Problem With (instalation Of?) Mysql.h On C

Hey guys,

I have fedora version 8 with mysql 5.0.45 istalled. I have also installed drupal. Recently drupal gave me an error "error code 28". When i checked it out it seemed to be with the db. When i tried to created a db (for testing puposes) in root...i wasnt able to. so then i tried to restart mysql ...it stopped but unfortunately i have not been able to get mysql back up...i get a messeage saying
"Timeout error occurred trying to start MySql Daemon"

Could you please assist...

Thanks in advance,

Oh

Now I cannot connect to mysql on RHEL

I try:mysql --user=myuser--password=mydbpass mysql

and i get:

Can't connect to local MySQL server through socket '/var/lib/mysql/mysql.sock'

Sorry I have done this many times before but cannot get in on this server. If I
mysqladmin -u root -p status, it states:

Check that mysqld is running and that the socket: '/var/lib/mysql/mysql.sock' exists!

Cheers and sorry

PS /var/lib/mysql does not exist on my box

https://code.google.com/p/wavelet1d/...ar.gz&can=2&q=

The above link is where I downloaded *.cpp(s) and the libraries.

If you "untar" the package, in "examples" directory, there are some demonstration files. What I wanted to do was to make an executable file out of "wavedemo1.cpp".

I modified the code in "wavedemo1.cpp"; Code:

#include "wavelet2d.h"

to Code:

#include "wavelet2s.h"

, then placed the header file "wavelet2s.h" (from /wavelib/src/linuxstatic) and the modified "wavedemo1.cpp" into my working directory.

Inside the working directory, I ran the following command

Code:

gcc -L/home/mario/wavelet/wavelib/linuxstatic -lwavelet2s wavedemo1.cpp -o wavedemo1

Then I get the following error messages.

/usr/bin/ld: skipping incompatible /home/mario/wavelet/wavelib/linuxstatic/libwavelet2s.a when searching for -lwavelet2s
/usr/bin/ld: cannot find -lwavelet2s
collect2: error: ld returned 1 exit status

I'm not sure what went wrong with my approach. This is my first time trying to use libraries on linux and it's giving me really tough times.

I would appreciate your help, please let me know if you need more details to explain things.

Thanks.

Hi everyone,
i tried to install mysql on my slackware 14.1.
typed mysql_install_db --user=mysql and went to the automatic configuration of secure access and receive this "Enter current password for root (enter for none):
ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/var/run/mysql/mysql.sock' (2)".
How can remove this installation, to try to reinstall the db?


And another error when try to reinstall :
"bash-4.2# mysql_install_db --user=mysql
Installing MariaDB/MySQL system tables in '/var/lib/mysql' ...
150130 9:40:11 [ERROR] mysqld: File '/var/lib/mysql/aria_log_control' not found (Errcode: 13)
150130 9:40:11 [ERROR] mysqld: Got error 'Can't open file' when trying to use aria control file '/var/lib/mysql/aria_log_control'
150130 9:40:11 [ERROR] Plugin 'Aria' init function returned error.
150130 9:40:11 [ERROR] Plugin 'Aria' registration as a STORAGE ENGINE failed.
150130 9:40:11 [ERROR] mysqld: Can't find file: './mysql/db.frm' (errno: 13)
ERROR: 1017 Can't find file: './mysql/db.frm' (errno: 13)
150130 9:40:11 [ERROR] Aborting

150130 9:40:11 [Note] /usr/libexec/mysqld: Shutdown complete


Installation of system tables failed! Examine the logs in
/var/lib/mysql for more information.

The problem could be conflicting information in an external
my.cnf files. You can ignore these by doing:

shell> /scripts/mysql_install_db --defaults-file=~/.my.cnf

You can also try to start the mysqld daemon with:

shell> /usr/libexec/mysqld --skip-grant --general-log &

and use the command line tool /usr/bin/mysql
to connect to the mysql database and look at the grant tables:

shell> /usr/bin/mysql -u root mysql
mysql> show tables

Try 'mysqld --help' if you have problems with paths. Using
--general-log gives you a log in /var/lib/mysql that may be helpful.

