URGENT: PHP shel_exec command

[Robert_85]

Hi guys,

I am sure this problem is very easy for you people to resolve.

basically i want to run an external program in PHP (via command line) and to do that i am using shel_exec command.

I understand that the shel_exec takes "String" parameter and my problem is exactly here:

I am trying to pass a variable (containing a file name) to the shel_exec, however it is not working. If I just type the content of my variable (i.e. file name) it will work but not when I am using the variable.

here is the part of the code i am working in:

$output = shell_exec('java -jar C:/htmltolatex-1.0.1/htmltolatex.jar -config C:/htmltolatex-1.0.1/config.xml -output C:/ConversionTest/output.tex -input' .$pAddress);

$pAddress is my variable which I want to pass.

I have tried to print out "$pAddress" to make sure it is actually containing what I think it contains and it was fine; it DOES contain the file name, However when it is used inside the shel_exec it is not working.

Please Please Help, I have spend 3 hours on this and could not resolve it 😕

[dagon]

try:

$output = shell_exec("java -jar C:/htmltolatex-1.0.1/htmltolatex.jar -config C:/htmltolatex-1.0.1/config.xml -output C:/ConversionTest/output.tex -input $pAddress");

[bradgrafelman]

Note that the difference between dagon's code and yours (at least, from what I noticed) is that in yours, you didn't insert a space between the "-input" parameter and its value, so instead of exeucting something like "-input foobar" you're telling PHP to pass "-inputfoobar".