Exec Command is not working in sql server - php

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));

Related

Laravel mysqldump create empty file, but same command work when execute via xampp shell

running Laravel 5.2 on Xampp 321 Win7 Mysql5.6 PHP 5.6.3 I cant create BD dump file .sql
$filename = "backup-".Carbon\Carbon::now()->format('Y-m-d_H-i-s').".sql 2>&1";
try{
$command = "mysqldump --user=" . env('DB_USERNAME') ." --password=" . env('DB_PASSWORD') . " --host=" . env('DB_HOST') . " " . env('DB_DATABASE') . " > " . storage_path() . "/" . $filename;
$returnVar = NULL;
$output = NULL;
//exec command allows you to run terminal commands from php
exec($command, $output, $returnVar);
//dd($command);
return 1;
}catch(Exception $e){
return $e->errorInfo; //some error
}
When loading script it generates an empty file! But by doing dd ($ command) and copying paste this text, this command works fine in the Xampp shell. Any ideas please?
Solved with this code: setting absolute path of mysqldump adn adding double backslash to url var
$filename = "backup-".date("d-m-Y-H-i-s").".sql";
$mysqlPath = "D:\\xampp/mysql/bin/mysqldump";
try{
$command = "$mysqlPath --user=" . env('DB_USERNAME') ." --password=" . env('DB_PASSWORD') . " --host=" . env('DB_HOST') . " " . env('DB_DATABASE') . " > " . storage_path() . "/" . $filename." 2>&1";
$returnVar = NULL;
$output = NULL;
exec($command, $output, $returnVar);
return 1;//ok
}catch(Exception $e){
return "0 ".$e->errorInfo; //some error
}

Why should I need to give the full mysqldump path when doing a database dump in php

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.

ffmpeg fix video orientation

A video can contain a meta info about the camera orientation. For example iPhone and other phones set this flag if you turn the device. Problem is while some player read this info and rotate the video accordingly, other players do not.
To fix this the video has to be rotated and the meta info needs to be set correctly.
Does ffmpeg provide a fix for this or do I have to go the hard way (Read rotation, rotate, set meta data)
I did go the hard way:
$ffmpeg == "path/to/ffmpeg";
$output_file_full = "file/after/normal/conversion";
// get rotation of the video
ob_start();
passthru($ffmpeg . " -i " . $output_file_full . " 2>&1");
$duration_output = ob_get_contents();
ob_end_clean();
// rotate?
if (preg_match('/rotate *: (.*?)\n/', $duration_output, $matches))
{
$rotation = $matches[1];
if ($rotation == "90")
{
echo shell_exec($ffmpeg . ' -i ' . $output_file_full . ' -metadata:s:v:0 rotate=0 -vf "transpose=1" ' . $output_file_full . ".rot.mp4 2>&1") . "\n";
echo shell_exec("mv $output_file_full.rot.mp4 $output_file_full") . "\n";
}
else if ($rotation == "180")
{
echo shell_exec($ffmpeg . ' -i ' . $output_file_full . ' -metadata:s:v:0 rotate=0 -vf "transpose=1,transpose=1" ' . $output_file_full . ".rot.mp4 2>&1") . "\n";
echo shell_exec("mv $output_file_full.rot.mp4 $output_file_full") . "\n";
}
else if ($rotation == "270")
{
echo shell_exec($ffmpeg . ' -i ' . $output_file_full . ' -metadata:s:v:0 rotate=0 -vf "transpose=2" ' . $output_file_full . ".rot.mp4 2>&1") . "\n";
echo shell_exec("mv $output_file_full.rot.mp4 $output_file_full") . "\n";
}
}
I used some ugly temp files. Sorry about that.

php shell_exec() interpolation

I have a php script that builds a dynamic command string (calls a perl script), then executes the command, like this:
$cmd_string = "perl $pushFile";
foreach($cmd_args AS $argName => $arg){
$cmd_string .= ' --' . $argName . '="' . $arg . '"';
}
$output = shell_exec('export PERL5LIB=/mnt/path/to/custom:$PERL5LIB && ' . $cmd_string . ' 2>&1');
I am getting failures that I think are being caused by interpolation of some of the arguments. For example is one of the arguments is '246+8GT>-', it gets turned into '246 8GT ' and an error that the string is unterminated. But, if I print_r $cmd_string to the screen and execute it via command line, or copy/paste it into the $cmd_string variable, it executes properly. I am stumped. How can I make sure these arguments are being passed properly? I tried this:
$output = shell_exec('export PERL5LIB=/mnt/path/to/custom:$PERL5LIB && ' . escapeshellcmd($cmd_string) . ' 2>&1');
but get the same result. Help?
You are escaping the comamand string, after it has been built.
Try this:
$cmd_string = "perl $pushFile";
foreach($cmd_args AS $argName => $arg){
$cmd_string .= ' --' . $argName . '="' . escapeshellarg($arg) . '"';
}
$output = shell_exec('export PERL5LIB=/mnt/path/to/custom:$PERL5LIB && ' . $cmd_string . ' 2>&1');

PHP LFTP data mirror output

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 );

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