SQL How to replace values of select return?

Question

In my database (MySQL) table, has a column with 1 and 0 for represent true and false respectively.

But in S

Answer 1

You have a number of choices:

1. Join with a domain table with TRUE, FALSE Boolean value.

2. Use (as pointed in this answer)

   SELECT CASE WHEN hide = 0 THEN FALSE ELSE TRUE END FROM
   

   Or if Boolean is not supported:

   SELECT CASE WHEN hide = 0 THEN 'false' ELSE 'true' END FROM
   

Answer 2

You can use casting in the select clause like:

SELECT id, name, CAST(hide AS BOOLEAN) FROM table_name;

Answer 3

I got the solution

   SELECT 
   CASE status
      WHEN 'VS' THEN 'validated by subsidiary'
      WHEN 'NA' THEN 'not acceptable'
      WHEN 'D'  THEN 'delisted'
      ELSE 'validated'
   END AS STATUS
   FROM SUPP_STATUS

This is using the CASE This is another to manipulate the selected value for more that two options.

Answer 4

I saying that the case statement is wrong but this can be a good solution instead. If you choose to use the CASE statement, you have to make sure that at least one of the CASE condition is matched. Otherwise, you need to define an error handler to catch the error. Recall that you don’t have to do this with the IF statement.

SELECT if(hide = 0,FALSE,TRUE) col FROM tbl; #for BOOLEAN Value return

or

SELECT if(hide = 0,'FALSE','TRUE') col FROM tbl; #for string Value return

Answer 5

You can do something like this:

SELECT id,name, REPLACE(REPLACE(hide,0,"false"),1,"true") AS hide FROM your-table

Hope this can help you.

Answer 6

Replace the value in select statement itself...

(CASE WHEN Mobile LIKE '966%' THEN (select REPLACE(CAST(Mobile AS nvarchar(MAX)),'966','0')) ELSE Mobile END)