Re: LIKE problem with characters 'å' (norwe gian) and 'a' (mysql bug?)
I get incorrect result when searching for the norwegian character 'å' using LIKE. I get rows with 'a' in it, and visa versa if I search for 'a', I get results which has 'å' in it in addition to the ones with 'a'. Make sure that your table has: charset=utf8 collation=utf8_norwegian_ci And that every column ALSO has: charset=utf8 collation=utf8_norwegian_ci Notice that I am making 'utf8_norwegian_ci' up. I looked for it using my MySQL Query Browser but couldn't find it. As I'm from Sweden I've had similar problems (åäöÅÄÖ matched åaäÅÄAÖO) and setting as above but using (the existing) 'utf8_swedish_ci' worked in my case. -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
LIKE problem with characters 'å' (norwe gian) and 'a' (mysql bug?)
Dear List,
I get incorrect result when searching for the norwegian character 'å'
using LIKE. I get rows with 'a' in it, and visa versa if I search for
'a', I get results which has 'å' in it in addition to the ones with 'a'.
Example:
CREATE TABLE names (
name VARCHAR(255)
)ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO names VALUES
('Foo'), ('Bar'), ('Båt'), ('Bør'), ('Ære');
Now, searching gives me the following results:
mysql SELECT * FROM names WHERE name LIKE '%å%';
+--+
| name |
+--+
| Bar |
| Båt |
+--+
mysql SELECT * FROM names WHERE name LIKE '%a%';
+--+
| name |
+--+
| Bar |
| Båt |
+--+
Searching for strings with other norwegian characters seams to work:
mysql SELECT * FROM names WHERE name LIKE '%ø%';
+--+
| name |
+--+
| Bør |
+--+
I found that I may use
mysql SELECT * FROM names WHERE LOWER(name) LIKE BINARY LOWER('%å%');
which returns correct results, but this disables me from letting the
user do case sensitive searches.
Am I doing something wrong or stupid? Could this be a MySQL bug?
How do I know this isn't a problem with other utf-8 characters in other
languages?
I've searched in bug reports, but cannot find this exact problem.
Some additional information that might be useful:
mysql SELECT VERSION();
+--+
| VERSION()|
+--+
| 5.0.45-Debian_1ubuntu3.1-log |
+--+
mysql SHOW VARIABLES LIKE '%character%';
+--++
| Variable_name| Value |
+--++
| character_set_client | utf8 |
| character_set_connection | utf8 |
| character_set_database | utf8 |
| character_set_filesystem | binary |
| character_set_results| utf8 |
| character_set_server | utf8 |
| character_set_system | utf8 |
| character_sets_dir | /usr/share/mysql/charsets/ |
+--++
Thanks,
Magne Westlie
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Re: LIKE problem part II
Exporting and then importing the table helped, now both the regexp and the like query produce the same result. I saved the table MYI file and will keep experimenting with it, maybe I can come up with something useful. Thanks for the idea. Zoltan On Fri, 18 Nov 2005, Scott Haneda wrote: on 11/18/05 7:18 AM, Peczöli Zoltán at [EMAIL PROTECTED] wrote: The result of the second query matches that of the corresponding LIKE query, but the first seems to be correct. Any ideas what the problem might be? You are getting strange results. At this point I would suggest dumping the data, and looking at it in a editor to see if you can see what may be wrong. Also, reimport it back into a new test table and run your tests again. -- - Scott HanedaTel: 415.898.2602 http://www.newgeo.com Novato, CA U.S.A. -- 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 problem part II
