I want to show first_name, last_name columns and make UPPER-CASE LETTERS for those entries who have the letter 's' in the last name.
I am getting
ORA-00933: SQL command not properly ended
error. Anyone has an idea where I go wrong and How do I fix it?
SELECT first_name, last_name FROM CUSTOMERS
WHERE last_name LIKE "S"
UPDATE CUSTOMERS
SET
first_name = UPPER(first_name)
last_name = UPPER(last_name)
You seem to want:
UPDATE CUSTOMERS
SET first_name = UPPER(first_name), last_name = UPPER(last_name)
WHERE last_name LIKE '%S%'
It is unclear whether you want to match on upper-case or lower-case "s" in the name. If you want both, then:
WHERE UPPER(last_name) LIKE '%S%'
try putting "%S" instead of "S". If you put "S" it will only show you results for surnames that are just the letter S
Try following with subquery:-
UPDATE CUSTOMERS
SET ( first_name , last_name ) = (SELECT UPPER(first_name) , UPPER(last_name) FROM CUSTOMERS )
WHERE UPPER(last_name) LIKE 'S'
However, if you want all last_name starting with S then add wildcard such as
WHERE UPPER(last_name) LIKE 'S%'
Related
In a table I have columns FIRST_NAME and LAST_NAME. I have to concatenate the two names and retrieve names containing more than 12 characters.
I tried the following query:
select *
from (select first_name, last_name
from customer as name
)
where length(NAME) = 12
Select *
from (
select first_name||last_name as name
from customer
)
where length(name)>12
You need only one SELECT
SELECT first_name || last_name AS name
FROM customer
WHERE LENGTH(first_name||last_name) > 12
Optionally TRIM name and last_name from spaces.
As an alternative, if you rewrite your question to check if the total length of first name and last name is more than 12, you probably get a different response from people with a more efficient code. It is not necessary to make Oracle actually concatenate the columns.
Programmers are too literal these days. :)
You are not concatenating the first name and last name.
Also no need to write an inline view, below is the query.
select FIRST_NAME || LAST_NAME as NAME
from CUSTOMER
where length(FIRST_NAME || LAST_NAME) > 12;
I have a phone_book table. It has columns named id, first_name, last_name and phone. Im trying to delete all contacts with the first name of Jonathan and the last name of Luna.
My code is this
DELETE FROM phone_book WHERE first_name, last_name = "Jonathan" , "Luna";
Im getting a syntax error near the comma. Please help thanks
Set the WHERE conditions separated by AND:
DELETE FROM phone_book WHERE first_name = "Jonathan" AND last_name = "Luna";
I know it was already answered but it doesn't work for me.
So quick introduce:
I have table called swimmers. I'll insert new record to it which fulfils all columns:
INSERT INTO swimmers(id, first_name, last_name, age, gender)
VALUES(9,'Maria','Spolsky',34,'Female');
Now I want to find records in swimmers table which first_name DOES NOT end with letter a and where gender is Female
So I wrote SQL query:
SELECT first_name, last_name
FROM swimmers
WHERE first_name NOT LIKE '%a' AND gender = 'Female'
But it does return Maria which we just added. It basically returns every female
I'm using ORACLE iSQL plus.
EDIT:
I tried to use substr(first_name, -1) = 'a' but it turned out that -1 is empty, because I use varchar(20) and names are usually smaller than 20 characters
EDIT2:
I tried to find issue in data type. I used char(20) for first_name.
I changed it into varchar2(20):
ALTER TABLE swimmers
MODIFY first_name varchar2(20);
but didn't solve the issue
EDIT 3:
Changing NOT LIKE to WHERE first_name LIKE '%a' returns no rows. I believe issues lies inside data type and empty spaces from the end of the first_name to the end of reserved space for string (names has ~10 characters and I use `varchar2(20))
Appearantly oracles is adding spaces behind the name and that's why '%a' doesn't work, either change your column definition to a varchar (so oracle doesn't add the extra spaces) or trim the spaces out of your name like this:
SELECT first_name, last_name
FROM swimmers
WHERE NOT trim(first_name) LIKE '%a' AND gender = 'Female';
The whole question is:
Rewrite the query so that the user is prompted to
enter a letter that the last name starts with. For example, if the user enters “H" (capitalized) when prompted for a letter, then the output should show all employees whose last name starts with the letter “H.”
SELECT last_name
FROM employees
WHERE last_name like '%' = &Start_Letter
This wont work :(
This query should work:
SELECT last_name FROM employees WHERE last_name like '&Start_Letter%';
I have tested it in Oracle11g.
SELECT last_name FROM employees WHERE last_name like &Start_Letter + '%';
Depending on the requirements of your particular application, a more efficient solution might be to create a function based index:
CREATE INDEX emp_start_letter_idx
ON employees ( substr( last_name, 1, 1 ) );
Then your query would be:
SELECT last_name
FROM employees
WHERE substr(last_name, 1, 1) = '&Start_Letter';
Likewise, depending on your requirements, this may be a lot of unnecessary fuss!
Here is the problem I am working on:
Select all the Oracle database employees whose last names end with “s” Change the
heading of the column to read Possible Candidates.
I tried
SELECT last_name AS possible_candidates FROM * WHERE last_name LIKE '%s';
That returned the error :
ORA-00903: invalid table name
Unless I am reading the question wrong how do I check the entire database for something like this?
To answer the question which is asking for employee names not just last names you would select all employees with a last name that ends in S.
SELECT *
FROM employees
WHERE last_name LIKE '%s'
If the table has more than a first_name and last_name column you can do
SELECT first_name, last_name
FROM employees
WHERE last_name LIKE '%s'
Now to combine your two parts (Select employees and rename column)
Run one of the two queries above but add an AS statement as show in the example below.
SELECT *
AS Possible_Candidates
FROM employees
WHERE last_name LIKE '%s'
Based on feedback: this appears to be your answer...
Is there a table called employees? if that I would interpret the question as from the employees table. Otherwise, you could select table_name from All_TAB_COLS where column_name like = 'LAST_NAME' and then build a union from the resulting tables all of which have a field called 'LAST_NAME'