输入一行命令 command --flag1 p1=a p2=b --flag2 p1=a p2=b,各个单词之间、单词和符号之间都有不确定数量的空格,请解析出命令,将各段命令放置于list中。当解析出
----ret----
command
--flag1
p1=a
p2=b
--flag2
p1=a
p2=b
时认为是正确解析结果。
我的方案是:
A:正则表达式
B:利用String类的split、replace、subString 等方法
下面说下方案B
public class ParseCommand {
public static void main(String[] args) {
String cmd = " command --flag1 p1 = a p2 = b --flag2 p1 = a p2 = b";
parseCmd(cmd);
}
private static List<String> parseCmd(String cmd) {
List<String> list = new ArrayList<>();
if (cmd == null || cmd.length() <= 0) {
return list;
}
int start = -1;
int end = -1;
cmd = cmd + " "; // 如果不加一位空格,下面的for循环无法进行最后一次的裁剪
int len = cmd.length();
System.out.println("len:" + len);
for (int i = 0; i < len; i++) {
if (Character.isWhitespace(cmd.charAt(i))) {
if (start > 0) {
end = i;
System.out.println("start>end: " + start + ">" + end);
String s = cmd.substring(start, end);
s = s.replace(" ", "");
if (!isEmpty(s)) {
int lastIndex = list.size() - 1;
String last = list.size() > 0 ? list.get(lastIndex) : "";
System.out.println(s + "," + last);
if (last.endsWith("=")) {
list.set(lastIndex, last + s);
} else if (s.startsWith("=")) {
list.set(lastIndex, last + s);
} else {
list.add(s);
}
}
start = -1;
end = -1;
}
} else {
if (start < 0) {
start = i;
}
}
}
System.out.println("----ret----");
for (int i = 0; i < list.size(); i++) {
System.out.println(list.get(i));
}
return list;
}
public static boolean isEmpty(String s) {
return s == null || s.length() <= 0;
}
}