I'm trying to restore MySQL dump created the following way:
$file = '/path/to/file.sql';
exec('mysqldump -u '.DB_USER.' -p'.DB_PASS.' '.DB_NAME.' > '.$file);
the above creates the dump as expected, then to restore I'm trying to use the following:
$file = '/path/to/file.sql';
exec('mysql -u '.DB_USER.' -p'.DB_PASS.' '.DB_NAME.' < '.$file);
but for some reason it doesn't do anything.
Please note that the constants contain the relevant database connection parameters.
Any idea what I'm doing wrong?
use mysql -e 'source $file' instead of redirection
$file = realpath('file.sql');
exec('mysqldump -u ' . DB_USER . ' -p' . DB_PASS . ' ' . DB_NAME . ' > ' . $file);
Perhaps try this.
Related
I am working on database migration. I have written a code for executing the command that retrieves the data from database and pushes into a csv file. This Works fine in MySQL but when I try to do the same in SQL Server it does not work. Infact when I copy paste the same command into command prompt it works fine. I double checked everything. I do not understand why its not working. It returns blank output. I have already tried many of the solutions provided before. None works. Any help on this is most appreciated.
Here is the code I am using:
//$sqlsrv is used to determine the database server type
$str_query = voc_get_query_string($query);
$output_uri = 'temporary://' . user_password();
$file_path = drupal_realpath($output_uri . '.csv');
if($sqlsrv){
$exec_path = drupal_realpath('private://Binn\sqlcmd');
}else{
$exec_path = drupal_realpath('private://mysql');
}
$sql_uri = 'temporary://' . user_password();
$sql_path = drupal_realpath($sql_uri);
$fp = fopen($sql_path, 'w');
fputs($fp, $str_query);
fclose($fp);
global $databases;
if ($sqlsrv) {
$cmd = ($exec_path .
' -S ' . $databases['default']['default']['host'] .
' -d ' . $databases['default']['default']['database'] .
' -U ' . $databases['default']['default']['username'] .
' -P ' . $databases['default']['default']['password'] .
' -i ' . $sql_path . '>>'. $file_path .
' -s '. '"," -W -m10 -r1');
}
else {
$cmd = ($exec_path . ' ' . $databases['default']['default']
['database'] .
' -h ' . $databases['default']['default']['host'] .
' -u ' . $databases['default']['default']['username'] .
' -p ' . $databases['default']['default']['password'] . ' < '
. $sql_path .
' > ' . $file_path);
}
exec($cmd);
watchdog('cmd', var_export($cmd, TRUE));
I am trying to create a mysql database dump in a php class.
I am executing the following code but only an empty dump file will create. But the same command works fine when I execute it in the command line.
$script = 'mysqldump -h ' . DB_HOST . ' -u ' . DB_USER . ' -p' . DB_PASSWORD . ' ' . DB_NAME . ' > ' . DB_DUMP_PATH . 'mysql-db-dump-' . date('Y-m-d') . '.sql';
exec($script);
But when I give the full path of mysqldump, the dump file create correctly.
$script = '/usr/local/mysql/bin/mysqldump -h ' . DB_HOST . ' -u ' . DB_USER . ' -p' . DB_PASSWORD . ' ' . DB_NAME . ' > ' . DB_DUMP_PATH . 'mysql-db-dump-' . date('Y-m-d') . '.sql';
exec($script);
But I saw most of the tutorials, they just use mysqldump command. Not the full path. What is the reason behind this? I am planing to use this in both unix and windows environment.
Is it possible to get the mysqldump path in php for both environments?
You can avoid path by copying mysqldump and mysql commands from /usr/local/mysql/bin/ to /usr/bin/ path.
My goal is to be able to submit a search query from a web form and have an AppleScript execute the search in DEVONagent. The AppleScript works fine in terminal but I get an error when having PHP do a shell_exec().
