I just noticed a rather annoying issue. Whenever I run a bash script, the first thing that it does is display the name of the script along with all the arguments passed to it (if there are any).
I am trying to pass arguments to a bash script and then to a php script, I have literally looked at 30+ links, and tried over a dozen examples, and I for whatever reason have not been able to get the following to work, I am seriously frustrated, any help is so very much appreciated.
For the sake of this question, lets say I have the following bash script (test.sh)
Write out a bash script called 'b.sh' that has a function that can be called recursively to scan the directory specified by its first argument ($1) for any file that is setuid and print out that file in ls -l format.
For any directory it encounters the function should call itself with the new computed path.
I've done very little programming in C++, and I'm not really understanding how to use code in Bash. The resources given to us by the professor has been little, so it's hard to work out. The two text books we have for class talks mostly about commands in variations of UNIX, though little to do with Bash.
Write a shell script that accepts exactly 1 argument which must be a positive integer.