使用Java中的regex提取excel中的所有单元格引用

Question

have excel data and formulars like:

• (S10+S14+S18+S22+S26+S30+S34)
• E10+E11+SUM(E10;E14:E17)*E18-IF(E19<1,E20, E21)
• SUM(E14:E19)
• S16*S15

and so on.

I want to get all the cell references out of the strings. Like in this Example: "E10+E11+SUM(E10;E14:E17)*E18-IF(E19<1,E20, E21)"

I want the output substring like "E10 E11 ... E21" or separated with ",".

I tested a lot with regex but can't get a valid result. I am using this code:

String formulaString = "E10+E11+SUM(E10;E14:E17)*E18-IF(E19<1,E20, E21)";
Pattern pattern = Pattern.compile("REGEX");
Matcher matcher = pattern.matcher(formulaString);

I have tried the following regex:

http://social.msdn.microsoft.com/Forums/en-US/815e819c-f0f2-4a53-8407-98b0f7f116e2/regex-to-extract-list-of-cell-references-from-excel-formula?forum=csharpgeneral

REGEX: (\\w+|)?\\$?(?:\\bXF[AD]|X[AE][AZ]|[AW][AZ]{2}|[AZ]{2}|[AZ])\\$?(?:104857[0-6]|10485[0-6]\\d|1048[0-4]\\d{2}|104[0-7]\\d{3}|10[0-3]\\d{4}|[1-9]\\d{1,5}|[1-9])d?\\b(:\\s?\\$?(?:\\bXF[AD]|X[AE][AZ]|[AW][AZ]{2}|[AZ]{2}|[AZ])\\$?(?:104857[0-6]|10485[0-6]\\d|1048[0-4]\\d{2}|104[0-7]\\d{3}|10[0-3]\\d{4}|[1-9]\\d{1,5}|[1-9])d?\\b)?

http://social.msdn.microsoft.com/Forums/en-US/dc179984-4fc8-4346-90e8-1649a23b6afe/regex-solution-to-id-excel-cell-references-in-an-excel-formula-string?forum=regexp

REGEX: \\$?\\b([AZ]|[AH][AZ]|I[AV])\\$?([1-9]\\d{0,3}|[1-5]\\d{4}|6[0-4]\\d{3}|65[0-4]\\d{2}|655[0-2]\\d|6553[0-6])\\b([:\\s]\\$?\\b([AZ]|[AH][AZ]|I[AV])\\$?([1-9]\\d{0,3}|[1-5]\\d{4}|6[0-4]\\d{3}|65[0-4]\\d{2}|655[0-2]\\d|6553[0-6])\\b)?

For some of the formular they are working, but not for all.

I hope anyone can help me or give me a tip :)

Answer 1

public static void main(String[]args){
String formula = "E10+E11+SUM(E10;E14:E17)*E18-IF(E19<1,E20, E21)";
String output="";


for(String c: formula.split("[^A-z0-9]+")){
    if(isCell(c)){
               output+=c+" ";
    }
}


}

private static boolean isCell(String current){
   boolean hasLetter = false;
   boolean hasNumber = false;

for(int i=0; i<current.length() && (!hasLetter || !hasNumber); i++){


    if(current.charAt(i)>=65 && current.charAt(i)<=90){


        hasLetter=true;
    }
    else if(current.charAt(i)>='0' && current.charAt(i)<='9'){

        hasNumber=true;
    }
}

return hasLetter && hasNumber;
}

Answer 2

On the first thread you linked to, the regex you put below the link is nowhere to be found on the page. Were you really using a regex from that page? The regex that was suggested was:

(\w+|)?\$?(?:\bXF[A-D]|X[A-E][A-Z]|[A-W][A-Z]{2}|[A-Z]{2}|[A-Z])\$?(?:104857[0-6]|10485[0-6]\d|1048[0-4]\d{2}|104[0-7]\d{3}|10[0-3]\d{4}|[1-9]\d{1,5}|[1-9])d?\b(:\s?\$?(?:\bXF[A-D]|X[A-E][A-Z]|[A-W][A-Z]{2}|[A-Z]{2}|[A-Z])\$?(?:104857[0-6]|10485[0-6]\d|1048[0-4]\d{2}|104[0-7]\d{3}|10[0-3]\d{4}|[1-9]\d{1,5}|[1-9])d?\b)?

Try that.

Also, it would help to know what specific strings don't get matched correctly, since you mentioned that some of them work.