو

  • الأعضــاء
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تاريخ المشاركة 12 January 2012 - 01:34 PM

إليكم الجزء الثانى
1. Write a query to display the last name and hire date of any employee in the same department as Zlotkey. Exclude Zlotkey.
005 select  last_name hire_date,department_id
006 FROM	 employees
007 WHERE  department_id =(select  department_id
008											 FROM	  employees
009											 WHERE   last_name like 'Zlotkey')
010 AND		 last_name not like 'Zlotkey';
011 
012 
013 2. create a query to display the employee numbers and last names of all employees who earn more than the average salary. Sort the results in ascending Order of salary.
014 select   last_name ,employee_id,salary
015 FROM	  employees
016 WHERE   salary > (select  avg(salary)
017								    FROM    employees)
018 
019 3. Write a query that displays the employee numbers and last names of all employees who work in a department with any employee whose last name contains a u.
020 select    employee_id,last_name
021 FROM	    employees
022 WHERE	 department_id in(select  department_id
023												  FROM	  employees
024												  WHERE   last_name like'%u%');
025 
026 
027 4. Display the last name, department number, and job ID of all employees whose department location ID is 1700
028 select    last_name,department_id,job_id
029 FROM	   employees
030 WHERE    department_id in(select   department_id
031												 FROM	  departments
032												 WHERE    location_id =1700);
033 
034 5. Display the last name and salary of every employee who reports to King.
035 
036 
037 
038 6. Display the department number, last name, and job ID for every employee in the Executive department.
039 select  department_ id , last _ name ,job_id
040 FROM	  employees
041 WHERE   job_id  like 'AD_%';
042 
043 7. Modify the query in lab6_3.sql to display the employee numbers, last names, and salaries of all employees who earn more than the average salary and who work in a department with any employee with a u in their name.
044 select   employee _ id, last_ name , salary , department _id
045 FROM	  employees
046 WHERE   salary >(select   avg(salary)
047								  FROM	  employees)
048 AND		 department _ id in (select    department _ id
049												   FROM	   employees
050												  WHERE	 last_name like '%u%');
051 
052 
053 
054 
055 Chapter 8
056 
057 Practice 8
058 1. Run the statement in the lab8_1.sql script to build the MY_EMPLOYEE table to be used for the lab.
059 create table my_employees
060				 (id number(4),last_name varchar2(25),
061				 first_name varchar2(25),userid varchar2(8),
062				 salary number(9,2));
063 
064 
065 2. Describe the structure of the MY_EMPLOYEE table to identify the column names.
066 DESC  my_employees
067 
068 3. Add the first row of data to the MY_EMPLOYEE table from the following sample data. Do not list the columns in the insert clause.
069 insert  INTO my_employees
070 VALUES		 (1,'Patel','Ralph', 'rpatel' ,895);
071 
072 4. Populate the MY_EMPLOYEE table with the second row of sample data from the preceding list.
073 This time, list the columns explicitly in the insert clause.
074 insert    INTO my_employees(id,last_name ,first_name,userid,salary)
075 VALUES (2,' Dancs', 'Betty',' bdancs', 860);
076 
077 5. Confirm your addition to the table.
078 select  *
079 FROM	 my_employees;
080 
081 6. Write an insert statement in a text file named loademp.sql to load rows into the MY_EMPLOYEE table. Concatenate the first letter of the first name and the first seven characters of the last name to produce the user ID.
082 insert    INTO  my_employees
083 VALUES			  (&id,' &frist_mame',
084								 '&last_name','lower(substr(&frist_name,1,1))||
085							    'lower( substr(&last_name,1,7)) ', &salary);
086 
087 
088 
089 7. Populate the table with the next two rows of sample data by running the insert statement in the
090 script that you created.
091 insert    INTO  my_employees
092 VALUES			  (&id,' &frist_mame',
093								 '&last_name','lower(substr(&frist_name,1,1))||
094							    'lower( substr(&last_name,1,7)) ', &salary);
095 
096 
097 8. Confirm your additions to the table.
098 select  *
099 FROM	 my_employees;
100 
101 9. Make the data additions permanent.
102 COMMIT;
103 
104 10. Change the last name of employee 3 to Drexler.
105 update   my_employees
106 SET		    last_name = 'Drexler'
107 WHERE	 id=3;
108 
109 11. Change the salary to 1000 for all employees with a salary less than 900.
110 update   my_employees
111 SET		    salary= 1000
112 WHERE    salary <900;
113 
114 12. Verify your changes to the table.
115 select *
116 FROM	 my_employees;
117 
118 13. delete Betty Dancs from the MY_EMPLOYEE table.
119 delete FROM  my_employees
120 WHERE				 first_name like 'Betty';
121 
122 14. Confirm your changes to the table
123 select *
124 FROM	 my_employees;
125 
126 
127 
128 15. Commit all pending changes.
129 COMMIT;
130 
131 16. Populate the table with the last row of sample data by modifying the statements in the script that you created in step 6.
132  Run the statements in the script.
133 insert    INTO  my_employees
134 VALUES			  (&id,' &frist_mame',
135								 '&last_name','lower(substr(&frist_name,1,1))||
136							    'lower( substr(&last_name,1,7)) ', &salary);
137 
138 17. Confirm your addition to the table.
139 select *
140 FROM	 my_employees;
141 
142 18. Mark an intermediate point in the processing of the transaction.
143 SAVEPOINT  b;
144 
145 19. Empty the entire table.
146 delete FROM  my_employees
147 
148 20. Confirm that the table is empty.
149 select *
150 FROM	 my_employees;
151 
152 21. Discard the most recent delete operation without discarding the earlier insert operation.
153 ROLLBACK  TO b;
154 
155 22. Confirm that the new row is still intact.
156 select *
157 FROM	 my_employees;
158 
159 23. Make the data addition permanent.
160 COMMIT;
161 
162 
163 
164 
165 Chapter 9
166 
167 Practice 9
168 1. create the DEPT table based on the following table instance chart.
169 Confirm that the table is created.
170 create  TABLE   dept
171					   (ID number(7),NAME varchar2(25));
172 DESC		  dept;
173 
174 2. Populate the DEPT table with data from the DEPARTMENTS table. Include only columns that you need.
175 create   TABLE  dept
176 AS				  select department_id,department_name
177					    FROM	 departments;
178 
179 3. create the EMP table based on the following table instance chart.
180 Confirm that the table is created
181 create  TABLE  emp (id number(7),last_name varchar2(25),
182				    first_name varchar2(25),dept_id number(7));
183 DESC emp;
184 
185 4. Modify the EMP table to allow for longer employee last names. Confirm your modification.
186 ALTER  TABLE  emp
187 MODIFY		 (last_name varchar2(50));
188 DESC   emp;
189 
190 5. Confirm that both the DEPT and EMP tables are stored in the data dictionary. (Hint: USER_TABLES)
191 select  table_name
192 FROM	 user_tables;
193 
194 
195 
196 
197 
198 
199 
200 6. create the EMPLOYEES2 table based on the structure of the EMPLOYEES table. Include only the EMPLOYEE_ID, FIRST_NAME, LAST_NAME, SALARY, and DEPARTMENT_ID columns.
201  Name the columns in your new table ID, FIRST_NAME, LAST_NAME, SALARY , and DEPT_ID, respectively.
202 
203 
204 create  TABLE employees2
205 AS	   select employee_id "ID",first_name "FIRST_NAME" ,
206			 last_name "LAST_NAME",
207			 salary "SALARY" ,department_id "DEPT_ID"
208 FROM   employees;
209 DESC    employees2
210 
211 7. drop the EMP table.
212 drop  TABLE  emp;
213 
214 8. Rename the EMPLOYEES2 table as EMP.
215 RENAME employees TO emp;
216 
217 9. Add a comment to the DEPT and EMP table definitions describing the tables. Confirm your additions in the data dictionary.
218 COMMENT  ON  TABLE emp
219 IS	 'employees information';
220 COMMENT  ON TABLE dept
221 IS	 'department information';
222 select *
223 FROM	 user_tab_comments;
224 
225 10. drop the FIRST_NAME column from the EMP table. Confirm your modification by checking the description of the table.
226 ALTER  TABLE    emp
227 drop  COLUMN  first_name;
228 DESC  emp
229 
230 
231 
232 11. In the EMP table, mark the DEPT_ID column in the EMP table as UNUSED. Confirm your modification by checking the description of the table.
233 ALTER  TABLE emp
234 SET    UNUSED (dept_id);
235 DESC emp;
236 
237 12. drop all the UNUSED columns from the EMP table. Confirm your modification by checking the description of the table.
238 ALTER  TABLE emp
239 drop  UNUSED columns;
240 DESC emp


