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1、传统的字符串split

【推荐】使用索引访问用String的split方法得到的数组时，需做最后一个分隔符后有无内容的检查，否则会有抛IndexOutOfBoundsException的风险。

说明：

String str = “a,b,,c,,”;

String[] ary = str.split(“,”);

// 预期大于4，结果是4

2、Splitter替代split

（1）、trimResults,从每个返回的子字符串中删除前导和尾随空格

（2）、omitEmptyStrings,从最终结果中忽略空字符串

（3）、withKeyValueSeparator转换为map

//传统的字符串split
        String strstr = "a,b,,c,,,";
        String[] ary = strstr.split(",");
        System.out.println(ary.length);
        for (int i = 0; i < ary.length; i++) {
            System.out.println(ary[i]);
        }
        System.out.println("====");
//trimResults,从每个返回的子字符串中删除前导和尾随空格
        String str = "Everyone. .  should. Learn. Data. Structures";
        List<String> myList1 = Splitter.on('.')
                .trimResults().splitToList(str);
        for (String temp:myList1) {
            System.out.println(temp);
        }
        System.out.println("====");
//omitEmptyStrings,从结果中忽略空字符串。例如，Splitter.on(',').omitEmptyStrings().split(“，a ,,, b，c ,,”)返回仅包含[“a”，“b”，“c”]
        List<String> myList2 = Splitter.on('.')
                .trimResults().omitEmptyStrings().splitToList(str);
        for (String temp:myList2) {
            System.out.println(temp);
        }
        System.out.println("====");
//withKeyValueSeparator转换为map
        Map<String, String> map1 = Splitter
                .on(",")
                .trimResults()
                .omitEmptyStrings()
                .withKeyValueSeparator("=")
                .split(EcologySwitch.insureTypeEnums);
        map1.entrySet().forEach(stringStringEntry -> {
            System.out.println(stringStringEntry.getKey()+":"+stringStringEntry.getValue());
        });

输出结果：
4
a
b

c
====
Everyone

should
Learn
Data
Structures
====
Everyone
should
Learn
Data
Structures
====
002006.flightaccident.air:16
002006.flightcompose.air:17
002006.flightisolate.air:32