MySQL Install on OS X Leoperd
Hi There, Has anyone been successful installing and running MySQL (current version) on the new Mac OS (Leopard)? __ Craig Hoffman iChat/AIM: m0untaind0g __ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: MySQL Install on OS X Leoperd
This worked -- To fix the mySQL socket error, start mysql in terminal by doing this... sudo /usr/local/mysql/bin/safe_mysqld close Terminal, then open it again and put in these two... sudo mkdir /var/mysql/ sudo ln -s /tmp/mysql.sock /var/mysql/mysql.sock Restart Apache Should work until they fix the pref pane. __ Craig Hoffman iChat/AIM: m0untaind0g __ On Oct 26, 2007, at 3:59 PM, Craig Hoffman wrote: Hi There, Has anyone been successful installing and running MySQL (current version) on the new Mac OS (Leopard)? __ Craig Hoffman iChat/AIM: m0untaind0g __ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/mysql? [EMAIL PROTECTED] -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
converting access (.mdb) files...
Anyone know a way to convert MS Access DB files (.mdb) into a format MySQL can import? I'm on a Mac hence I don't have Access. Best, CH ___ Craig Hoffman www.eclimb.net [EMAIL PROTECTED] iChat / AIM: m0untaind0g ___ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Error - #1251
Thanks - I got it working. On Jan 18, 2005, at 12:35 AM, Mattias J wrote: At 2005-01-18 04:53, Craig Hoffman wrote: I upgrade to 4.1.9 last night and I keep getting this error in PhpMyAdmin after I changed the root password (yes I changed both pw's). Does anyone have a solution on how to fix this? I'm running a Mac 10.3.x #1251 - Client does not support authentication protocol requested by server; consider upgrading MySQL client If you try searching for 1251 authentication on mysql.com, the first hit is this page, which will provide the answer http://dev.mysql.com/doc/mysql/en/Old_client.html -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED] -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Like statement help
Hi There, I have web form where a user can search certain fields and then have them displayed aggregated. For example, find all the routes I climbed with partner A in area(s) ALL (% - wildcard) between date1 and date2 -- so on. See below for the complete query. I'm using pull down menu's and when a the user does not select something the default is ALL or option value='%' ALL /option for all. Then all the other options are listed. Should I be using the % as a wildcard? I would like it to work if one, two, three... or all fields are selected. Obviously, the more options you select the more detailed your search becomes and vice versa. Any thoughts on what could be wrong with my query? Any help would be appreciated. Thanks - Craig Here is my query query = SELECT routes.*, users.email, users.fname, users.lname, users.user_id, ranking.* FROM routes, users, ranking WHERE email='$email' AND area LIKE '%$area%' AND partner LIKE '%$partner%' AND id BETWEEN '$rating1' AND '$rating2' AND additional_rating IS NOT NULL LIKE '%$additional_rating%' AND pitchs LIKE '%$pitchs%' AND `type` LIKE '%$type%' AND style LIKE '%$style%' AND stars LIKE '%$stars%' AND fall LIKE '%$fall%' AND popular LIKE '%$popular%' AND date_climbed BETWEEN '$date_climbed1' AND '$date_climbed2' AND routes.rating = ranking.rating AND routes.user_id = users.user_id GROUP BY route_count ORDER BY area, date_climbed DESC; -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Error - #1251
I upgrade to 4.1.9 last night and I keep getting this error in PhpMyAdmin after I changed the root password (yes I changed both pw's). Does anyone have a solution on how to fix this? I'm running a Mac 10.3.x #1251 - Client does not support authentication protocol requested by server; consider upgrading MySQL client -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: AVG Function
Mark,
Yes its close and thank you. The problem I am having is I am able to
generate the correct ranking.id for that particular user but I can't
seem to make it equal the ranking.rating.
ID ranking.rating
9 = 5.6 (example)
Here's my query:
SELECT routes.user_id, ranking.rating, ROUND(AVG(ranking.id), 0)
avg_ranking, users.username, users.user_id, routes.rating FROM ranking,
routes, users WHERE username='$username' AND routes.user_id =
users.user_id AND ranking.rating = routes.rating GROUP BY
routes.user_id
//echo some stuff out
echo(td align='right'.$row[ranking.rating]. /td);
I know I need to make the avg_ranking or the ranking.id =
ranking.rating but I can't seem to get it work. Any more suggestions?