<?php
$theQuery = $_GET["Query"];
$cmd = "theSearch=\"$theQuery\" osascript -e \"set theSearch to system attribute " . "\\" . "\"theSearch" . "\\" . "\"\" -e \"tell application " . "\\" . "\"DEVONagent" . "\\" . "\"\" -e \"search theSearch using set " . "\\" . "\"Web (Deep Link)" . "\\" . "\"\" -e \"end tell\" 2>&1";
echo "<pre>$cmd</pre><BR><BR>";
$theResponse = shell_exec ( $cmd );
echo "Your search for \"$theQuery\" has started and the results will be emailed to you once complete.";
echo "<pre>$theResponse</pre>";
?>
I end up with the following error from the $theResponse echo:
83:92: syntax error: Expected end of line but found identifier. (-2741)
I'm thinking maybe a permissions thing but I just cannot figure it out.
I'm using a Linux local computer and need to backup/mirror some very large file structures regularly. I only have access to SFTP.
I was after a simple one click solution. I originally tried to write the little script in BASH but I've never used it before and am not up to scratch with the syntax so I resorted to PHP. (I do understand PHP is not designed for this kind of work, but I'm on a tight time scale and don't have the time to get into BASH atm)
<?php
//init
parse_str(implode('&', array_slice($argv, 1)), $_GET);
$error = array();
$lPrefix = '/home/hozza/Sites/';
$archiveLocation = '/home/hozza/Backups/';
$lDir = isset($_GET['l']) ? $_GET['l'] : $error[] = 'Local Directory Required';
$rDir = isset($_GET['r']) ? $_GET['r'] : $error[] = 'Remote Directory Required';
$bookmark = isset($_GET['b']) ? $_GET['b'] : $error[] = 'lftp Bookmark Required';
//Check for args
if(count($error) == 0) {
$archiveName = end(explode('/', $lDir)) . '_' . date('Y-m-d_H-i');
//Validate local dir
if(is_dir($lPrefix . $lDir)) {
//preserve Sublime Text 2 config SFTP files
$ST2_SFTP_conf = false;
if(file_exists($lPrefix . $lDir . '/sftp-config.json')) {
$ST2_SFTP_conf = file_get_contents($lPrefix . $lDir . '/sftp-config.json');
unlink($lPrefix . $lDir . '/sftp-config.json');
}
//Start mirror
$lftOutput = explode("\n", shell_exec('lftp -e "mirror -e -p --parallel=10 --log=' . $archiveLocation . 'logs/' . $archiveName . '.txt ' . $rDir . '/ ' . $lPrefix . $lDir . '/; exit top" ' . $bookmark));
//Tar regardless of lftp error or success
$tarOutput = shell_exec('cd ' . $lPrefix . ' && tar -czf ' . $archiveLocation . $archiveName . '.tar.gz ' . $lDir);
//Output completion or errors
shell_exec('notify-send -i gnome-network-properties -t 0 "Mirror & Archive Complete" "' . $archiveName . '\n\n' . implode('\n', $lftOutput) . $tarOutput . '"');
//put back ST2 SFTP conf
if($ST2_SFTP_conf != false) file_put_contents($lPrefix . $lDir . '/sftp-config.json', $ST2_SFTP_conf);
exit;
}
else shell_exec('notify-send -i error -t 0 "Mirror & Archive Error" "' . date('Y-m-d') . ' ' . date('H-i') . '\n' . $lDir . ' \n Does not exist! D:"');
}
else shell_exec('notify-send -i error -t 0 "Mirror & Archive Error" "' . date('Y-m-d') . ' ' . date('H-i') . '\n' . implode('\n', $error) . '"');
?>
It can be run for many sites via a short-cut like so...
terminator -T "Mirror & Archive" -e "php ~/Programs/mirror.php l=local-dir_path r=./ b=lftp-bookmark-name"
If no password is in the LFTP bookmark (there shouldn’t be as it's stored in plain text) the terminal prompts for a password, after the script has run, a nice notification is given with some info about files/folders/speed etc.
