无法将命令行参数传递给运行PHP CLI的NSTask

Question

Edit: [Solved] see below.

I'm launching PHP as an NSTask and trying to pass some command line options to it:

NSTask *php = [[NSTask alloc] init];
[php setLaunchPath: phpPath];
NSArray *args = [[NSArray alloc] initWithObjects: @"-r 'echo 123;'", nil ];
[php setArguments: args];
[php launch];

When I run this from Objective-C I always get the following PHP errors:

For:
@"-r 'echo 123;'"
@"-r \'echo 123;\'"
@"-r \"echo 123;\""
Parse error: syntax error, unexpected $end in Command line code on line 1

For:
@"-r 'echo \'hello\';'" Parse error: syntax error, unexpected T_STRING in Command line code on line 1

For:
@"--run 'echo 123;'"
Usage: php [options] ...

It should be noted that all errors are reported from the PHP interpreter when called from Objective-C, but when I try the exact same option string on the command line, it works as expected.

What's wrong here? how should I write arguments to be correctly passed to the NSTask?

PS: My real goal is to execute some include() lines instead of the echo, which doesn't work either, and instead I get an error of an unexpected Encapsed String.

Answer 1

Oh I saw it on another question.
I'm treating -r 'echo 123;' as a single argument, when in fact they're two, it should be:
NSArray *args = [[NSArray alloc] initWithObjects: @"-r", @"echo 123;", nil ];
Sorry! and thanks.