Java字符串split方法的坑及解决
目录
- Java字符串split方法的坑
- Java字符串split方法的探究
- 总结
Java字符串split方法的坑
先来看几行简单的Java代码,如下:
System.out.println("1,2".split(",").length); System.out.println("1,2,".split(",").length); System.out.println("".split(",").length); System.out.println(",".split(",").length);
接下来,猜一下各行的输出结果。OK,下面给出真正的运行结果:
2
2
1
0
这里先给出jdk相关源码,再来对应分析各自的输出:
public String[] split(String regex, int limit) { /* fastpath if the regex is a (1)one-char String and this character is not one of the RegEx's meta characters ".$|()[{^?*+\\", or (2)two-char String and the first char is the backslash and the second is not the ascii digit or ascii letter. */ char ch = 0; if (((regex.value.length == 1 && ".$|()[{^?*+\\".indexOf(ch = regex.charAt(0)) == -1) || (regex.length() == 2 && regex.charAt(0) == '\\' && (((ch = regex.charAt(1))-'0')|('9'-ch)) < 0 && ((ch-'a')|('z'-ch)) < 0 && ((ch-'A')|('Z'-ch)) < 0)) && (ch < Character.MIN_HIGH_SURROGATE || ch > Character.MAX_LOW_SURROGATE)) { int off = 0; int next = 0; boolean limited = limit > 0; ArrayList<String> list = new ArrayList<>(); while ((next = indexOf(ch, off)) != -1) { if (!limited || list.size() < limit - 1) { list.add(substring(off, next)); off = next + 1; } else { // last one //assert (list.size() == limit - 1); list.add(substring(off, value.length)); off = value.length; break; } } // If no match was found, return this if (off == 0) return new String[]{this}; // Add remaining segment if (!limited || list.size() < limit) list.add(substring(off, value.length)); // Construct result int resultSize = list.size(); if (limit == 0) { while (resultSize > 0 && list.get(resultSize - 1).length() == 0) { resultSize--; } } String[] result = new String[resultSize]; return list.subList(0, resultSize).toArray(result); } return Pattern.compile(regex).split(this, limit); }
1.第一行代码的输出结果肯定没什么问题,字符串 "1,2" 以 "," 分隔,结果很直观的是 ["1", "2"],length=2。
2.第二行代码的输出结果,可能大家有人认为是length=3才对,因为字符串 "1,2," 以 "," 分隔,结果应该是 ["1", "2", ""],length=3;其实不然,jdk在split处理的时候,确实会先生成一个集合list = ["1", "2", ""],但之后却会循环判断末位元素是否为空字符串(即末位元素length=0),因此集合最终会变成 ["1", "2"],length=2。具体判断如下:
while (resultSize > 0 && list.get(resultSize - 1).length() == 0) { resultSize--; }
3.第三行代码的输出结果,数组 [""],length=1。与其他三种情况不同,空字符串 "" 中不包含regex字符串 ",",所以代表没有匹配上的子串(off=0),则返回字符串本身。具体处理如下:
// If no match was found, return this if (off == 0) return new String[]{this};
4.第四行代码的输出结果,可能也有部分人认为结果应是length=2,因为字符串 "," 以 "," 分隔,结果应该是 ["", ""],length=2;其实亦不然,与第2行同样的原理,最终将list=["", ""] 处理为空集合 [],length=0。
以上,系本文分享的split的一个小坑;除此之外,另一个需要注意的地方,split方法的参数是正则表达式而非一般字符串,所以在处理正则转义字符和特殊字符时留意即可。
Java字符串split方法的探究
今天在使用split分割字符串时突然想到一种情况,如下:
String str="aaaaaaaab"; String arr[]=str.split("aa");
问,arr数组的长度是多少?
