i have this file :
~/abd.php
<?php
echo 5+5;
?>
i run this command on terminal
php5 ~/abd.php
i expect to get 10 . However, No output found.
Does it output before the prompt, e.g:
10user#server:~#
If so, try adding a newline.
Related
I have a shell script in Linux that performs SFTP to get some files. It works fine when I execute it from a terminal.
I am trying to call the script from PHP. It seems to work until the echo, and then it doesn't do anything.
The script and the PHP file are in the same folder.
This is the PHP code:
<?php
$comando = "sh ftpgesdoc.sh";
$result=exec($comando);
echo $result;
?>
And this is shell script. When I execute from the web, I can see the echo "ejecutando sftp", but nothing happens after this point.
#!/bin/sh
echo "ejecutando sftp"
folder="/aaa/bbb"
file="xxx.PDF"
sftp UserXX#nnn.nn.n.nn << EOF
cd $folder
get $file test.pdf
EOF
exec returns only the last line from the command output. If you want to capture entire output, use proc_open. See this answer, for instance.
you have to give the full path to file
and use this 2>&1 and know the error
try something like this
$comando = "sh pathTofile/location/ftpgesdoc.sh";
if(exec("$comando 2>&1", $output, $return_var))
{
print_r($output);
echo "<br>";
print_r($return_var);
}
I've got the following script:
#!/bin/sh
export DISPLAY=:0
phantomjs --version
It try to run it from the following PHP script:
<?php
$result = shell_exec('sh test.sh');
echo $result;
?>
This script return the following error:
[Thu Jun 19 10:31:31 2014] [error] [client] test.sh: line 3: phantomjs: command not found
I tried to run phantomjs -v by hand in a console, and it runs fine. I checked the PATH, and phantomjs is correctly defined and found.
The execution environment is a virtual Server with LiveConfig.
Can someone help me understand what I'm doing wrong ?
It could be an issue with shell_exec() and line breaks,
try adding "2>&1" to the string you are passing:
$result = shell_exec('sh test.sh 2>&1');
this worked for me, found it in the top comment here, naturally ;)
Your PATH probably lacks the location for the phantomjs executable. PhantomJS is probably installed in /usr/local/bin so you need to add this to your PATH variable:
#!/bin/sh
export DISPLAY=:0
PATH=$PATH:/usr/local/bin
phantomjs --version
To check what the current PATH is, you could begin the shell script with:
#!/bin/sh
echo $PATH
<?php
exec('/usr/local/bin/phantomjs path/somescript.js');
?>
Yes. Sometimes phantomjs don't need full path in some environment without generate any error. However, sometimes it does.
Always use the full path for all argument in the php command.
Did you use the fullpath for hello.js?
Do not use exec(). Never. It's a bad way.
Use the php-phantomjs and PhantomJS Runner instead.
I'm trying to execute a shell command from php using the exec command
The following command works
echo exec('whoami');
but I need to execute the following shell command
cat > /var/www/myfolder/abc.txt
hi
hello
welcome
the problem is that the above shell command has to be executed in multiplelines. The words "hi", "hello", "welcome" are dynamic and come from php variables. Is there a workaround for this problem.
To execute a multiline shell command in php:
passthru("bash <<'END'
echo \"executed in new bash session\";
pwd;
whoami;
exit;
END
");
This way you could for example connect to a ssh server using php and then execute some commands there
I don't understand what you mean by cat > abc.txt. This command will empty the file.
I have a command to be run like this
$command="java -jar ".dirname(__FILE__)."\gmksplit.jar"." ".$input_path." ".$output_path;
I have echoed the $command variable and I get the output as
java -jar X:\wamp\www\moodle\gmksplit.jar X:\wamp\www\moodle/upload/maze_4.gmk X:\wamp\www\moodle/outputs/maze_4;
which is exactly I want to run..
I am trying to run it as
echo $exec($command);
it is not running. I have tried all the functions like shell_exec() and system()
It gives the output as
Java Version: 10700 (1.7.0_01)
when i run the same line in command prompt I get the output as
Java Version: 10700 (1.7.0_01)
Loading lib files in X:\wamp\www\moodle\gmksplit.jar
01_move.lgl 02_main1.lgl 03_main2.lgl 04_control.lgl
05_score.lgl 06_extra.lgl 07_draw.lgl
time taken to load file: 254 ms
so, as you see my php code is giving only the first line as output. The command is not running properly and I am not getting the intended output.
please help me
I am using the wampp server
Can you try this:
<?php
$command="java -jar ".dirname(__FILE__)."\gmksplit.jar"." ".$input_path." ".$output_path;
$out = array();
exec = ($command, $out);
print_r($out);
?>
I have a .jar file which has a command line interface. I want to call the jar file through command prompt and capture the output of the Jar file.
I have tried with the exec() command.
The command I have used is:
<?php
exec('java -jar D:\\Development\\Filehandler\\dist\\Filehandler.jar \ getConfigLang', $result);
echo $result;
echo $count = count($result);
for($i=0; $i<$count;$i++){
print($result[$i]);
}
?>
The output for this was just '0 0'
Should something else be done before executing this command? like adding path etc???
I am using WAMP server.
Please help me...
Well, you can try two approaches:
1) change current directory in PHP via function http://php.net/manual/en/function.chdir.php
<?php
chdir('D:\Development\Filehandler\dist');
exec('java -jar ./Filehandler.jar \ getConfigLang', $result);
...
?>
2) change .jar file: I don't know if it is possible but try to add these additional libraries with absolute paths.