I tried to investigate my previous problem with statements containing LIKE clause on a specific table. The problem was basically the following: mysql SELECT count(*) FROM user WHERE username LIKE 'o%'; +--+ | count(*) | +--+ |0 | +--+ 1 row in set (0.00 sec) mysql SELECT count(*) FROM user WHERE username LIKE 'ors%'; +--+ | count(*) | +--+ | 91 | +--+ 1 row in set (0.01 sec) I checked the table: mysql CHECK TABLE user; +---+---+--+--+ | Table | Op| Msg_type | Msg_text | +---+---+--+--+ | database.user | check | status | OK | +---+---+--+--+ 1 row in set (0.77 sec) So the table seems to be healthy. I queried the same as above but with REGEXP instead of LIKE: mysql SELECT count(*) FROM user WHERE username REGEXP ^o.*; +--+ | count(*) | +--+ | 801 | +--+ 1 row in set (0.19 sec) mysql SELECT count(*) FROM user WHERE username REGEXP ^ors.*; +--+ | count(*) | +--+ | 91 | +--+ 1 row in set (0.23 sec) The result of the second query matches that of the corresponding LIKE query, but the first seems to be correct. Any ideas what the problem might be? Zoltan -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: LIKE problem part II
on 11/18/05 7:18 AM, Peczöli Zoltán at [EMAIL PROTECTED] wrote: The result of the second query matches that of the corresponding LIKE query, but the first seems to be correct. Any ideas what the problem might be? You are getting strange results. At this point I would suggest dumping the data, and looking at it in a editor to see if you can see what may be wrong. Also, reimport it back into a new test table and run your tests again. -- - Scott HanedaTel: 415.898.2602 http://www.newgeo.com Novato, CA U.S.A. -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: LIKE problem?
Hello. Really, seems a bit weird for me. I suggest you to check your 'character_set_xxx' variables to be sure that there're no unnecessary translations from one encoding to another. If you're able to make a small repeatable test case, install on your Debian server second instance of MySQL (use official binaries) and check out if the problem remains with a new copy. [EMAIL PROTECTED] wrote: Recently I ran into a problem with 'LIKE' in mysql on Debian Sarge: mysql select VERSION(); +---+ | VERSION() | +---+ | 4.1.11-Debian_4sarge2-log | +---+ 1 row in set (0.00 sec) with the following table: CREATE TABLE `user` ( `id` int(10) unsigned NOT NULL auto_increment, `username` varchar(64) collate latin2_hungarian_ci default NULL, ... ) ENGINE=MyISAM DEFAULT CHARSET=latin2 COLLATE=latin2_hungarian_ci I get the following outputs: mysql select count(*) from user where username like 'o%'; +--+ | count(*) | +--+ |0 | +--+ 1 row in set (0.00 sec) mysql select count(*) from user where username like 'or%'; +--+ | count(*) | +--+ |0 | +--+ 1 row in set (0.00 sec) mysql select count(*) from user where username like 'ors%'; +--+ | count(*) | +--+ | 89 | +--+ 1 row in set (0.00 sec) So the number of usernames which match like 'o%' is zero, while the number of matching lines for like 'ors%' is 89. Moreover, the sum of the results of these two queries select count(*) from user where username like 'a%'; select count(*) from user where username not like 'a%' or username is null; is not the same for all letters of the alphabet: letter like not-like sum n 2304 59317 61621 o 0 60797 60797 p 3048 58573 61621 Any ideas? Zoltan -- For technical support contracts, goto https://order.mysql.com/?ref=ensita This email is sponsored by Ensita.NET http://www.ensita.net/ __ ___ ___ __ / |/ /_ __/ __/ __ \/ /Gleb Paharenko / /|_/ / // /\ \/ /_/ / /__ [EMAIL PROTECTED] /_/ /_/\_, /___/\___\_\___/ MySQL AB / Ensita.NET ___/ www.mysql.com -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: LIKE problem?
select count(*) from user where username like 'a%'; select count(*) from user where username not like 'a%' or username is null; is not the same for all letters of the alphabet: letter like not-like sum n 2304 59317 61621 o 0 60797 60797 p 3048 58573 61621 Sounds like a corrupt index. Try CHECK TABLE and REPAIR TABLE. Tried that, tried myisamchk as well, everything seems to be healthy, still the problem exists. Zoltan -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
LIKE problem?
