How to pass arguments in php exec()?

Question

I am trying to run a .exe application with input file and argument.

With cmd I can successfully start the executable like this...

C:\Program Files\MyApp.exe "path\to\input file" argument

However, nothing happens when I simply copy paste the string above into the exec() function like this..

exec("C:\Program Files\MyApp.exe "path\to\input file" argument")

Do I need to escape parts of the string? How should I proceed?

Please show us your whole statement including the `exec()` call. — TiMESPLiNTER, Oct 31 '13 at 13:52

score 3 · Answer 1 · answered Oct 31 '13 at 13:56

3

Just pass the arguments like a normal calling from shell

ex:

exec("C:\Program Files\MyApp.exe \"path to\input file\" argument")

answered Oct 31 '13 at 13:56

Carlos

2

Try `exec("C:\\Program Files\\MyApp.exe \"path to\\input file\" argument")` – TiMESPLiNTER Oct 31 '13 at 14:28

score 0 · Answer 2 · answered Jan 29 '18 at 22:28

0

I had to use this format

php -q "./yii.php" migrate/up --interactive=0 --migrationPath=@vendor/pheme/yii2-settings/migrations

answered Jan 29 '18 at 22:28

FantomX1

2 Answers2