However, when the script is running in a terminal, only the "input password" bit is output to the terminal, I would like all the output displayed in the terminal (normally that would display what file/folder is currently working with etc.)
Anyone know how to do that?
IIRC the reason that you see the password prompt output to the terminal is that it is using stderr. You could try redirecting stdout to stderr for your commands which should show you the 'real-time' progress. Tack this on to the end of the shell_exec() command: 1>&2
ie:
shell_exec('lftp -e "mirror -e -p --parallel=10 --log=' . $archiveLocation . 'logs/' . $archiveName . '.txt ' . $rDir . '/ ' . $lPrefix . $lDir . '/; exit top" ' . $bookmark . ' 1>&2')
However, this will preclude you from having anything returned by shell_exec for logging purposes. What I would suggest is something like:
$log_stem = '/tmp/' . time() . '_'; // ie: /tmp/1357581737_
$lfOutfile = $log_stem . 'lftp.log';
$tarOutfile = $log_stem . 'tar.log';
shell_exec('lftp -blah | tee ' . $lfOutfile ' 1>&2' );
shell_exec('tar -blah | tee ' . $tarOutfile ' 1>&2' );
$lfOut = file_get_contents($lfOutfile);
$tarOut = file_get_contetns(tarOutfile);
// remove tmp files
unlink($lfOutfile);
unlink($tarOutfile);
Which will capture a copy of the output to a file before redirecting the output to stderr so you can watch it live.
However, if you want to run this via cron I would recommend against writing anything to stderr that is not an error, otherwise cron will send a warning email every time it is run.
I think the last answer was close:
Either this:
shell_exec('lftp -blah |& tee ' . $lfOutfile );
shell_exec('tar -blah |& tee ' . $tarOutfile );
Or if that still doesn't work try this:
shell_exec('lftp -blah 2>&1 | tee ' . $lfOutfile );
shell_exec('tar -blah 2>&1 | tee ' . $tarOutfile );
I am using Windows 7 and using php i am backing up my database with the help of mysqldump. The file gets successfully written but the size remains always 0. Any idea why is this happening?
Note :- If i type the same command in command line, it works. This is my function :-
public static function BackupDatabase($backupPath){
$fileName = uniqid() .'.sql';
$backupCommand = 'mysqldump -u ' . DBUsername .' -p' . DBPassword .' abc >' . $backupPath . $fileName ;
$retVal = '';
$feedback = system($backupCommand, $retVal);
if($feedback == NULL || $feedback == '')
return 'Database backed up successfully by name ' . $fileName;
else
return $feedback;
}
EDIT :-
public static function BackupDatabase($backupPath){
$fileName = uniqid() .'.sql';
$backupCommand = 'mysqldump -u ' . DBUsername . ' abc > ' . $backupPath . $fileName .' 2>&1' ;
echo $backupCommand;
$retVal = '';
$feedback = system($backupCommand, $retVal);
echo $retVal;
if($feedback == NULL || $feedback == '')
return 'Database backed up successfully by name ' . $fileName;
else
return $feedback;
}
Thanks in advance :)
I believe is caused by the environment setting.
You can execute mysqldump successfully in the command line as you probably has mysqldump register in your environment.
While running in a PHP, the path to mysqldump might not recognized by user that running the web server.
And
$feedback = system('unknown program > file', $retval);
Will always set $feeback = ''; even PHP cannot find where is your mysqldump program, it still pipe a single character to backup file.
You can put in absolute path to mysqldump to test again.
Check your $backupCommand string. It seems to me that there are some spaces missing (e.g. in the vicinity of '-p').
After much headache, this syntax worked for me:
cmd /c " C:\mysqldump.exe -h $mysql_server -u $mysql_user --password=$mysql_password $dbname > C:\Dump.sql "