Again thanks for all your help.
-- Craig
On Aug 25, 2004, at 12:48 AM, Mark C. Stafford wrote:
Hi Craig,
It sounds to me as though you're practically there. My syntax is a
little different when I do groupings. Here's what I got from your
question. Is it what you wanted...or close enough to get you where
you're going?
Good luck,
Mark
drop table if exists test.ranking;
create table test.ranking (
id int(3) unsigned auto_increment primary key
, rating varchar(5)
);
insert into test.ranking(rating) values('5.0');
insert into test.ranking(rating) values('5.1');
insert into test.ranking(rating) values('5.2');
insert into test.ranking(rating) values('5.3');
insert into test.ranking(rating) values('5.3a');
insert into test.ranking(rating) values('5.3b');
drop table if exists test.routes;
create table test.routes (
user_id int(3) unsigned
, rating varchar(5)
);
insert into test.routes(user_id, rating) values(1, '5.2');
insert into test.routes(user_id, rating) values(1, '5.3');
insert into test.routes(user_id, rating) values(1, '5.3a');
SELECT routes.user_id
, @avg:=ROUND(AVG(ranking.id), 0) avg_ranking
FROM test.ranking
, test.routes
WHERE routes.user_id = 1
AND ranking.rating = routes.rating
GROUP BY routes.user_id
;
SELECT *
FROM test.ranking
WHERE id = @avg
;
+++
| id | rating |
+++
| 4 | 5.3|
+++
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AVG Function
Hey Everyone, I can you some assistance on this query. I have three tables one is called ranking and the other is called routes and finally the users table. The users table is not really important. The ranking table looks like this: id rating 1 5.0 2 5.1 3 5.2 4 5.3 5 5.3a 6 5.3b and so on... The routes table looks like this: user_id route rating 1 somename5.2 1 5.3 1 5.3a Here's my query: SELECT ranking.rating, AVG(id), users.username, users.user_id, routes.rating, routes.user_id FROM ranking, routes, users WHERE username='$username' AND users.user_id = routes.user_id AND ranking.rating = routes.rating GROUP BY username What I am trying to do is find the average rating for this user. For example: 5.2 = 3 5.3 = 4 5.3a = 5 ___ 3 + 4 + 5 = 12 / 3 = 4 So 4 = 5.3 The average for this user would be 5.3. Any help would be most appreciated. Craig -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
ORDER BY rand()
Hey Folks, I have a query where it pulls random data and display's it. SELECT route_photo, route, route_count, area FROM routes WHERE ORDER BY RAND() LIMIT 1 The query works fine, however, the route_photo field is partially populated. This results in just a route name appearing but no photo. How can I change the query to only pull up routes that have a route_photo listed in the DB? Thanks, CH -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Between Operator
Hey Everyone, I have query where one selects an style then a area and finally rating. When some selects a rating they select a range of ratings. For example: Style: Traditional Area: Yosemite Rating: From: 5.5 To: 5.10c This should pull up all the rock climbs that are in Yosemite, that are traditional style and are between the rating 5.5 to 5.10c. Here is my query: SELECT * FROM routes, users WHERE area='$area' AND style='$style' BETWEEN rating='[$rating1]' AND rating='[$rating2]' GROUP BY route ORDER BY rating ASC ; For some reason which I am not seeing, this query is not doing what it should be doing. Does anyone have any suggestions? Thanks, Craig -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Between Operator
Thanks everyone for helping out.I took Michael's advice and made a new table called ranking and two columns. It defiantly cleared some things up but I am still having issues using the BETWEEN operator. I just need to pull up everything BETWEEEN 10 and 18 and it keeps adding additional rows. Suggestions? What am I doing wrong? Here is my query: SELECT area, style, route, stars, date_climbed, ranking.