如何将字符串解析为map

Question

有一个像A = B&C = D&E = F的字符串,如何将其解析成地图？

Answer 1

我会用拆分

String text = "A=B&C=D&E=F";
Map<String, String> map = new LinkedHashMap<String, String>();
for(String keyValue : text.split(" *& *")) {
   String[] pairs = keyValue.split(" *= *", 2);
   map.put(pairs[0], pairs.length == 1 ? "" : pairs[1]);
}

EDIT允许填充空格和具有=或没有值的值.例如

A = minus- & C=equals= & E==F

Answer 2

只需使用番石榴分离器

\n\n

String src="A=B&C=D&E=F";\nMap map= Splitter.on(\'&\').withKeyValueSeparator(\'=\').split(src)\xef\xbc\x9b\n

\n

Answer 3

public class TestMapParser {
    @Test
    public void testParsing() {
        Map<String, String> map = parseMap("A=B&C=D&E=F");
        Assert.assertTrue("contains key", map.containsKey("A"));
        Assert.assertEquals("contains value", "B", map.get("A"));

        Assert.assertTrue("contains key", map.containsKey("C"));
        Assert.assertEquals("contains value", "D", map.get("C"));

        Assert.assertTrue("contains key", map.containsKey("E"));
        Assert.assertEquals("contains value", "F", map.get("E"));
    }

    private Map<String, String> parseMap(final String input) {
        final Map<String, String> map = new HashMap<String, String>();
        for (String pair : input.split("&")) {
            String[] kv = pair.split("=");
            map.put(kv[0], kv[1]);
        }
        return map;
    }
}