code: public static int ordinalIndexOf(String str, String searchStr, int ordinal) {
          return ordinalIndexOf(str, searchStr, ordinal, false);
      }
docs: <p>Finds the n-th index within a String, handling <code>null</code>.
  This method uses {@link String#indexOf(String)}.</p>
  
  <p>A <code>null</code> String will return <code>-1</code>.</p>
  
  <pre>
  StringUtils.ordinalIndexOf(null, *, *)          = -1
  StringUtils.ordinalIndexOf(*, null, *)          = -1
  StringUtils.ordinalIndexOf("", "", *)           = 0
  StringUtils.ordinalIndexOf("aabaabaa", "a", 1)  = 0
  StringUtils.ordinalIndexOf("aabaabaa", "a", 2)  = 1
  StringUtils.ordinalIndexOf("aabaabaa", "b", 1)  = 2
  StringUtils.ordinalIndexOf("aabaabaa", "b", 2)  = 5
  StringUtils.ordinalIndexOf("aabaabaa", "ab", 1) = 1
  StringUtils.ordinalIndexOf("aabaabaa", "ab", 2) = 4
  StringUtils.ordinalIndexOf("aabaabaa", "", 1)   = 0
  StringUtils.ordinalIndexOf("aabaabaa", "", 2)   = 0
  </pre>
  
  <p>Note that 'head(String str, int n)' may be implemented as: </p>
  
  <pre>
  str.substring(0, lastOrdinalIndexOf(str, "\n", n))
  </pre>
  
  @param str  the String to check, may be null
  @param searchStr  the String to find, may be null
  @param ordinal  the n-th <code>searchStr</code> to find
  @return the n-th index of the search String,
  <code>-1</code> (<code>INDEX_NOT_FOUND</code>) if no match or <code>null</code> string input
  @since 2.1