Remove parameters from a uri. : URI « Network Protocol « Java

Remove parameters from a uri.

import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map;
import java.util.StringTokenizer;


 Derby - Class org.apache.derby.iapi.util.PropertyUtil

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public class Main {

   * Remove parameters from a uri.
   * Passed in parameters map will be populated with parameter names as keys and
   * parameter values as map values. Values are of type String array
   * (similarly to {@link javax.servlet.ServletRequest#getParameterMap()}).
   * @param uri The uri path to deparameterize.
   * @param parameters The map that collects parameters.
   * @return The cleaned uri
  public static String deparameterize(String uri, Map parameters) {
      int i = uri.lastIndexOf('?');
      if (i == -1) {
          return uri;

      String[] params = uri.substring(i + 1).split("&");
      for (int j = 0; j < params.length; j++) {
          String p = params[j];
          int k = p.indexOf('=');
          if (k == -1) {
          String name = p.substring(0, k);
          String value = p.substring(k + 1);
          Object values = parameters.get(name);
          if (values == null) {
              parameters.put(name, new String[]{value});
          } else {
              String[] v1 = (String[])values;
              String[] v2 = new String[v1.length + 1];
              System.arraycopy(v1, 0, v2, 0, v1.length);
              v2[v1.length] = value;
              parameters.put(name, v2);

      return uri.substring(0, i);



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