Question

Can I include a php variable in shell_exec?

I’m tring to make this line work $Dname is a php variable

shell_exec(cat . $Dname.txt . | sed "s#,]}#]}#g" > $Dname2.txt);

Answer 1

You have not quoted your shell command, and you should escape $Dname with escapeshellarg(). Note I have also escaped the quotes in the sed expression, though in this case they could be changed to single quotes without negative effects.

$Dname = escapeshellarg($Dname);
shell_exec("cat $Dname.txt . | sed \"s#,]}#]}#g\" > {$Dname}2.txt");