how to convert a javascript function using charCodeAt to Coldfusion?

Question

I have a javascript function, which I need to run serverside using Coldfusion8.

The javascript function uses charCodeAt to check entered values (ILN, 13-digit values, for example 1234567891231) and computes a checkVal:

checker=0;
for (i=11; i>0; i=i-2){
    checker += (iln.charCodeAt(i)-48)*3;
    checker += iln.charCodeAt(i-1)-48;
    }
checkVal = 10 - (checker%10);
if(checkVal==10){
    checkVal=0;
    }
if(checkVal != iln.charAt(12)){
    alert("error, should be" + checkVal);
    }

In Coldfusion I'm trying to do this with a loop, mid() and asc(), but I can't get it to work. Here is what I have:

<cscript>
    var checkValue = 1234567891231;        
    var done = "";
</cscript>
<cfloop from="11" to="1" index="i" step="-2">    
    <cfscript>
        check = check +   (ASC(MID(checkValue,#i#,1))-48)*3;
        check = check +   ASC(MID(checkValue,(#i#-1),1))-48;
    </cfscript>
</cfloop>
<cfset done = 10 - check/10>
<cfif done EQ 10><cfset done = 0></cfif>
<cfif done NEQ mid(checkValue,12,1)>
    <cfscript>
// error handler 
    </cfscript>
</cfif>

The values are passed in correctly but my ASC(MID()) is producing an error and I have not found out what I'm doing wrong.

Can someone give me a pointer?

Thanks!

EDIT: updated CF loop parameters
EDIT2: pinned it down to the 2nd mid-function getting 1-1=0 on the last loop iteration.

Peter Boughton · Accepted Answer · 2012-06-03 12:00:36Z

2

I don't quite understand the algorithm here, but the error is because you're doing Mid(string,0,1) - which is one character before the start of the string, and thus doesn't work. (JS indexes strings from zero, CF indexes from one.)

Simple solution:

<cfloop from="11" to="1" index="i" step="-2">
    <cfset check += mid(checkValue,i,1) * 3 />
</cfloop>
<cfloop from="10" to="2" index="i" step="-2">
    <cfset check += mid(checkValue,i,1) />
</cfloop>

It doesn't waste time doing Asc since the values are already numbers (if they might not be, validate the input before this step), and also doesn't have unnecessary hashes.

answered Jun 3, 2012 at 12:00

Peter Boughton

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1 Comment

frequent Over a year ago

Just found that, too. Plus I also did not now that 10%3=1 and just did 10/3.

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