Here you can find the source of split(String str, String delims)
| Parameter | Description |
|---|---|
| str | the string to split. Must not be null. |
| delims | the delimiter characters. Each character in the string is individually treated as a delimiter. |
@Deprecated public static String[] split(String str, String delims)
//package com.java2s; /**//w w w .jav a 2 s .c o m * Copyright (c) 2000, Google Inc. * * Licensed under the Apache License, Version 2.0 (the "License"); * you may not use this file except in compliance with the License. * You may obtain a copy of the License at * * http://www.apache.org/licenses/LICENSE-2.0 * * Unless required by applicable law or agreed to in writing, software * distributed under the License is distributed on an "AS IS" BASIS, * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. * See the License for the specific language governing permissions and * limitations under the License. */ import java.util.StringTokenizer; public class Main { /** * @param str the string to split. Must not be null. * @param delims the delimiter characters. Each character in the * string is individually treated as a delimiter. * @return an array of tokens. Will not return null. Individual tokens * do not have leading/trailing whitespace removed. * @deprecated see the detailed instructions under * {@link #split(String, String, boolean)} */ @Deprecated public static String[] split(String str, String delims) { return split(str, delims, false); } /** * This method is deprecated because it is too inflexible, providing * only a very specific set of behaviors that almost never matches exactly * what you intend. Prefer using a {@link Splitter}, which is more flexible * and consistent in the way it handles trimming and empty tokens. * * <ul> * <li>Create a {@link Splitter} using {@link Splitter#on(CharMatcher)} such * as {@code Splitter.on(CharMatcher.anyOf(delims))}. * <li><i>If</i> you need whitespace trimmed from the ends of each segment, * adding {@code .trimResults()} to your splitter definition should work * in most cases. To match the exact behavior of this method, use * {@code .trimResults(CharMatcher.inRange('\0', ' '))}. * <li>This method silently ignores empty tokens in the input, but allows * empty tokens to appear in the output if {@code trimTokens} is * {@code true}. Adding {@code .omitEmptyStrings()} to your splitter * definition will filter empty tokens out but will do so <i>after</i> * having performed trimming. If you absolutely require this method's * behavior in this respect, Splitter is not able to match it. * <li>If you need the result as an array, use {@link * com.google.common.collect.Iterables#toArray(Iterable, Class)} on the * {@code Iterable<String>} returned by {@link Splitter#split}. * </ul> * * @param str the string to split. Must not be null. * @param delims the delimiter characters. Each character in the string * is individually treated as a delimiter. * @param trimTokens if true, leading/trailing whitespace is removed * from the tokens * @return an array of tokens. Will not return null. * @deprecated */ @Deprecated public static String[] split(String str, String delims, boolean trimTokens) { StringTokenizer tokenizer = new StringTokenizer(str, delims); int n = tokenizer.countTokens(); String[] list = new String[n]; for (int i = 0; i < n; i++) { if (trimTokens) { list[i] = tokenizer.nextToken().trim(); } else { list[i] = tokenizer.nextToken(); } } return list; } }