Recently I ran into a problem with 'LIKE' in mysql on Debian Sarge: mysql select VERSION(); +---+ | VERSION() | +---+ | 4.1.11-Debian_4sarge2-log | +---+ 1 row in set (0.00 sec) with the following table: CREATE TABLE `user` ( `id` int(10) unsigned NOT NULL auto_increment, `username` varchar(64) collate latin2_hungarian_ci default NULL, ... ) ENGINE=MyISAM DEFAULT CHARSET=latin2 COLLATE=latin2_hungarian_ci I get the following outputs: mysql select count(*) from user where username like 'o%'; +--+ | count(*) | +--+ |0 | +--+ 1 row in set (0.00 sec) mysql select count(*) from user where username like 'or%'; +--+ | count(*) | +--+ |0 | +--+ 1 row in set (0.00 sec) mysql select count(*) from user where username like 'ors%'; +--+ | count(*) | +--+ | 89 | +--+ 1 row in set (0.00 sec) So the number of usernames which match like 'o%' is zero, while the number of matching lines for like 'ors%' is 89. Moreover, the sum of the results of these two queries select count(*) from user where username like 'a%'; select count(*) from user where username not like 'a%' or username is null; is not the same for all letters of the alphabet: letter like not-like sum n 2304 59317 61621 o 0 60797 60797 p 3048 58573 61621 Any ideas? Zoltan -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Re: LIKE problem?
Hi, Moreover, the sum of the results of these two queries select count(*) from user where username like 'a%'; select count(*) from user where username not like 'a%' or username is null; is not the same for all letters of the alphabet: letter like not-like sum n 2304 59317 61621 o 0 60797 60797 p 3048 58573 61621 Sounds like a corrupt index. Try CHECK TABLE and REPAIR TABLE. Regards, Jeremy -- Jeremy Cole MySQL Geek, Yahoo! Inc. Desk: 408 349 5104 -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
Like - Problem
Hello there,
I was wondering how I could make a specific type of search when the string
has more than one word.
Ex.-
String = Green Apple
Select * from fruits
where
(fruits.color like '%Green Apple%'
or fruits.type like '%Green Apple%')
What I thought was breaking the string in 2 words and compares each word
with the fields. The problem is that I can't control how many fields should
be compared.
Also don't know how to compare each word. The following syntax doesn't work:
---
Select * from fruits
where
(fruits.color like in ('%Green%', '%Apple%')
or fruits.type like in ('%Green%', '%Apple%')
---
Any ideas would be very thankful.
Cheer's
Rui Monteiro
Re: Like - Problem
This is the third time you have asked this, and it has been answered twice. Once is enough. Rui Monteiro wrote: Hello there, I was wondering how I could make a specific type of search when the string has more than one word. Ex.- String = Green Apple -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe:http://lists.mysql.com/[EMAIL PROTECTED]
RE: Like - Problem
Try the following sintax:
Select * from fruits
where
(fruits.color like '%Green%' or
fruits.color like '%Apple%')
or
(fruits.type like '%Green%' or
fruits.type like '%Apple%');
Regards,
Laercio.
-Original Message-
From: Rui Monteiro [mailto:[EMAIL PROTECTED]
Sent: quarta-feira, 22 de setembro de 2004 04:27
To: [EMAIL PROTECTED]
Subject: Like - Problem
Hello there,
I was wondering how I could make a specific type of search when the string
has more than one word.
Ex.-
String = Green Apple
Select * from fruits
where
(fruits.color like '%Green Apple%'
or fruits.type like '%Green Apple%')
What I thought was breaking the string in 2 words and compares each word
with the fields. The problem is that I can't control how many fields should
be compared.
