Algorithm to search and replace delimited parameters

Question

I have a string that contains multiple parameters delimited by #, like this :

.... #param1# ... #param2# ... #paramN# ...

And I want to replace the parameter placeholders by values.

The current algorithm looks like this:

    //retrieve place holder into this SQL select
    Pattern p = Pattern.compile(DIMConstants.FILE_LINE_ESCAPE_INDICATOR);
    Matcher m = p.matcher(sqlToExec); // get a matcher object
    int count = 0;
    int start = 0;
    int end = 0;

    StringBuilder params = new StringBuilder();
    while (m.find()) {
        count++;

        if (count % 2 == 0) {

            // Second parameter delimiter

            String patternId = sqlToExec.substring(start, m.end());

            //Clean value (#value#->value)
            String columnName = patternId.substring(1, patternId.length() - 1);

            //Look for this column into preLoad row ResultSet and retrieve its value
            String preLoadTableValue = DIMFormatUtil.convertToString(sourceRow.get(columnName));

            if (!StringUtils.isEmpty(preLoadTableValue)) {
                aSQL.append(loadGemaDao.escapeChars(preLoadTableValue).trim());
            } else {
                aSQL.append(DIMConstants.COL_VALUE_NULL);
            }
            params.append(" " + columnName + "=" + preLoadTableValue + " ");

            end = m.end();

        } else {
            // First parameter delimiter
            start = m.start();
            aSQL.append(sqlToExec.substring(end, m.start()));
        }
    }

    if (end < sqlToExec.length()) {
        aSQL.append(sqlToExec.substring(end, sqlToExec.length()));
    }

I'm looking for a simplest solution, using regexp or another public API. Input parameters will be the source string, a delimiter and a map of values. Output parameter will be the source string with all the parameters replaced.

Are there ellipses between the parameters or are they back to back? #param1#...#param2#...#param3# or #param1##param2##param3? — Kevin
– Kevin, Commented Jan 20, 2010 at 14:57

Chad Okere · Accepted Answer · 2010-01-20 15:02:32Z

3

If this is for a normal SQL query, you might want to look into using PreparedStatements

Beyond that, am I missing something? Why not just use String.replace()? Your code could look like this:

for(int i = 0; i < n; i++){
   String paramName = "#param" + i + "#"
   sqlToExec = sqlToExec.replace(paramName,values.get(paramName));
}

That assumes you have a map called "values" with string mappings between parameters in the form "#paramN#"

answered Jan 20, 2010 at 15:02

Chad Okere

4,6081 gold badge23 silver badges19 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Lluis Martinez Over a year ago

The name of the parameters is free, but yes I could read all the keys in the map and to the replace.

Lluis Martinez Over a year ago

The queries are retrieved from an XML and using PreparedStatements implies changing the format of the parameters (to use ?).

Kylar · Accepted Answer · 2010-01-20 15:05:03Z

0

If you need it more generic, this will find and return the whole param including the #'s:

public class ParamFinder {

    public static void main(String[] args) {
        String foo = "#Field1# #Field2# #Field3#";

        Pattern p = Pattern.compile("#.+?#");
        Matcher m = p.matcher(foo);
        List matchesFound = new ArrayList();
        int ndx = 0;
        while(m.find(ndx)){
            matchesFound.add(m.group());
            ndx = m.end();
        }

        for(Object o : matchesFound){
            System.out.println(o);
        }


    }

}

answered Jan 20, 2010 at 15:05

Kylar

9,37410 gold badges46 silver badges76 bronze badges

Collectives™ on Stack Overflow

Algorithm to search and replace delimited parameters

2 Answers 2

2 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related