RE to remove ,

Question

Hi given a string of 
 inPutString = "signal_A{2,4,6,8:10},signal_B,signal_C[10,12:14,16]_E" 
 is it possible to remove the 2 commas between the 3 signals? 
 e.g outPutList => (" signal_A{2,4,6,8:10} " " signal_B" "signal_C[10,12:14,16]_E" )

mstaub · Answer

Some parts of this are a hack, but it works. You might want to explore regular expressions for this as well.

Text version of the code: 
 
 axlCmdRegister("parsenetstring" 'ParseNetString) 
 procedure(ParseNetString( ) let(((b_validIndex t) (s_inPutString "") (l_output list()) (l_temp list()) (s_curChar "") (b_indexFound nil) ) s_inPutString = "signal_A{2,4,6,8:10},signal_B,signal_C[10,12:14,16]_E" 
 for(i 1 strlen(s_inPutString) b_indexFound = nil s_curChar = buildString(list(getchar(s_inPutString i))) 
 ;Ignore any commas inside brackets if(s_curChar == "{" || s_curChar == "[" then b_validIndex = nil ) ;Character is outside of brackets, so it could be a valid comma if(s_curChar == "}" || s_curChar == "]" then b_validIndex = t ) ;The current character could be a valid comma if(b_validIndex then ;Valid comma found. Store the if(s_curChar == "," then b_indexFound = t l_output = cons(buildString(reverse(l_temp) "") l_output) l_temp = list() ) ) ;Couldn't find an easy way to remove a portion of a string by a known index so we are ; just creating a list of chars which we will build a new string from when a valid ; comma is found if(!b_indexFound then l_temp = cons(s_curChar l_temp) ) ;We are at the end of the string, so add the final value to the list if(i==strlen(s_inPutString) then l_output = cons(buildString(reverse(l_temp) "") l_output) ) i++ ) l_output = reverse(l_output) println(l_output) 
 ) )

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RE to remove ,

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