那如果str为”baaaaaaaa”呢
String str="baaaaaaaa";
如果str=”aaaaaaaab”呢
String str="aaaaaaaab";
如果str=”baaaaaaaab”呢
String str="baaaaaaaab";
好,我们先在程序中验证一下:
public class Test { public static void main(String[] args) { String str="aaaaaaaa"; String [] arr=str.split("aa"); System.out.println("字符串aaaaaaaa分割的数组长度为:"+arr.length); str="baaaaaaaa"; arr=str.split("aa"); System.out.println("字符串baaaaaaaa分割的数组长度为:"+arr.length); str="aaaaaaaab"; arr=str.split("aa"); System.out.println("字符串aaaaaaaab分割的数组长度为:"+arr.length); str="baaaaaaaab"; arr=str.split("aa"); System.out.println("字符串baaaaaaaab分割的数组长度为:"+arr.length); } }
运行以上代码输出结果
看到结果的你是不是有点小小的惊讶,如果有的话那就继续往下看。
通过split方法查看源码可知又调用了split(regex, 0)方法并且传入一个0:
public String[] split(String regex) { return split(regex, 0); }
继续查看源码
public String[] split(String regex, int limit) { /* fastpath if the regex is a (1)one-char String and this character is not one of the RegEx's meta characters ".$|()[{^?*+\\", or (2)two-char String and the first char is the backslash and the second is not the ascii digit or ascii letter. */ char ch = 0; if (((regex.value.length == 1 && ".$|()[{^?*+\\".indexOf(ch = regex.charAt(0)) == -1) || (regex.length() == 2 && regex.charAt(0) == '\\' && (((ch = regex.charAt(1))-'0')|('9'-ch)) < 0 && ((ch-'a')|('z'-ch)) < 0 && ((ch-'A')|('Z'-ch)) < 0)) && (ch < Character.MIN_HIGH_SURROGATE || ch > Character.MAX_LOW_SURROGATE)) { int off = 0; int next = 0; boolean limited = limit > 0; ArrayList<String> list = new ArrayList<>(); while ((next = indexOf(ch, off)) != -1) { if (!limited || list.size() < limit - 1) { list.add(substring(off, next)); off = next + 1; } else { // last one //assert (list.size() == limit - 1); list.add(substring(off, value.length)); off = value.length; break; } } // If no match was found, return this if (off == 0) return new String[]{this}; // Add remaining segment if (!limited || list.size() < limit) list.add(substring(off, value.length)); // Construct result int resultSize = list.size(); if (limit == 0) { while (resultSize > 0 && list.get(resultSize - 1).length() == 0) { resultSize--; } } String[] result = new String[resultSize]; return list.subList(0, resultSize).toArray(result); } return Pattern.compile(regex).split(this, limit); }
有其中关系可知最终会执行 Pattern.compile(regex).split(this, limit)这一段代码,基础往下扒代码:
public String[] split(CharSequence input, int limit) { int index = 0; boolean matchLimited = limit > 0; ArrayList<String> matchList = new ArrayList<>(); Matcher m = matcher(input); // Add segments before each match found while(m.find()) { if (!matchLimited || matchList.size() < limit - 1) { if (index == 0 && index == m.start() && m.start() == m.end()) { // no empty leading substring included for zero-width match // at the beginning of the input char sequence. continue; } String match = input.subSequence(index, m.start()).toString(); matchList.add(match); index = m.end(); } else if (matchList.size() == limit - 1) { // last one String match = input.subSequence(index, input.length()).toString(); matchList.add(match); index = m.end(); } } // If no match was found, return this if (index == 0) return new String[] {input.toString()}; // Add remaining segment if (!matchLimited || matchList.size() < limit) matchList.add(input.subSequence(index, input.length()).toString()); // Construct result int resultSize = matchList.size(); if (limit == 0) while (resultSize > 0 && matchList.get(resultSize-1).equals("")) resultSize--; String[] result = new String[resultSize]; return matchList.subList(0, resultSize).toArray(result); }
通过代码我们可以发现最终matchList集合中会有值,不过都是空值,然后在
while (resultSize > 0 && matchList.get(resultSize-1).equals("")) resultSize--;
这一段代码中,首先判断最后一个是不是空,如果没有值的话就减一位,依次类推,所以看到这大家对以上程序出现的结果是不是就不奇怪了。
所以我们可以大胆的总结一下,使用split方法分割字符串,如果最后几位是空的话,会将空的位置去掉。
总结
以上为个人经验,希望能给大家一个参考,也希望大家多多支持我们。