#3 عزيز القادري

عزيز القادري

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تاريخ المشاركة 12 January 2012 - 04:21 PM

جزاك الله خير وزادك الله علماً

صورة


#4 mohamed hosary

mohamed hosary

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  • المنصب الحالي:Oracle Developer

تاريخ المشاركة 12 January 2012 - 11:00 PM

جزاك الله كل خير

Oracle Developer


#5 محمد بطاله

محمد بطاله

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تاريخ المشاركة 13 January 2012 - 12:54 AM

جزاك الله خيرااا

يَا أَيُّهَا الَّذِينَ آمَنُوا اتَّقُوا اللَّهَ وَآمِنُوا بِرَسُولِهِ يُؤْتِكُمْ كِفْلَيْنِ مِن رَّحْمَتِهِ وَيَجْعَل لَّكُمْ نُورًا تَمْشُونَ بِهِ وَيَغْفِرْ لَكُمْ وَاللَّهُ غَفُورٌ رَّحِيمٌ


#6 medani

medani

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تاريخ المشاركة 13 January 2012 - 11:34 AM

مشاركة اكثر من راااااااااااااااااااائعة
جزاك الله خير وزادك علما وان شاء الله فى ميزان حسناتك

#7 oracle9

oracle9

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تاريخ المشاركة 15 January 2012 - 10:24 AM

هل من مشاكل تواجهكم يا أخوانى فى جمل الإستعلام
أنا تحت أمركم

#8 Ahmad.Hasan

Ahmad.Hasan

    مشرف قسم الـ Forms & Reports والمشاريع

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تاريخ المشاركة 15 January 2012 - 10:37 AM

جزاك الله كل خير...

.Oracle E-Business Suite R12 HCM Certified Implementation Specialist

 

.Oracle Fusion Human Capital Management 11g Human Resources Certified Implementation Specialist