Display Employees Names Who Have 2L's, SQL Query

My code so far is:

select ename from emp where ename =ll%;

The question开发者_JS百科 is, display all the names of all employees who have 2Ls in their name and are in department 30 or their manager is 7782;

I tried my code but it is giving me errors, I'm practicing for my test.

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You want to use the like command and sandwich your ll's w/ % wild cards:

select ename from emp where ename like '%ll%'

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Do you mean "two l's like hello", or "two l's like lala"? If the second, this might work:

SELECT 
  ename FROM emp 
WHERE 
  ename LIKE '%l%l%' 
AND  
  (department = 30 AND manager = 7782) 

If the first, change the LIKE to %ll% instead.

If department and manager are CHAR/VARCHAR instead of numeric, you'll need single quotes around them as well.

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Select * from emp
Where instr(ename, 'l', 2, 2)>1 and
(Department=30 or manager=7782);

Here in instr function I start search from 2nd position to make it efficient. N no. Of occurance will be 2 as his query all about the 2nd 'l'.

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SELECT ename 
  FROM emp 
 WHERE (ename LIKE '%l%l%' 
    OR ename LIKE '%ll%')
   AND (department = 30 AND manager = 7782) ;

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SELECT ename FROM emp
WHERE instr(ename,'L',1,2)>1 AND department =30 AND manager = 7782

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Try this:

SELECT *
FROM EMP
WHERE LENGTH(ENAME)-LENGTH(REPLACE(ENAME,'L'))>=2;

It will work.