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LawrenceS18-Opal
18-Opal
November 14, 2019
Solved

BOM Table Relations: Convert String to Number (ston?)?

  • November 14, 2019
  • 4 replies
  • 4257 views

Previous erroneous Subject Heading: 2D Detailing / MBD: New Text and Symbol Fonts

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Does anyone know of a good way to convert a string to a number so number operations can be performed?

For example, a simplified version of what I want to do is:

 

pn_end2=extract(asm_mbr_name,5,2)

/******e.g. pn_end2="12" (they will *always* be numbers but could also be something like 02)

pn_compare="07"

 

/*Convert to number using my made up string to number function ston (instead of ntos)

pn_end2_num=ston(pn_end2)

pn_compare_num=ston(pn_compare)

 

IF pn_compare_num<pn_end2_num

   <do something>

ELSE

   <do something else>

ENDIF

-------------------------------------------------

Any ideas on how to do this?

Best answer by KenFarley

Here's my bit of code that converts a string to a number. It's only for integers, so adding code for real numbers would be some additional fun to extract the numerals before and after the decimal marker. In the code, txtNumber is the string for the number, intNumber is the resultant integer.

intNumber = 0
numString = "0123456789"
index = 1
position = string_length ( txtNumber )

charDigit = EXTRACT ( txtNumber, position, 1 )
intNumber = intNumber + 10^(index - 1) * ( SEARCH ( numString, charDigit ) - 1 )
IF position > 1
 index = index + 1 
 position = position - 1
 charDigit = EXTRACT ( txtNumber, position, 1 )
 intNumber = intNumber + 10^(index - 1) * ( SEARCH ( numString, charDigit ) - 1 )
ENDIF

/* Repeat the IF--THEN block above as many times as you think you'll need, one for each
/* digit beyond the first.

 

4 replies

T
23-Emerald IV
TomU23-Emerald IV
23-Emerald IV
November 14, 2019

PTC has been unwilling to do anything about this for as long as I can remember.  It would be so easy for them to just provide some industry standard type casting functions, but they won't.  Instead you're going to have to manually convert each string character to its corresponding number.

 

STRING1 = "37"
VALUE1 = -1

VAL = -1
CHAR = EXTRACT(STRING1, 1, 1)
IF CHAR == "0"
 VAL = 0
ENDIF
IF CHAR == "1"
 VAL = 1
ENDIF
IF CHAR == "2"
 VAL = 2
ENDIF
IF CHAR == "3"
 VAL = 3
ENDIF
IF CHAR == "4"
 VAL = 4
ENDIF
IF CHAR == "5"
 VAL = 5
ENDIF
IF CHAR == "6"
 VAL = 6
ENDIF
IF CHAR == "7"
 VAL = 7
ENDIF
IF CHAR == "8"
 VAL = 8
ENDIF
IF CHAR == "9"
 VAL = 9
ENDIF
IF VAL <> -1
 VAL1 = VAL
ENDIF

VAL = -1
CHAR = EXTRACT(STRING1, 2, 1)
IF CHAR == "0"
 VAL = 0
ENDIF
IF CHAR == "1"
 VAL = 1
ENDIF
IF CHAR == "2"
 VAL = 2
ENDIF
IF CHAR == "3"
 VAL = 3
ENDIF
IF CHAR == "4"
 VAL = 4
ENDIF
IF CHAR == "5"
 VAL = 5
ENDIF
IF CHAR == "6"
 VAL = 6
ENDIF
IF CHAR == "7"
 VAL = 7
ENDIF
IF CHAR == "8"
 VAL = 8
ENDIF
IF CHAR == "9"
 VAL = 9
ENDIF
IF VAL <> -1
 VAL2 = VAL
ENDIF

IF VAL1 <> -1 & VAL2 <> -1
 VALUE1 = ((VAL1*10) + VAL2)
ENDIF

 

    M
    MartinHanak24-Ruby III
    24-Ruby III
    November 15, 2019

    Hi,

    see https://community.ptc.com/t5/Routed-Systems/Getting-part-parameter-for-component-into-cabling/m-p/627912#M4499 

    You can find there following solution

    intNumber = 0
numString = "0123456789"
index = 1
position = string_length ( txtNumber )

charDigit = EXTRACT ( txtNumber, position, 1 )
intNumber = intNumber + 10^(index - 1) * ( SEARCH ( numString, charDigit ) - 1 )
IF position > 1
 index = index + 1 
 position = position - 1
 charDigit = EXTRACT ( txtNumber, position, 1 )
 intNumber = intNumber + 10^(index - 1) * ( SEARCH ( numString, charDigit ) - 1 )
ENDIF

/* Repeat the IF--THEN block above as many times as you think you'll need, one for each
/* digit beyond the first.

     

      K
      KenFarley21-Topaz II
      21-Topaz II
      November 15, 2019

      Hey, that was me. I wish I'd looked before posting my own code again....

      K
      KenFarley21-Topaz IIAnswer
      21-Topaz II
      November 15, 2019

      Here's my bit of code that converts a string to a number. It's only for integers, so adding code for real numbers would be some additional fun to extract the numerals before and after the decimal marker. In the code, txtNumber is the string for the number, intNumber is the resultant integer.

      intNumber = 0
numString = "0123456789"
index = 1
position = string_length ( txtNumber )

charDigit = EXTRACT ( txtNumber, position, 1 )
intNumber = intNumber + 10^(index - 1) * ( SEARCH ( numString, charDigit ) - 1 )
IF position > 1
 index = index + 1 
 position = position - 1
 charDigit = EXTRACT ( txtNumber, position, 1 )
 intNumber = intNumber + 10^(index - 1) * ( SEARCH ( numString, charDigit ) - 1 )
ENDIF

/* Repeat the IF--THEN block above as many times as you think you'll need, one for each
/* digit beyond the first.

       

        L
        LawrenceS18-OpalAuthor
        18-Opal
        November 18, 2019

        @KenFarley  & @MartinHanak,

         

        Thank you very much for posting that very clever code.  I took some doing to even figure out what was happening and why it worked but I do understand it now.  I have successfully tested it and it works great...although I agree that PTC really needs to give us some extra commands as it could take one line instead of 30+ lines (depending on how many characters in the string that needs to be converted to a number.  Actually it would have been nice if Creo just recognized that it was a number when it first extracted it.

         

        Thank you also @TomU  for posting the other method.  That was more efficient then the method I was attempting to do too prior to posting this.  Although I did find someone who created a code generator to do the job too.  I mention here to be comprehensive and because it is both brilliant and sad that a code generator was created to compensate for such a strong Creo relations deficiency.  Although like I said I am now using @KenFarley's method... above.

        http://bl.ocks.org/veggielane/raw/449293ab850608d73a93/ 

        T
        23-Emerald IV
        TomU23-Emerald IV
        23-Emerald IV
        November 18, 2019

        Glad to help.  That code of @KenFarley's is slick.  I'll definitely be using that in the future as well.  Always fun to learn new stuff!

          L
          LawrenceS18-OpalAuthor
          18-Opal
          November 15, 2019

          I saw some great replies so thank you and will reply to those shortly.  however I noticed that I apparently posted this under the subject name which was supposedly 'restored' by the website before posting.  I will correct that now.  Sorry for the confusion but you guys still posted some really great answers!

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