Dear all,
I have a JSP program whish is using MySQL database engine.
It was working successfully,Now it doesn't work.
Note:
When i use mysqlshow command , i see test bank only.But if i use mysql
shell use show databases,I can see all my databases.
It's emergency.
Please help me
--
Mohsen Pahlevanzadeh wrote:
Dear all,
I have a JSP program whish is using MySQL database engine.
It was working successfully,Now it doesn't work.
Note:
When i use mysqlshow command , i see test bank only.But if i use mysql
shell use show databases,I can see all my databases.
It's
Mohsen,
First off, what version of MySQL are you running and on what platform?
Now when i use --skip-grant-tables i can see my databases.
Please help me...
When you use MySQL via a shell I would assume you're logging in as root?
You also say you can only see a test database, can you not see
Mohsen,
First off, what version of MySQL are you running and on what platform?
Now when i use --skip-grant-tables i can see my databases.
Please help me...
When you use MySQL via a shell I would assume you're logging in as root?
You also say you can only see a test database, can you not see a
Mohsen,
I'm not sure you're receiving any of this as you also seem to have a
rather over-eager spam filter as well...
Symantec Mail Security detected prohibited content in a message sent from your
address (SYM:40763633734165155763)
Subject of the message: Re: MySQL My program
Recipient
I think it's your PHP application; how did you debug your application?
All this is doing is letting our customer add their contacts to the database.
This is on the quick add form and asks them to enter a first and last name
and an email.
I debugged by re-writing it temporarily to do this...
Did you make a mistake when showing us your PHP code? There's a missing
single quote after the period on the second line of you $SQL =
Regards,
Jerry Schwartz
Global Information Incorporated
195 Farmington Ave.
Farmington, CT 06032
860.674.8796 / FAX: 860.674.8341
-Original Message-
Matt,
$SQL = 'INSERT INTO Contacts (ContFirst,ContLast,ContEmail,UserID)
VALUES ('
.$_POST[ContFirst].','.$_POST[ContLast]
.','.$_POST[ContEmail].','.$MyID.')';
echo $SQL;
$result = mysql_query($SQL,$db);
If userid is an int, why quote the $myid value?
Did you check what's echoed
I see it was my mistake, not yours. I'm going to shut up now.
Regards,
Jerry Schwartz
Global Information Incorporated
195 Farmington Ave.
Farmington, CT 06032
860.674.8796 / FAX: 860.674.8341
-Original Message-
From: Matt Neimeyer [mailto:[EMAIL PROTECTED]
Sent: Thursday, December
Matt Neimeyer wrote:
I think it's your PHP application; how did you debug your application?
All this is doing is letting our customer add their contacts to the
database.
This is on the quick add form and asks them to enter a first and last
name
and an email.
I debugged by re-writing it
Thanks to Chris yesterday, I managed to figure some things out by
myself. Now I'm faced with another problem. Given the same database again:
++---+--+-+---++
| Field | Type | Null | Key |
Hi Juan,
The default (and recommended) is 2. The log files, save the trasactions
into
file in circular order. This files are like a redolog files in oracle. This
log file are useful when you recover your database after some crash for
example or when you use a replication mysql.
Ashley M. Kirchner wrote:
?php
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$mins[] = $row['the_minute'];
$temp_f[] = $row['avg_temp_f'];
$temp_c[] = $row['avg_temp_c'];
}
?
I'd try php here, something like:
?php
while($row =
Chris White wrote:
I'd try php here, something like:
Problem is, PHP doesn't know which record is blank. I select for 60
records and MySQL returns 55. How is PHP supposed to know which 5 are
blank?
--
W | It's not a bug - it's an undocumented feature.
In the last episode (Dec 28), Ashley M. Kirchner said:
Chris White wrote:
I'd try php here, something like:
Problem is, PHP doesn't know which record is blank. I select for 60
records and MySQL returns 55. How is PHP supposed to know which 5 are
blank?
Mysql doesn't know either. All
Hello,
I noticed the error messages below in my MySQL error log and found
them a bit perplexing. Can't find anything on them in the MySQL
documentation. If anyone has any clue what they mean it is greatly
appreciated. As a sidenote, SHOW INNODB STATUS completes, but only
shows through the
Dan Nelson wrote:
If not, you'll probably have to read the records and store them in an
array indexed by minute. Then when you're done reading from mysql,
walk the array from 0 to 59 and write out each element. That way
you're guaranteed 60 output rows.
Eh, did something different. Now
Hey everyone
I have more of a general question regarding your experience with large
tables.
I currently have a table (MyISAM, 6 columns, lots of reading access, some
writing) with about 70.000.000 records, using 2.5GB of diskspace. I am
running MySQL 5.0.* on a RedHat Enterprise AS 4 system (2
Hi, I'm getting the following error message:
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 ' enabled = '1''
at line 3INSERT INTO clf_cities SET cityname = 'Santa Barbara', countryid =
, enabled = '1'
and
Hi,
have you checked the 'enabled' field datatype or can you give the query.
Thanks
ViSolve DB Team.
- Original Message -
From: Scott Yamahata [EMAIL PROTECTED]
To: mysql@lists.mysql.com
Sent: Friday, December 29, 2006 11:59 AM
Subject: SQL syntax
Hi, I'm getting the following
Hi Scott,
at line 3INSERT INTO clf_cities SET cityname = 'Santa Barbara', countryid
= , enabled = '1'
The error is because you havent specified the value for the column
countryid. If you do not want to insert the value to the column
countryid then use the following query.. do not leave the
Hi,
The maximum effective table size for MySQL databases is usually determined by
operating system constraints on file sizes, not by MySQL internal limits.
If you need a MyISAM table that is larger than 4GB in size (and your operating
system supports large files), the CREATE TABLE statement
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