Also don't know how to compare each word. The following syntax doesn't work:
---
Select * from fruits
where
(fruits.color like in ('%Green%', '%Apple%')
or fruits.type like in ('%Green%', '%Apple%')
---
Any ideas would be very thankful.
Cheer's
Rui Monteiro
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like problem
i need this consult select a.name from table1 a, table2 b where a.name like '%b.name%'; but i have a problem because Mysql not recognize the value of b.name - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: like problem
that would be because b.name is inside the ' ' and it is taken as a constant. -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:10 AM To: [EMAIL PROTECTED] Subject: like problem i need this consult select a.name from table1 a, table2 b where a.name like '%b.name%'; but i have a problem because Mysql not recognize the value of b.name - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: like problem
yes, but select a.name from table1 a, table2 b where a.name like '%'b.name'%'; is an error - Original Message - From: Mary Stickney [EMAIL PROTECTED] To: Jorge Martinez [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Thursday, August 29, 2002 1:26 PM Subject: RE: like problem that would be because b.name is inside the ' ' and it is taken as a constant. -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:10 AM To: [EMAIL PROTECTED] Subject: like problem i need this consult select a.name from table1 a, table2 b where a.name like '%b.name%'; but i have a problem because Mysql not recognize the value of b.name - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: like problem
ok try this select a.name from table1 a, table2 b where a.name like '%'+ b.name + '%'; -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:48 AM To: Mary Stickney; [EMAIL PROTECTED] Subject: RE: like problem yes, but select a.name from table1 a, table2 b where a.name like '%'b.name'%'; is an error - Original Message - From: Mary Stickney [EMAIL PROTECTED] To: Jorge Martinez [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Thursday, August 29, 2002 1:26 PM Subject: RE: like problem that would be because b.name is inside the ' ' and it is taken as a constant. -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:10 AM To: [EMAIL PROTECTED] Subject: like problem i need this consult select a.name from table1 a, table2 b where a.name like '%b.name%'; but i have a problem because Mysql not recognize the value of b.name - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
Re: like problem
Hi,
Mary gives you a sugestion about how can you use field(b.name) in this case.
You can try this :
select a.name from table1 a, table2 b where a.name like
CONCAT('%',b.name,'%');
Regards,
Gelu
_
G.NET SOFTWARE COMPANY
Permanent e-mail address : [EMAIL PROTECTED]
[EMAIL PROTECTED]
- Original Message -
From: Jorge Martinez [EMAIL PROTECTED]
To: Mary Stickney [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Thursday, August 29, 2002 7:48 PM
Subject: RE: like problem
yes, but select a.name from table1 a, table2 b where a.name like
'%'b.name'%';
is an error
- Original Message -
From: Mary Stickney [EMAIL PROTECTED]
To: Jorge Martinez [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Thursday, August 29, 2002 1:26 PM
Subject: RE: like problem
that would be because b.name is inside the ' ' and it is taken as a
constant.
