if statement - How to use options in bash for skipping parts of code -


I have a bidding script and I want to be able to leave it (which - is not executed) / P>

To make it clear what I want to accomplish, I have given the statement if which basically fulfills the same thing. 

 <["$ 1"! = "SkipB"] then ["$ 1"!] = "SkipC"] then "" 1 "C" Fi if ["$ 1"! = "SkipD"] then "D" file resonance   
 no arguments are passed when using the script, then executing all four  echo  statements Are there. If I want to leave one of those echo statements, then I use the  skipLETTER  as the first code:  
 . All letters are printed except /script.sh skipA   
 and  A . It works, but it has many limitations - for one, I can not leave many parts using the same logic ( $ 1 ). I want to achieve the same thing with the options, so I can write something like this:  
  ./ script - skipA -skipD   
 And as a result, print only  B  and  C  letters.  
 I know I can use the  getopt  or  getopts  to get it, but I can not find any simple usage example I've received.   
 
 
 I code with the following algorithm:  
     
  1.  All in file 1 Capture executable codes  
  2.  Call each function according to logic in file 2  
  3.  When calling File 2, each call will be limited to one liner and the steps  
  4.  Call file 2 with the option of skipping  
  5.  If the steps to read lines as a statement If the next step will continue, then the steps and check of the current phase are less than the required stage. And use eval to work for it.

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