解析内容 if else太多 请问优化代码？

Question

//List<String> 内容
[ti:我本闺中一钗裙]
[oti:我本闺中一钗裙]
[jz:黄梅戏]
[jm:女驸马]
[ojm:女驸马]
[offset:-1]
[by:制作者]
[ver:驸马公主合唱]
[keys:民女，洞房]
[00:04.99]（驸马）：我本闺中一钗裙
[00:31.93]（公主白）：此话当真
[00:34.19]（驸马）：公主请看耳环痕
[00:59.33]<结束>

/**
     * 解析唱段内容
     *
     * @param changDuan  唱段对象
     * @param line       当前需要解析的内容 即上面 List<String> 每一行内容
     * @param patternMap 唱段正则表达式 可以正则匹配上面 List<String> 内容
     * @throws ParseLrcException
     */
private static void parseLrcContent(ChangDuan changDuan, String line, Map<String, Pattern> patternMap) throws ParseLrcException {

        if (matcher(line, patternMap, LrcConstants.LrcRegEnum.TITLE.name())) {
            changDuan.setName(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.TITLE.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.ORIGIN_TITLE.name())) {
            changDuan.setOriginName(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.ORIGIN_TITLE.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.JUMU.name())) {
            changDuan.setJuMu(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.JUMU.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.ORIGIN_JUMU.name())) {
            changDuan.setOriginJuMu(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.ORIGIN_JUMU.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.OFFSET_REG.name())) {
            changDuan.setOffset(Integer.parseInt(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.OFFSET_REG.getStartIndex())));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.JU_ZHONG.name())) {
            changDuan.setJuZhong(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.JU_ZHONG.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.BEIZHU.name())) {
            changDuan.setBeiZhu(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.BEIZHU.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.VERSION.name())) {
            changDuan.setVersion(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.VERSION.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.KEYS.name())) {
            changDuan.setSearchKeys(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.KEYS.getStartIndex()));
        } else if (matcher(line, patternMap, LrcConstants.LrcRegEnum.BY.name())) {
            changDuan.setMaker(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.BY.getStartIndex()));
        } else {
            throw new ParseLrcLineException(line);
        }
    }

上段代码是用于将List<String>解析成ChangDuan

想请教一下这个代码写的咋样，能否优化，我就是觉得if else太多了。

Answer 1

抽象一个接口：

interface IMatchParser{
    boolean matcher(String line, Map<String, Patter> map);
    handle(ChangDuan changduan);
}

每个if分支抽象成一个类继承IMatchParser;

public class MatchParser1 implements IMatchParser{
    public void handle(ChangDuan changduan) {
        changDuan.setName(getChangDuanInfoContent(line, LrcConstants.LrcRegEnum.TITLE.getStartIndex()));
    }

    public boolean matcher(String line, Map<String, Patter> map) {
        return matcher(line, patternMap, LrcConstants.LrcRegEnum.TITLE.name());
    }
}

然后放到一个list中：

List<IMatchParser> list = new ArrayList();
list.add(new MatchParser1());
list.add(new MatchParser2());
...

然后实现方法变成：

private static void parseLrcContent(ChangDuan changDuan, String line, Map<String, Pattern> patternMap) throws ParseLrcException {
    for (IMatchParser parser: list) {
        if (parser.match(line, patternMap)) {
            parser.handle(changDuan);
            break;    
        }
    }
}

后续加新的if分支只需要实现一个新的class插入到list即可

Answer 2

使用表驱动设计模式即可