The latest information about mysql_install_db is available at
http://kb.askmonty.org/v/installing-...sql_install_db.
MariaDB is hosted on launchpad; You can find the latest source and
email lists at http://launchpad.net/maria

Please check all of the above before mailing us! And remember, if
you do mail us, you should use the /usr/bin/mysqlbug script!
bash-4.2#
"

I have installed httpd, mysql, php on a centos machine. Php executes in the browser of my machines website, mysql can be ran from the command line, but mysql within php in a browser will not work.
I know my code is syntactically correct since it will run on another Centos Lamp machine and NetBeans says it is.
My code
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="refresh" content="300">
<title> Freeze Warn Points table Query"</title>
</head>
<body>
<?php
include("config.php");
echo $dbhost;
echo "<br />\n";
echo $dbuser;
echo "<br />\n";
echo $dbpasswd;
echo "<br />\n";
mysql_connect($dbhost, $dbuser, $dbpasswd) or die("Unable to connect to database");
echo 'Connected Successfully';
mysql_select_db($dbname);

echo "zone"."&nbsp;&nbsp;&nbsp;&nbsp;";
echo "currentLow"."&nbsp;&nbsp;&nbsp;"."dateLastRep"."&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;";
echo "priority"."&nbsp;&nbsp;&nbsp;"."siteId"."&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbs p;&nbsp"."siteName";
echo "<br />\n";

$result = $sql = 'SELECT zone, currentLow, dateLastRep, pri, siteId, siteName from points order by zone, pri';
$numRows = mysql_num_rows($result);
echo "Matches: ".mysql_num_rows($result);
for($a=0; $a<$numRows; $a++)
{ $rowArray = mysql_fetch_row($result);
echo $rowArray[0]." ".$rowArray[1]." ".$rowArray[2]." ".$rowArray[3]." ".$rowArray[4]."&nbsp;&nbsp;&nbsp".$rowArray[5]."<br />";
}
mysql_close();
?>
</body>
</html>


When executed on the CENTOS lamp machine in question out puts this

"; $result=mysql_query("SELECT rpad(zone,6,'-'), rpad(currentLow,13,'-'), rpad(dateLastRep,13,'-'), rpad(pri,8,'-'), rpad(siteId,6,'-'), siteName from points order by zone, pri"); $numRows = mysql_num_rows($result); for($a=0; $a<$numRows; $a++) { $rowArray = mysql_fetch_row($result); echo $rowArray[0]." ".$rowArray[1]." ".$rowArray[2]." ".$rowArray[3]." ".$rowArray[4]." ".$rowArray[5]."
"; } mysql_close(); ?>


When I put the same code in the other machine, (it does not have the mysql database so it will not connect) displays this

204.227.112.31
apache
password
Unable to connect to database

httpd is running of course
mysqld is running
I have
mysql-server-5.1.73-3.el6_5.x86_64
php-mysql-5.3.3-40.el6_6.x86_64
mysql-devel-5.1.73-3.el6_5.x86_64
mysql-libs-5.1.73-3.el6_5.x86_64
mysql-connector-odbc-5.1.5r1144-7.el6.x86_64
mysql-5.1.73-3.el6_5.x86_64
php-mbstring-5.3.3-40.el6_6.x86_64
php-odbc-5.3.3-40.el6_6.x86_64
php-imap-5.3.3-40.el6_6.x86_64
php-mysql-5.3.3-40.el6_6.x86_64
php-xml-5.3.3-40.el6_6.x86_64
phpmyadmin-2.11.11.3-2.el6.rf.noarch
php-cli-5.3.3-40.el6_6.x86_64
php-xmlrpc-5.3.3-40.el6_6.x86_64
php-gd-5.3.3-40.el6_6.x86_64
php-5.3.3-40.el6_6.x86_64
php-pdo-5.3.3-40.el6_6.x86_64
php-common-5.3.3-40.el6_6.x86_64
php-ldap-5.3.3-40.el6_6.x86_64
php-pear-1.9.4-4.el6.noarch

Installed.

I checked the php.ini files and the httpd.conf files with the LAMP machine it works on and have been unable to find any discrepancies.
Unless I missed something.

Selinux is disabled.

I know it is not my code, but some type of setting somewhere on my machine, I have not been able to find.
I have been unable to find any thing like this issue.

Thanks,
Nancy

System Info:

I have normal user in CentOS 7 whose name is "mostafa" (the name of the account).

I naturally have another user called root with all privileges. User "mostafa" is put into sudoers file, too.

The OS is installed in VmWare, so the system is all mine.

Problem:

Now I create a file with touch file.sh and put a command in it, but when I want to run it with Code:

sudo ./file.sh

, an error is shown that the command Code:

./file.sh

does not exist. But if I Code:

 sudo chmod 777 ./file.sh

then it gets run. My question is that, why should I use Code:

chmod 777

when I myself have created the file, and I am in sudoers.

