Thanks all.
And after these two days I have understood how to create a index to help my
query to run more faster,also I found that use 'explain' before a query just
get a result only provide reference cause it doesn't show the really result
how the query is run,for example the 'key' item.
That is
Hello,
I'm not very experienced with MySql and I'm building a website that deal with
user profiles. User profiles involves lists of preferences to choose from (for
example, your contry, your profession, etc.), so the user chooses from a
drop-down lists. The most natural solution for this is to
This is makes more sense because it causes a SQL developer
to group the joined tables in a logical manner.
- Original Message -
From: wang shuming <[EMAIL PROTECTED]>
To: mysql@lists.mysql.com
Sent: Friday, November 3, 2006 8:04:20 PM GMT-0500 US/Eastern
Subject: left join , Unknown column
Thanks for the help. As often happens in these situations, a little
searching and experimentation got me a workable answer before any replies
arrived.
In my case, I was able to use the LOCATE operator, i.e.
select idx, lname,schedalonerscs from usrs where LOCATE(",7,",
schedalonerscs) != 0;
I'
Dave, you could just use a LIKE statement
SELECT *
FROM table
WHERE mystring LIKE "%,2,%"
Of course if it is at the beginning or end, or the only item, it may
look like '2' '2,' or ',2' and not ',2,' so this would work in all
cases I think:
WHERE (mystring LIKE "%,2,%" OR mystring LIKE "2,%" OR
Hi,
I'm a little bit lost on MySQL statements.
I am implementing a many:many relationship using a string (wrong way, I
know, but I'm trying to keep the design simple and this is a trivial
application).
In a VARCHAR field, I will store a comma-delimited set of integers
corresponding to the indic
http://linux.inet.hr/poll_favorite_database.html
I don't think so. Do something!
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