12 окт. 2014 г. · This function allow you to extract a part from a string: SELECT first_name FROM table WHERE SUBSTR(last_name, 2, 1) = 'o'

20 сент. 2015 г. · SELECT ename AS "Employee Name" FROM emp WHERE UPPER(SUBSTR(ename, 2, 1)) BETWEEN 'A' AND 'M'; The UPPER function is not necessary if your data is in all caps.

9 апр. 2013 г. · My theory was to locate the row number of the first "ProjTemp" and then insert a new row where the project is 20.

26 июн. 2015 г. · I want to put a condition in my query where I have a column that should contain second position as an alphabet. How to achieve this?

14 июн. 2021 г. · sql - Display all the names of the employees whose names having letter A as 2nd occurrence without using like operator - Stack Overflow.

6 июл. 2012 г. · =MID(A1,3,1) = "i" Should work, you don't need to use IF , the evaluation using the equals will return either TRUE or FALSE.

27 мар. 2012 г. · I need a solution so that I can find, lets say names, between A-F, INCLUDING all names that start with F. If you use BETWEEN or A >= value <= F ...

18 апр. 2018 г. · Iam trying to find names, whose second character is an 'A'. I wrote the following SELECT for it: SELECT name FROM student WHERE REGEXP_LIKE(name,'[az]a.*','i');

21 мая 2012 г. · I want select the records where the column1 contains '_'(underscore) as the second character.For example, I want to select the records with column1 like A_John ...

3 мар. 2023 г. · My table in SQL has a column with persons name and surname like "Mike Baldwin", "Sara Larson" etc. and need to output all the persons whose name starts with "M ...