* FROM routes, ranking WHERE ranking.id = ranking.rating BETWEEN ranking.id '10' AND ranking.id = '18' AND routes.rating = ranking.rating AND area = 'Eldorado Canyon' AND style = 'Traditonal' GROUP BY route ORDER BY id DESC Thanks, Craig On Jul 9, 2004, at 1:17 PM, Pete Harlan wrote: On Fri, Jul 09, 2004 at 09:39:02AM -0500, Craig Hoffman wrote: Style: Traditional Area: Yosemite Rating: From: 5.5 To: 5.10c ... SELECT * FROM routes, users WHERE area='$area' AND style='$style' BETWEEN rating='[$rating1]' AND rating='[$rating2]' GROUP BY route ORDER BY rating ASC ; For some reason which I am not seeing, this query is not doing what it should be doing. Does anyone have any suggestions? For starters your between syntax isn't correct (but is parsable in ways you didn't want). You probably want: select * fromroutes, users where area = '$area' and style = '$style'and rating between '$rating1' and '$rating2' group by route order by rating As others have pointed out, your ratings aren't something MySQL will know how to order. That's a separate problem (and more difficult to solve), but the between syntax is also one. --Pete -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED] -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Long numbers
Hi There: I have a query where I get an average of weekly miles. For example: Average Weekly Miles: 3.4285714285714 How would I get this number into a more read format? Such as 3.42. I have tried the truncate but I can't seem to get it to work. Does any one have any suggestions? SELECT TURNCATE(AVG(distance)), rest of the query Thanks - CH __ Craig Hoffman - eClimb Media v: (847) 644 - 8914 f: (847) 866 - 1946 e: [EMAIL PROTECTED] w: www.eclimb.net _ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Long numbers
that worked - Thanks a bunch!
CH
__
Craig Hoffman - eClimb Media
v: (847) 644 - 8914
f: (847) 866 - 1946
e: [EMAIL PROTECTED]
w: www.eclimb.net
_
On Apr 11, 2004, at 3:14 PM, Mike Blezien wrote:
I think your looking for the ROUND(N,D) function, IE:
select round('3.4285714285714',2) as miles;
3.43 miles
MikemickaloBlezien
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Craig Hoffman wrote:
Hi There:
I have a query where I get an average of weekly miles.
For example:
Average Weekly Miles:
3.4285714285714
How would I get this number into a more read format? Such as 3.42. I
have tried the truncate but I can't seem to get it to work. Does any
one have any suggestions?
SELECT TURNCATE(AVG(distance)), rest of the query
Thanks - CH
__
Craig Hoffman - eClimb Media
v: (847) 644 - 8914
f: (847) 866 - 1946
e: [EMAIL PROTECTED]
w: www.eclimb.net
_
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Truncating long numbers
Hi there, I have this query (see below) SELECT SUM(distance + date), DATE_FORMAT(time_upload, '%M') FROM traininglog_client, users WHERE username='$username' AND DATE_SUB(CURDATE(),INTERVAL 1 MONTH) = time_upload AND users.user_id = traininglog_client.user_id GROUP BY username and it returns this: (see below) SUM( distance + date ) DATE_FORMAT( time_upload, '%M' ) 140282467.4 March How can I make the 140282467.4 into a more readable number? Such as 14.54. My distance field in MySQL is a double, if that helps. __ Craig Hoffman - eClimb Media v: (847) 644 - 8914 f: (847) 866 - 1946 e: [EMAIL PROTECTED] w: www.eclimb.net _ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: Truncating long numbers
Yes your right - I thought the numbers in the DB were something else. __ Craig Hoffman - eClimb Media v: (847) 644 - 8914 f: (847) 866 - 1946 e: [EMAIL PROTECTED] w: www.eclimb.net _ On Apr 2, 2004, at 11:11 AM, Paul DuBois wrote: At 10:44 -0600 4/2/04, Craig Hoffman wrote: Hi there, I have this query (see below) SELECT SUM(distance + date), DATE_FORMAT(time_upload, '%M') FROM traininglog_client, users WHERE username='$username' AND DATE_SUB(CURDATE(),INTERVAL 1 MONTH) = time_upload AND users.user_id = traininglog_client.user_id GROUP BY username and it returns this: (see below) SUM( distance + date ) DATE_FORMAT( time_upload, '%M' ) 140282467.4 March How can I make the 140282467.4 into a more readable number? Such as 14.54. Huh? How can 140282467.4 become 14.54? -- Paul DuBois, MySQL Documentation Team Madison, Wisconsin, USA MySQL AB, www.mysql.com MySQL Users Conference: April 14-16, 2004 http://www.mysql.com/uc2004/ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED] -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
