Linux Help keep getting new line error for line 6 dont know what the problem is

Question

#!/bin/sh if ($#argv == 0) echo There are no arguments if ($#argv != 0) echo There are $#argv arguments

Answer 1

I'm not surprised you're getting errors; you have several syntax errors and incorrect constructs that would prevent your code from running on the Bourne shell (the interpreter indicated by your shebang line). I've fixed your code below so sh will actually run it:

#!/bin/sh
if [ "$#" == "0" ]; then
    echo "There are no arguments"
fi
if [ "$#" != "0" ]; then
    echo "There are $# arguments"
fi

That being said, you're testing for the same condition twice ("$#" == "0" / "$#" != "0"); I would refactor this to just a simple if/else instead of the two separate conditionals you have currently.