shell - /usr/bin/perl -e "statement" doesn't accept variables -

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I would like to execute perl script as the following: 

  perl -e 'print 'My $ var = "Hello"; Print $ var "World \ n"; '  
 This ends in the following error message:  
  Bearded found that the operator is on -e line 1, $ var = "hello" Near the pass (unavailable) operator "hello", the operator is expected, with the print $ var "" (Do you need to predeceate Hello?) Backslash found that the operator will be found on "e-line 1", "near the world" string found that at the end of the operator's expected line-line (Missing semicolon on previous row?) Syntax error, "near $ var" = "hello", "can not find string terminator" 'before eof - e line 1. first   
 The following statement works perfectly:  
  perl -e ' Rint 'Hello'; Print "World \ n"; '  
 It seems that perl -e variable does not accept whether this is true and if so, why?   
 
 
 You have an additional  ":  
 < Code> perl -e 'print' my $ var = "hello"; print $ var. "World \ n"; '# here __ ^   
 delete it.  < P> But it's not really clear what you want.

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