-Original Message-
From: Jorge Martinez [mailto:[EMAIL PROTECTED]]
Sent: Thursday, August 29, 2002 11:10 AM
To: [EMAIL PROTECTED]
Subject: like problem
i need this consult
select a.name from table1 a, table2 b where a.name like '%b.name%';
but i have a problem because Mysql not recognize the value of b.name
-
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Re: like problem
How about select a.name from table1 a, table2 b where a.name like concat(%,b.name,%); You might want to read chapter 6.3 in the mysql manual: http://www.mysql.com/doc/en/Functions.html HTH, Jed On the threshold of genius, Jorge Martinez wrote: yes, but select a.name from table1 a, table2 b where a.name like '%'b.name'%'; is an error - Original Message - From: Mary Stickney [EMAIL PROTECTED] To: Jorge Martinez [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Thursday, August 29, 2002 1:26 PM Subject: RE: like problem that would be because b.name is inside the ' ' and it is taken as a constant. -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:10 AM To: [EMAIL PROTECTED] Subject: like problem i need this consult select a.name from table1 a, table2 b where a.name like '%b.name%'; but i have a problem because Mysql not recognize the value of b.name - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: like problem
I'm doing this in the command line of mysql for linux. I tried your suggestion and the results was bads. it assume that I write a.nombre like '%' and shows all registers. any suggestion?? thanks - Original Message - From: Mary Stickney [EMAIL PROTECTED] To: Jorge Martinez [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Thursday, August 29, 2002 1:52 PM Subject: RE: like problem ok try this select a.name from table1 a, table2 b where a.name like '%'+ b.name + '%'; -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:48 AM To: Mary Stickney; [EMAIL PROTECTED] Subject: RE: like problem yes, but select a.name from table1 a, table2 b where a.name like '%'b.name'%'; is an error - Original Message - From: Mary Stickney [EMAIL PROTECTED] To: Jorge Martinez [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Thursday, August 29, 2002 1:26 PM Subject: RE: like problem that would be because b.name is inside the ' ' and it is taken as a constant. -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:10 AM To: [EMAIL PROTECTED] Subject: like problem i need this consult select a.name from table1 a, table2 b where a.name like '%b.name%'; but i have a problem because Mysql not recognize the value of b.name - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: like problem
muchas gracias!! you are a genius. it's all, I can to continue working thanks Sandra - Original Message - From: Jed Verity [EMAIL PROTECTED] To: Jorge Martinez [EMAIL PROTECTED]; Mary Stickney [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Thursday, August 29, 2002 2:08 PM Subject: Re: like problem How about select a.name from table1 a, table2 b where a.name like concat(%,b.name,%); You might want to read chapter 6.3 in the mysql manual: http://www.mysql.com/doc/en/Functions.html HTH, Jed On the threshold of genius, Jorge Martinez wrote: yes, but select a.name from table1 a, table2 b where a.name like '%'b.name'%'; is an error - Original Message - From: Mary Stickney [EMAIL PROTECTED] To: Jorge Martinez [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Thursday, August 29, 2002 1:26 PM Subject: RE: like problem that would be because b.name is inside the ' ' and it is taken as a constant. -Original Message- From: Jorge Martinez [mailto:[EMAIL PROTECTED]] Sent: Thursday, August 29, 2002 11:10 AM To: [EMAIL PROTECTED] Subject: like problem i need this consult select a.name from table1 a, table2 b where a.name like '%b.name%'; but i have a problem because Mysql not recognize the value of b.name - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
RE: a LIKE problem ecc.
Use SELECT * FROM Table1 WHERE BINARY name LIKE K% It will probably slow the search... You can also define the table a BINARY but 'Most' will be different than 'most' Butch Bean -Original Message- From: savaidis [mailto:[EMAIL PROTECTED]] Sent: Saturday, February 09, 2002 3:07 AM To: MySQL list (E-mail) Subject: a LIKE problem ecc. Hi! I use a query : SELECT * FROM Table1 WHERE name LIKE K% (K = Greek char = ASCII 137) but it selects also k (small K Greek) and L (Greek , ASCII 138) Why so and what I have to do? Something is in Apache setup? Also is there a way to print a counter at first row intead of ID (autoincrement) that shows the increment of each record displayed? (1,2,3, ecc) Also something strange with Apache (on WIN98) I have a root directory for root of localhost and Php directory on it for testing various php files. I changed Php to php becouse the browsers IE and Netscape did NOT recognize the capital P (!!!) and now even I reboot it continues to accept only Php as valid directory. I change it to Php1 and still see only the old Php. What happens? Thanks Makis - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php - Before posting, please check: http://www.mysql.com/manual.php (the manual) http://lists.mysql.com/ (the list archive) To request this thread, e-mail [EMAIL PROTECTED] To unsubscribe, e-mail [EMAIL PROTECTED] Trouble unsubscribing? Try: http://lists.mysql.com/php/unsubscribe.php
