Creating a bash file with variables with PHP and phpseclib

Question

I need to create a bash file with PHP using the phpseclib library. This is the code I am using now:

   $ssh->exec("cat > $sPath$sSavingCode <<EOF
screen -dmS $1 java -Xincgc -Xmx200M -jar craftbukkit-1.4.7.jar nogui
        ");

The code works, and the file saves, but it skips the "$1". instead, it makes a file with two spaces between -dmS and java.

How can I make it so the $1 is written to the bash file?

Thanks

EDIT

This is the whole function:

    {
        $sPath = "minecraft/servers/".$user."/";
        $sSavingCode = "start.sh";
        $ssh->exec("cat > $sPath$sSavingCode <<EOF
screen -dmS $1 java -Xincgc -Xmx200M -jar craftbukkit-1.4.7.jar nogui
        ");
    }

$sPath and $sSavingCode are PHP variables, and the $1 is a bash variable that needs to be in the script.

I have tried escaping, making the $1 into a variable like $foo = '$1', and everything I have tried or tried to find has not worked. — Runner, Mar 31 '13 at 03:09
I have added the whole function. The $1 needs to be in the bash file as a bash variable, and the $sPath and $sSavingCode are PHP variables — Runner, Mar 31 '13 at 12:40
OK. I see your point. Anyway there.is no way bash gets.$1 with this expression. $1 is fed through function calling — fedorqui, Mar 31 '13 at 13:51
So there is no way I can create a bash file with PHP that has a $1 in it? — Runner, Mar 31 '13 at 13:55
From @Shal answer I see it was possible. In fact it is strange that it is working this way and not by changing your code to `\$1` directly. — fedorqui, Mar 31 '13 at 21:56

Th. Ma. · Accepted Answer · 2013-03-31T15:46:56.773

3

Why not assigning your command to a simple variable?

    $command = ' screen -dmS \$1 java -Xincgc -Xmx200M -jar craftbukkit-1.4.7.jar nogui'

    $ssh->exec("cat > $sPath$sSavingCode <<EOF
$command
        ");

edited Mar 31 '13 at 15:46

answered Mar 30 '13 at 20:56

Th. Ma.

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I'm assuming $sPath and $sSavingCode were declared in your PHP script, weren't they? Besides, is $1 actually supposed to be a shell variable? Could you please give a try to the snippet added to my original answer? – Th. Ma. Mar 31 '13 at 08:16
That code did not work, it creates the file, but does not print any text in it – Runner Mar 31 '13 at 12:44
Are you willing to obtain the first argument passed to your script using $1? If so, please consider using native PHP global variable $argv as shown in my updated codeblock. – Th. Ma. Mar 31 '13 at 13:46
I tried your code, and it creates the file still, but does not print any text to it. I also tried `$ssh->exec("cat > $sPath$sSavingCode < – Runner Mar 31 '13 at 13:54
1

I remove the first one which might be misleading then. Thanks. – Th. Ma. Mar 31 '13 at 15:46

Creating a bash file with variables with PHP and phpseclib

1 Answers1