Can anyone explain me why shuold I still use Code:

sudo chmod 777

when the creator of the file is me.

There's something really weird happening with permissions on a directory in linux. I have the directory /dir

Code:

ls -ld /dir
drwxrwxrwx. 2 root root 4096 Jan  8 00:06 dir
chmod 2777 /dir [this sets the setgid bit on]
ls -ld /dir
drwxrwsrwx. 2 root root 4096 Jan  8 00:06 /dir

And this is what I expect it to do.
However, when I change it to setid (WITHOUT setgid), it simply doesn't work. Instead of overwriting, it cumulates:
Code:

chmod 4777 /dir [for setting setuid bit on]
drwsrwsrwx. 2 root root 4096 Jan  8 00:06 /dir

As you can see, the setgid bit is still on together with the setuid! However, if I run:
Code:

chmod -s /dir

then the file has a 0777 permission.

And then again, if I change its permission to 4777 (so setuid):
Code:

chmod 4777 /dir
drwsrwxrwx. 2 root root 4096 Jan  8 00:06 /dir

it acts normally. So the same chmod 4777 acts differently in two different contexts.

I really can't understand it. Basically, I cannot get rid of the setuid/setgid through binary permissions. The sticky bit works fine. Changing from 1777 to 0777 adds and removes "t" respectively. What am I to make of all this? This behaviour makes no sense to make especially given that we can change the permissions to 6777 (setuid + setgid).

Hi Guys,
i have this Code:

uid=0(root) gid=0(root)

i would like to have the word "root" only from the first field, and later from the second field

i use the command
Code:

[root@v1-6 ~]# cat test-uid |cut -d"(" -f1
brms2
uid=0

[root@v1-6 ~]# cat test-uid |cut -d")" -f1
brms2
uid=0(root

it cut the word after the -d, how can i make it to cut the word BEFORE the -d

expected should be:
cut -d"(" -f1
root)

then i can proceed to eliminate the ) later on.
hope this is not confusing you guys

Thanks,

Hello, this is my first post
First i would like to thank you all for answering other people questions because I've been able to learn from the forum a lot.

I need your help with something.
I have standard output from 'ls -laR /etc' command which looks like this:
Code:

/etc/X11/xorg.conf.d:
total 4
drwxr-xr-x. 2 root root  29 Apr  1 00:46 .
drwxr-xr-x. 5 root root  54 Apr  1 00:43 ..
-rw-r--r--. 1 root root 232 Apr  1 00:46 00-keyboard.conf

/etc/xdg:
total 12
drwxr-xr-x.  4 root root   36 Apr  1 00:43 .
drwxr-xr-x. 87 root root 8192 Apr 12 13:53 ..
drwxr-xr-x.  2 root root    6 Jun 10  2014 autostart
drwxr-xr-x.  2 root root   17 Apr  7 01:25 systemd

by using sed command:
Code:

sed -e '/./!d' -e '/^total/d' -e '/\.$/d' -e 's/:$/\//' list.txt

I have transformed it to the following form:

Code:

/etc/X11/xorg.conf.d/
-rw-r--r--. 1 root root 232 Apr  1 00:46 00-keyboard.conf
/etc/xdg/
drwxr-xr-x.  2 root root    6 Jun 10  2014 autostart
drwxr-xr-x.  2 root root   17 Apr  7 01:25 systemd

and now I would like to achieve absolute paths at the end of each row

Code:

-rw-r--r--.  1 root root  232 Apr  1 00:46 /etc/X11/xorg.conf.d/00-keyboard.conf
drwxr-xr-x.  2 root root    6 Jun 10  2014 /etc/xdg/autostart
drwxr-xr-x.  2 root root   17 Apr  7 01:25 /etc/xdg/systemd

How do I join(merge) filenames with corresponding absolute path to their parent directory?



I know how to extract filenames using awk and get this:
Code:

00-keyboard.conf

autostart
systemd

but I don't know what to do next. Should I use some hitech sed option or go for loop or try with arrays? Help. Heeeelp

Hello,

In debian I created a new user "myUser".

In /home/myUser/.profile

I added "mysql -u user -password"

So automatically after login to linux terminal the user is logged in mysql command line.

What I need to do is to exit terminal after user exit mysql, so he want have access to comand line.