Please note: when you write ‘@var’ (with the apostrophes) it means a string with the letters ‘@’,'v’,'a’,'r’.

The SELECT already returns the hashed value, so if I don´t use “BY PASSWORD”, MySQL sets the hashed value itself as the password:

SELECT PASSWORD FROM USER WHERE USER = ‘jon’ INTO @var;
SELECT @var; –> *81F5E21E35407D884A6CD4A731AEBFB6AF209E1B
CREATE USER ‘jane’@’127.0.0.1′ IDENTIFIED BY ‘@var’;
SELECT PASSWORD FROM USER WHERE USER = ‘jane’ ; –> *33EF0CD2D187BCD7DBFFDE2943960E094684DAE6

As you can see, the hash values of the two passwords differ, but they should be the same. Not using apostrophes with @var only gives a MySQL error…

I want to create a user with the password from another user like this:
SELECT PASSWORD FROM USER WHERE USER = ‘jon’ INTO @var;
CREATE USER ‘jane’@'localhost’ IDENTIFIED BY PASSWORD @var;

Unfortunately, this only gives me “You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ‘@var’ at line 1″.

I have tried other ways but didn´t succeed. Is what I´m trying to do even possible?

Lars

As a general rule, you should never assign a value to a user variable and read the value within the same statement. You might get the results you expect, but this is not guaranteed. The order of evaluation for expressions involving user variables is undefined and may change based on the elements contained within a given statement; in addition, this order is not guaranteed to be the same between releases of the MySQL Server. In SELECT @a, @a:=@a+1, …, you might think that MySQL will evaluate @a first and then do an assignment second. However, changing the statement (for example, by adding a GROUP BY, HAVING, or ORDER BY clause) may cause MySQL to select an execution plan with a different order of evaluation.

Does anyone know a good way to fix this?

1.so how to insert variable in mysql using java..

for ex: int a=20

stmt.executeUpdate(“SET @b:=a INSERT INTO product VALUES (b);”);

it gives error…so how to insert..

2.Similarly how to retrieve and assign data from database to int , string variable and applet label in JAVA…”

SELECT *
FROM (SELECT ( @counter := @counter + 1 ) & 1 AS x,
w.*
FROM (SELECT @counter := 0) v,
(SELECT tag, COUNT(*)AS c
FROM categories_tags
INNER JOIN categories USING(categories_id)
WHERE `status` AND `list`
GROUP BY tag
ORDER BY c)w)x
ORDER BY x, IF(x, -c, c), IF(x, -CHAR_LENGTH(tag), CHAR_LENGTH(tag))

To add to the answer Hazan provided, it’s more of a phpmyadmin ‘thing’ getting the [BLOB] response than MySQL itself. You can use type casting to get the desired results.

asjain01,
wrapping your select statement in brackets removes the error.

set @h = (select count(*) from reportx);